Potential
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- Intro To Potential
- Techniques To Calculate Potential
- Potential Of A Point Charge
- First Method - Potential Of A Finite Wire
- Potential Of a Ring
- Conductors
- Difference Between Conductors and Insulators
- Symmetry
- Second Method - Gauss Law
- Exercise 1
- Exercise 2
- Exercise 3
- Third Method - Potential Difference
- Exercise 4

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[{"Name":"Potential","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Intro To Potential","Duration":"11m 12s","ChapterTopicVideoID":12146,"CourseChapterTopicPlaylistID":157337,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12146.jpeg","UploadDate":"2018-06-28T04:54:18.8800000","DurationForVideoObject":"PT11M12S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.875","Text":"Hello. In this lesson,"},{"Start":"00:01.875 ","End":"00:04.950","Text":"we\u0027re going to be speaking about the electric potential."},{"Start":"00:04.950 ","End":"00:11.175","Text":"A lot of people symbolize the electric potential with a capital V_E,"},{"Start":"00:11.175 ","End":"00:12.915","Text":"so that\u0027s the electric potential."},{"Start":"00:12.915 ","End":"00:17.055","Text":"However, I am going to symbolize it with the Greek letter Phi,"},{"Start":"00:17.055 ","End":"00:19.890","Text":"and I\u0027ll explain why soon."},{"Start":"00:19.890 ","End":"00:26.910","Text":"The electric potential is equal to negative the integral on our"},{"Start":"00:26.910 ","End":"00:34.065","Text":"E field.dr where dr is across our trajectory."},{"Start":"00:34.065 ","End":"00:40.655","Text":"What this integral means is that I\u0027m going along my trajectory and summing up all"},{"Start":"00:40.655 ","End":"00:47.260","Text":"of my tangential values for my E field across my trajectory."},{"Start":"00:47.260 ","End":"00:49.190","Text":"Let\u0027s explain what that means."},{"Start":"00:49.190 ","End":"00:54.350","Text":"Let\u0027s say I have some trajectory in 3D space, like so."},{"Start":"00:54.350 ","End":"00:57.110","Text":"Let\u0027s say that over here,"},{"Start":"00:57.110 ","End":"00:58.955","Text":"I have my origin."},{"Start":"00:58.955 ","End":"01:03.140","Text":"At any point from on my trajectory,"},{"Start":"01:03.140 ","End":"01:04.625","Text":"so let\u0027s say over here,"},{"Start":"01:04.625 ","End":"01:07.235","Text":"I\u0027m going to have some vector r,"},{"Start":"01:07.235 ","End":"01:13.535","Text":"which goes from my origin and points to this point on my trajectory."},{"Start":"01:13.535 ","End":"01:15.590","Text":"Imagine a moment later,"},{"Start":"01:15.590 ","End":"01:20.660","Text":"I move further along my trajectory and I\u0027m now at this point over here."},{"Start":"01:20.660 ","End":"01:26.315","Text":"Then my r vector is going to be from the origin until my new point,"},{"Start":"01:26.315 ","End":"01:30.410","Text":"so now this is my i vector a moment later."},{"Start":"01:30.410 ","End":"01:35.630","Text":"Because the time difference between these 2 points is tiny,"},{"Start":"01:35.630 ","End":"01:38.780","Text":"the vector connecting these 2 points,"},{"Start":"01:38.780 ","End":"01:46.425","Text":"this blue arrow over here is going to be called the dr vector and that\u0027s this over here."},{"Start":"01:46.425 ","End":"01:51.300","Text":"We can see that our dr is tangential to our trajectory."},{"Start":"01:51.300 ","End":"01:56.510","Text":"What we\u0027ll see is that our trajectory is in fact made up of lots"},{"Start":"01:56.510 ","End":"02:01.685","Text":"of these dr vectors that are just going along,"},{"Start":"02:01.685 ","End":"02:03.995","Text":"and they build up our trajectory."},{"Start":"02:03.995 ","End":"02:06.995","Text":"Then if I sum up all of my dr\u0027s,"},{"Start":"02:06.995 ","End":"02:10.480","Text":"I\u0027ll get the displacement of my trajectory."},{"Start":"02:10.480 ","End":"02:13.400","Text":"Now let\u0027s see what our E field is."},{"Start":"02:13.400 ","End":"02:16.955","Text":"Let\u0027s say that we\u0027re located at this point over here,"},{"Start":"02:16.955 ","End":"02:19.190","Text":"and let\u0027s say that at this point over here,"},{"Start":"02:19.190 ","End":"02:22.930","Text":"our E field is pointing in a direction like so."},{"Start":"02:22.930 ","End":"02:24.770","Text":"What is my integral do?"},{"Start":"02:24.770 ","End":"02:29.135","Text":"It\u0027s going to sum on my component for my E field,"},{"Start":"02:29.135 ","End":"02:32.690","Text":"which is tangential to a trajectory."},{"Start":"02:32.690 ","End":"02:34.130","Text":"How does it do that?"},{"Start":"02:34.130 ","End":"02:38.835","Text":"It uses the help of this dot product over here."},{"Start":"02:38.835 ","End":"02:44.455","Text":"The dot product between any 2 vectors is going to"},{"Start":"02:44.455 ","End":"02:51.179","Text":"give the component of this vector in the direction of this vector,"},{"Start":"02:51.179 ","End":"02:55.745","Text":"so the projection of this vector on this vector."},{"Start":"02:55.745 ","End":"02:59.840","Text":"This integral is going to sum up all of"},{"Start":"02:59.840 ","End":"03:06.400","Text":"the tangential components of our E field along our trajectory."},{"Start":"03:06.400 ","End":"03:07.820","Text":"This is E, E,"},{"Start":"03:07.820 ","End":"03:11.820","Text":"E. Of course, when we\u0027re tangential,"},{"Start":"03:11.820 ","End":"03:17.210","Text":"so it\u0027s parallel, so the parallel component of our E field to our trajectory."},{"Start":"03:17.210 ","End":"03:23.585","Text":"If our E field has no component which is parallel to our trajectory at that point,"},{"Start":"03:23.585 ","End":"03:27.440","Text":"so our E field is always perpendicular to our trajectory."},{"Start":"03:27.440 ","End":"03:30.265","Text":"Then we\u0027re going to be summing up on 0."},{"Start":"03:30.265 ","End":"03:33.260","Text":"We\u0027re going to explain why we\u0027re doing"},{"Start":"03:33.260 ","End":"03:36.815","Text":"this soon and the motivation for the electric potential."},{"Start":"03:36.815 ","End":"03:44.960","Text":"But in the meantime, the last basic explanation is going to be that if our E field is,"},{"Start":"03:44.960 ","End":"03:48.590","Text":"or the parallel component of our E field is a uniform,"},{"Start":"03:48.590 ","End":"03:50.180","Text":"if it\u0027s not changing,"},{"Start":"03:50.180 ","End":"03:53.610","Text":"so we can write this integral of E.dr"},{"Start":"03:54.100 ","End":"04:00.110","Text":"with a minus simply as if we just have a parallel component"},{"Start":"04:00.110 ","End":"04:04.550","Text":"and it\u0027s uniform as the parallel component dr."},{"Start":"04:04.550 ","End":"04:10.335","Text":"Then we can just simply write that as the E-field multiplied by L,"},{"Start":"04:10.335 ","End":"04:14.010","Text":"where L is the length of the trajectory,"},{"Start":"04:14.010 ","End":"04:19.850","Text":"and this is only if E is equal to a constant."},{"Start":"04:19.850 ","End":"04:21.830","Text":"In our mechanics course,"},{"Start":"04:21.830 ","End":"04:24.410","Text":"we saw a similar integral to this,"},{"Start":"04:24.410 ","End":"04:28.685","Text":"where we worked out that the work that a system does is"},{"Start":"04:28.685 ","End":"04:34.005","Text":"equal to the integral of our force vector dot dr."},{"Start":"04:34.005 ","End":"04:38.035","Text":"It\u0027s exactly the same just for our electric potential,"},{"Start":"04:38.035 ","End":"04:43.850","Text":"we have a minus and we\u0027re integrating along the E rather than F. Now we know"},{"Start":"04:43.850 ","End":"04:50.150","Text":"that work is simply the change in potential energy,"},{"Start":"04:50.150 ","End":"04:52.910","Text":"and of course, there\u0027s a minus over here."},{"Start":"04:52.910 ","End":"04:56.675","Text":"We\u0027ve already seen something similar to this integration,"},{"Start":"04:56.675 ","End":"04:58.745","Text":"and not just that,"},{"Start":"04:58.745 ","End":"05:04.060","Text":"what this integral does is exactly what our work integral does."},{"Start":"05:04.060 ","End":"05:09.905","Text":"We already know from our first chapter on Coulomb\u0027s law that our force is"},{"Start":"05:09.905 ","End":"05:15.840","Text":"equal to the charge multiplied by our electric field."},{"Start":"05:15.840 ","End":"05:19.670","Text":"Our electric field is simply a mathematical function"},{"Start":"05:19.670 ","End":"05:25.280","Text":"that tells me that if I place some kind of charge in some area of space,"},{"Start":"05:25.280 ","End":"05:30.745","Text":"then I can know what the force between that charge and my original charge will be."},{"Start":"05:30.745 ","End":"05:36.065","Text":"Or we can say that our E field is the force per unit charge."},{"Start":"05:36.065 ","End":"05:43.500","Text":"We can say that this equation for electric potential is the work done per unit charge."},{"Start":"05:44.180 ","End":"05:47.255","Text":"As we saw in our mechanics chapter,"},{"Start":"05:47.255 ","End":"05:55.950","Text":"that our energy or our potential energy is equal to negative this integral, so F.dr."},{"Start":"05:56.240 ","End":"06:01.940","Text":"Now we can see that our electric potential is exactly"},{"Start":"06:01.940 ","End":"06:07.630","Text":"like our potential energy per unit charge."},{"Start":"06:07.630 ","End":"06:11.120","Text":"In that case, why do we even need electric potential?"},{"Start":"06:11.120 ","End":"06:12.725","Text":"Why do we calculate it?"},{"Start":"06:12.725 ","End":"06:19.555","Text":"We do that in order to find out what our potential energy is in the system."},{"Start":"06:19.555 ","End":"06:24.335","Text":"Our potential energy of the system is going to be equal to"},{"Start":"06:24.335 ","End":"06:30.070","Text":"our charge q multiplied by our electric potential."},{"Start":"06:30.070 ","End":"06:32.645","Text":"Just like when we have an electric field,"},{"Start":"06:32.645 ","End":"06:36.080","Text":"if we place some kind of point charge in the electric field,"},{"Start":"06:36.080 ","End":"06:38.840","Text":"we can know what the force is going to be between"},{"Start":"06:38.840 ","End":"06:42.775","Text":"our original charge and our newly placed point charge."},{"Start":"06:42.775 ","End":"06:45.545","Text":"Similarly, with our electric potential,"},{"Start":"06:45.545 ","End":"06:48.830","Text":"we know we can have some kind of equation for"},{"Start":"06:48.830 ","End":"06:52.955","Text":"the potential and if we put a point charge or some kind of charge,"},{"Start":"06:52.955 ","End":"06:57.830","Text":"a test charge in this function for our electric potential,"},{"Start":"06:57.830 ","End":"07:02.585","Text":"we can know what the potential energy is going to be."},{"Start":"07:02.585 ","End":"07:07.130","Text":"Our electric potential is almost the same as our potential energy,"},{"Start":"07:07.130 ","End":"07:13.720","Text":"just without knowing what kind of charge I have in my system."},{"Start":"07:13.720 ","End":"07:18.815","Text":"Why do people sometimes write this out as V_E?"},{"Start":"07:18.815 ","End":"07:21.395","Text":"And why do people sometimes write it as Phi?"},{"Start":"07:21.395 ","End":"07:25.235","Text":"I prefer to symbolize the potential as PHI,"},{"Start":"07:25.235 ","End":"07:27.230","Text":"but people use this V_E."},{"Start":"07:27.230 ","End":"07:31.950","Text":"V_E is for voltage,"},{"Start":"07:31.950 ","End":"07:35.650","Text":"so let\u0027s write volts, and our voltage"},{"Start":"07:35.650 ","End":"07:40.505","Text":"is the difference in potential, Delta is difference."},{"Start":"07:40.505 ","End":"07:42.935","Text":"It\u0027s the difference in our potential,"},{"Start":"07:42.935 ","End":"07:50.350","Text":"so that is equal to our final potential minus our initial potential."},{"Start":"07:50.350 ","End":"07:53.015","Text":"Our voltage is the difference,"},{"Start":"07:53.015 ","End":"07:54.740","Text":"or the potential difference."},{"Start":"07:54.740 ","End":"07:58.990","Text":"Then I can say that if I have the potential at any point,"},{"Start":"07:58.990 ","End":"08:00.935","Text":"so the potential difference,"},{"Start":"08:00.935 ","End":"08:03.560","Text":"or our voltage, is going to be the potential at"},{"Start":"08:03.560 ","End":"08:07.340","Text":"that point minus the potential at point 0,"},{"Start":"08:07.340 ","End":"08:08.920","Text":"which is equal to 0."},{"Start":"08:08.920 ","End":"08:10.760","Text":"This voltage that we know,"},{"Start":"08:10.760 ","End":"08:12.770","Text":"which is our potential difference,"},{"Start":"08:12.770 ","End":"08:15.320","Text":"we have that in our plugs."},{"Start":"08:15.320 ","End":"08:19.455","Text":"Imagine that you have a plug that looks like so."},{"Start":"08:19.455 ","End":"08:23.705","Text":"We know that here we have a positive and here"},{"Start":"08:23.705 ","End":"08:28.205","Text":"we have a negative and we have a voltage difference of,"},{"Start":"08:28.205 ","End":"08:34.380","Text":"let\u0027s say, 220 volts between our plus and minus."},{"Start":"08:34.380 ","End":"08:37.880","Text":"Let\u0027s try and find out what this potential difference"},{"Start":"08:37.880 ","End":"08:41.495","Text":"or what this voltage in our plugs is equal to."},{"Start":"08:41.495 ","End":"08:44.630","Text":"We already know that our work is equal"},{"Start":"08:44.630 ","End":"08:50.040","Text":"to negative our difference in our potential energy."},{"Start":"08:50.040 ","End":"08:53.960","Text":"That means that it\u0027s equal to our potential energy at"},{"Start":"08:53.960 ","End":"08:58.240","Text":"the end minus our initial potential energy."},{"Start":"08:58.240 ","End":"09:01.700","Text":"Then we can say that this is equal to negative."},{"Start":"09:01.700 ","End":"09:03.470","Text":"Now, our potential energy,"},{"Start":"09:03.470 ","End":"09:07.370","Text":"we can see that it\u0027s q multiplied by our electric potential."},{"Start":"09:07.370 ","End":"09:14.735","Text":"There\u0027ll be negative q and then we have our electric potential at the end,"},{"Start":"09:14.735 ","End":"09:18.770","Text":"minus our electric potential at the beginning."},{"Start":"09:18.770 ","End":"09:24.395","Text":"Then our Phi_f minus Phi_i is what we have over here,"},{"Start":"09:24.395 ","End":"09:26.885","Text":"is equal to our voltage."},{"Start":"09:26.885 ","End":"09:29.920","Text":"That\u0027s equal to negative q,"},{"Start":"09:29.920 ","End":"09:33.105","Text":"our charge, multiplied by our voltage."},{"Start":"09:33.105 ","End":"09:36.470","Text":"What we can see is that voltage multiplied by"},{"Start":"09:36.470 ","End":"09:41.080","Text":"charge with a negative sign is going to give us work."},{"Start":"09:41.080 ","End":"09:47.005","Text":"The voltage that I get is my work per unit charge."},{"Start":"09:47.005 ","End":"09:52.535","Text":"What we have for our plug system is how much voltage we need"},{"Start":"09:52.535 ","End":"09:58.380","Text":"in order to move 1 coulomb of charge from here to here."},{"Start":"09:58.380 ","End":"10:04.610","Text":"The most important equations to take from this lesson and first of all, all of them."},{"Start":"10:04.610 ","End":"10:10.040","Text":"But the 3 most important are really these that our electric potential is"},{"Start":"10:10.040 ","End":"10:16.235","Text":"also sometimes known as our voltage and it\u0027s equal to the negative integral of E.dr."},{"Start":"10:16.235 ","End":"10:20.600","Text":"Also, to remember that our voltage is defined as"},{"Start":"10:20.600 ","End":"10:23.180","Text":"our electric potential difference"},{"Start":"10:23.180 ","End":"10:26.510","Text":"and to remember that our potential energy is equal to q,"},{"Start":"10:26.510 ","End":"10:30.470","Text":"or some kind of charge multiplied by our electric potential."},{"Start":"10:30.470 ","End":"10:35.060","Text":"Now, another important equation to know is what if we want to find the opposite?"},{"Start":"10:35.060 ","End":"10:37.445","Text":"Here, if we have our electric field,"},{"Start":"10:37.445 ","End":"10:38.990","Text":"we can find our potential."},{"Start":"10:38.990 ","End":"10:43.705","Text":"But what if we have our potential and we want to find our electric field?"},{"Start":"10:43.705 ","End":"10:52.185","Text":"Our electric field is going to be equal to negative grad Phi."},{"Start":"10:52.185 ","End":"10:55.610","Text":"Remember that Phi is some scalar quantity,"},{"Start":"10:55.610 ","End":"10:59.105","Text":"and we\u0027re taking the negative grad of this scalar quantity,"},{"Start":"10:59.105 ","End":"11:02.080","Text":"and then we find our electric field."},{"Start":"11:02.080 ","End":"11:04.145","Text":"Write this in as well. This is"},{"Start":"11:04.145 ","End":"11:08.675","Text":"the fourth very important equation to take from this lesson."},{"Start":"11:08.675 ","End":"11:11.670","Text":"That\u0027s the end of this lesson."}],"ID":14199},{"Watched":false,"Name":"Techniques To Calculate Potential","Duration":"5m 26s","ChapterTopicVideoID":12147,"CourseChapterTopicPlaylistID":157337,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12147.jpeg","UploadDate":"2018-06-28T04:55:13.7600000","DurationForVideoObject":"PT5M26S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.815","Text":"Hello. In this lesson,"},{"Start":"00:01.815 ","End":"00:05.970","Text":"we\u0027re going to be learning how to calculate the potential."},{"Start":"00:05.970 ","End":"00:12.360","Text":"I divide the techniques to calculate the potential into 3 different types."},{"Start":"00:12.360 ","End":"00:19.110","Text":"The first subcategory are shapes that don\u0027t really have symmetry in them."},{"Start":"00:19.110 ","End":"00:22.515","Text":"That if we liken this to Coulomb\u0027s law,"},{"Start":"00:22.515 ","End":"00:25.965","Text":"it\u0027s all the weird shapes where we had a finite wire."},{"Start":"00:25.965 ","End":"00:28.890","Text":"Not an infinite, but rather a finite wire."},{"Start":"00:28.890 ","End":"00:34.510","Text":"Or we had some bent wire or some weird shape like so."},{"Start":"00:34.510 ","End":"00:40.528","Text":"Then we had to find our electric field via using Coulomb\u0027s law."},{"Start":"00:40.528 ","End":"00:46.307","Text":"Here, our technique to calculate the potential is going to be pretty similar."},{"Start":"00:46.307 ","End":"00:48.215","Text":"For our shapes with no symmetry,"},{"Start":"00:48.215 ","End":"00:51.060","Text":"we\u0027re going to work this out with Coulomb\u0027s law."},{"Start":"00:51.060 ","End":"00:56.758","Text":"We split up our shape into lots of tiny pieces where each piece has some kind of charge,"},{"Start":"00:56.758 ","End":"01:02.285","Text":"dq, and then from this charge will work out our d Phi."},{"Start":"01:02.285 ","End":"01:06.195","Text":"A small potential for that tiny piece."},{"Start":"01:06.195 ","End":"01:08.820","Text":"Then how we work out the total potential,"},{"Start":"01:08.820 ","End":"01:13.170","Text":"it\u0027s going to be like for a point charge, p.c point charge."},{"Start":"01:13.170 ","End":"01:18.350","Text":"The potential for a point charge is equal to kq divided"},{"Start":"01:18.350 ","End":"01:23.630","Text":"by r. Just like in Coulomb\u0027s law as kq divided by r squared,"},{"Start":"01:23.630 ","End":"01:26.840","Text":"here is just divided by r. Of course,"},{"Start":"01:26.840 ","End":"01:29.615","Text":"our potential is a scalar function."},{"Start":"01:29.615 ","End":"01:31.760","Text":"We only have size,"},{"Start":"01:31.760 ","End":"01:33.985","Text":"if you will, and we don\u0027t have direction."},{"Start":"01:33.985 ","End":"01:36.440","Text":"Here, now I\u0027m just going to speak"},{"Start":"01:36.440 ","End":"01:39.485","Text":"quickly and then later we\u0027re going to see how we do this in detail."},{"Start":"01:39.485 ","End":"01:43.745","Text":"We split up our piece into tiny little lengths where each 1 has charge dq."},{"Start":"01:43.745 ","End":"01:47.735","Text":"Then we find the potential for each of these little charges."},{"Start":"01:47.735 ","End":"01:51.440","Text":"We can consider each of these little charges as point charges."},{"Start":"01:51.440 ","End":"01:55.490","Text":"The potential for every point charge is going to be this,"},{"Start":"01:55.490 ","End":"01:57.800","Text":"or given by this equation over here."},{"Start":"01:57.800 ","End":"02:02.225","Text":"Then in order to find the total electric potential,"},{"Start":"02:02.225 ","End":"02:07.570","Text":"it\u0027s simply going to be an integral on our d Phi,"},{"Start":"02:07.570 ","End":"02:10.095","Text":"which is simply this."},{"Start":"02:10.095 ","End":"02:14.030","Text":"Notice that our integral is going to be super easy because we don\u0027t"},{"Start":"02:14.030 ","End":"02:18.470","Text":"need to do the integration along different directions,"},{"Start":"02:18.470 ","End":"02:21.205","Text":"because it\u0027s a scalar quantity."},{"Start":"02:21.205 ","End":"02:25.295","Text":"The next technique that we can use in order to calculate the potential"},{"Start":"02:25.295 ","End":"02:30.805","Text":"is on similar questions to which we can use our Gauss\u0027s law."},{"Start":"02:30.805 ","End":"02:35.650","Text":"We have 4 types of shapes that we can use our Gauss\u0027s law."},{"Start":"02:35.650 ","End":"02:38.745","Text":"Like Gauss, let\u0027s just write that."},{"Start":"02:38.745 ","End":"02:42.080","Text":"That\u0027s when we have a shape for an infinite wire and we\u0027re asked to"},{"Start":"02:42.080 ","End":"02:45.610","Text":"calculate the potential or an infinite cylinder or cylindrical shell,"},{"Start":"02:45.610 ","End":"02:48.109","Text":"an infinite plane, or a sphere,"},{"Start":"02:48.109 ","End":"02:49.615","Text":"or a spherical shell."},{"Start":"02:49.615 ","End":"02:51.290","Text":"Just like with Gauss\u0027s law,"},{"Start":"02:51.290 ","End":"02:54.445","Text":"we\u0027re going to use this technique."},{"Start":"02:54.445 ","End":"02:56.435","Text":"How do we solve this?"},{"Start":"02:56.435 ","End":"03:00.605","Text":"First, we calculate our electric field via Gauss\u0027s law,"},{"Start":"03:00.605 ","End":"03:07.320","Text":"and then we calculate the potential via integrating along our electric field.dr."},{"Start":"03:07.550 ","End":"03:15.550","Text":"In this case, we have a further 2 options on how to do this."},{"Start":"03:15.550 ","End":"03:19.999","Text":"The first option is when we have an indefinite integral,"},{"Start":"03:19.999 ","End":"03:23.255","Text":"then what we have to do is we have to use continuity."},{"Start":"03:23.255 ","End":"03:25.715","Text":"I\u0027ll go into that later."},{"Start":"03:25.715 ","End":"03:30.878","Text":"The second type is when we have a definite integral,"},{"Start":"03:30.878 ","End":"03:36.080","Text":"and then we\u0027re going to have to figure out what our bounds are for the definite integral."},{"Start":"03:36.080 ","End":"03:40.340","Text":"So 2 of these options are correct and we can use them."},{"Start":"03:40.340 ","End":"03:45.953","Text":"I personally think that this option is slightly easier,"},{"Start":"03:45.953 ","End":"03:48.179","Text":"so maybe you\u0027ll want to pay a bit more attention to this."},{"Start":"03:48.179 ","End":"03:51.149","Text":"However, for some reason in most of the courses,"},{"Start":"03:51.149 ","End":"03:55.399","Text":"they focus more on this option of how to solve this."},{"Start":"03:55.399 ","End":"03:56.750","Text":"Pay attention to both."},{"Start":"03:56.750 ","End":"03:59.390","Text":"I personally think that this is a lot easier to work with."},{"Start":"03:59.390 ","End":"04:05.270","Text":"The final step, whatever we do is we have to calibrate our system."},{"Start":"04:05.270 ","End":"04:08.846","Text":"We\u0027ll speak about what this means a little bit later."},{"Start":"04:08.846 ","End":"04:14.964","Text":"Now, the third option is if I want to find the potential difference."},{"Start":"04:14.964 ","End":"04:24.405","Text":"Let\u0027s say if I want to find the potential at Point B and the potential at Point A."},{"Start":"04:24.405 ","End":"04:34.350","Text":"This is going to be equal to the line integral between A and B, E.dr."},{"Start":"04:35.570 ","End":"04:41.315","Text":"What does that mean? Let\u0027s say I know my potential at Point A,"},{"Start":"04:41.315 ","End":"04:46.220","Text":"this is known, and that\u0027s equal to Phi."},{"Start":"04:46.220 ","End":"04:51.755","Text":"I want to find my potential here at Point B because I don\u0027t know what that is,"},{"Start":"04:51.755 ","End":"04:55.055","Text":"but what I do know is I know my electric field,"},{"Start":"04:55.055 ","End":"04:57.680","Text":"what my electric field is between these 2 points."},{"Start":"04:57.680 ","End":"05:03.290","Text":"What I\u0027m going to do is I\u0027m going to sum up my electric field between these 2 points,"},{"Start":"05:03.290 ","End":"05:08.830","Text":"and then I\u0027m going to get the potential difference between Point A and Point B."},{"Start":"05:08.830 ","End":"05:16.235","Text":"These techniques are more for the general potential due to my shape with a charge."},{"Start":"05:16.235 ","End":"05:23.605","Text":"This third technique is if I want to find the electric potential at a certain point."},{"Start":"05:23.605 ","End":"05:26.770","Text":"That\u0027s the end of the lesson."}],"ID":14200},{"Watched":false,"Name":"Potential Of A Point Charge","Duration":"3m 51s","ChapterTopicVideoID":12148,"CourseChapterTopicPlaylistID":157337,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12148.jpeg","UploadDate":"2018-06-28T04:55:46.6130000","DurationForVideoObject":"PT3M51S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.040","Text":"Hello. In this lesson,"},{"Start":"00:02.040 ","End":"00:07.680","Text":"we\u0027re going to see technique number one for finding our potential."},{"Start":"00:07.680 ","End":"00:10.883","Text":"We\u0027re going to be using the superposition principle,"},{"Start":"00:10.883 ","End":"00:14.010","Text":"and we\u0027re going to be finding the potential of a point charge."},{"Start":"00:14.010 ","End":"00:20.055","Text":"First of all, we know that the electric potential is given by negative,"},{"Start":"00:20.055 ","End":"00:24.810","Text":"the integral of E.dr."},{"Start":"00:24.810 ","End":"00:34.725","Text":"The electric field of a point charge is given as kq divided by r^2 in the r direction."},{"Start":"00:34.725 ","End":"00:38.165","Text":"Now, if I\u0027m working in spherical coordinates,"},{"Start":"00:38.165 ","End":"00:47.060","Text":"my dr vector is going to be equal to dr in my r or radial direction."},{"Start":"00:47.060 ","End":"00:51.140","Text":"Now, what I can do is I can substitute in these values."},{"Start":"00:51.140 ","End":"00:54.935","Text":"I have the negative integral of my E field,"},{"Start":"00:54.935 ","End":"01:00.780","Text":"which is equal to kq divided by r^2 in the r direction,"},{"Start":"01:00.780 ","End":"01:04.590","Text":"in the radial direction.dr vector,"},{"Start":"01:04.590 ","End":"01:10.064","Text":"which is equal to dr in the r or radial direction."},{"Start":"01:10.064 ","End":"01:11.960","Text":"Because we have a dot product,"},{"Start":"01:11.960 ","End":"01:14.285","Text":"the dot product between r hat."},{"Start":"01:14.285 ","End":"01:16.760","Text":"An r hat is simply equal to 1,"},{"Start":"01:16.760 ","End":"01:20.685","Text":"so we can just cross them out because multiplication by 1 doesn\u0027t change anything."},{"Start":"01:20.685 ","End":"01:28.690","Text":"We\u0027ll get that this is equal to the negative integral of kq divided by r^2 dr."},{"Start":"01:28.690 ","End":"01:32.300","Text":"This is great because then we\u0027re going to get a scalar quantity"},{"Start":"01:32.300 ","End":"01:36.745","Text":"out of this and we know that our potential is a scalar quantity."},{"Start":"01:36.745 ","End":"01:39.869","Text":"Now, let\u0027s do this integral."},{"Start":"01:39.869 ","End":"01:44.247","Text":"I\u0027m going to stick with an indefinite integral because it\u0027s slightly easier,"},{"Start":"01:44.247 ","End":"01:46.525","Text":"so I\u0027m not going to put in any bounds."},{"Start":"01:46.525 ","End":"01:49.490","Text":"When we integrate this r minus from"},{"Start":"01:49.490 ","End":"01:52.735","Text":"the integration will cancel out with this minus over here,"},{"Start":"01:52.735 ","End":"01:58.685","Text":"and we\u0027ll be left with kq divided by r plus c,"},{"Start":"01:58.685 ","End":"02:04.465","Text":"where c is some integrating constant because we did an indefinite integral."},{"Start":"02:04.465 ","End":"02:10.510","Text":"Now, what I want to do is I want to find what my value of c is equal to."},{"Start":"02:10.510 ","End":"02:12.640","Text":"In order to find my value of c,"},{"Start":"02:12.640 ","End":"02:15.109","Text":"I have to calibrate my potential."},{"Start":"02:15.109 ","End":"02:16.805","Text":"What does that mean?"},{"Start":"02:16.805 ","End":"02:23.945","Text":"That means that I set my potential at infinity to be equal to 0."},{"Start":"02:23.945 ","End":"02:29.390","Text":"What does that mean that my potential at infinity is equal to 0."},{"Start":"02:29.390 ","End":"02:35.530","Text":"That means that we\u0027re located in infinite distance away from my point charge."},{"Start":"02:35.530 ","End":"02:38.415","Text":"That means that our distance,"},{"Start":"02:38.415 ","End":"02:43.155","Text":"small r is much larger than 1."},{"Start":"02:43.155 ","End":"02:53.945","Text":"Then we\u0027ll get that our potential at infinity is equal to kq divided by infinity plus c,"},{"Start":"02:53.945 ","End":"02:57.370","Text":"and that this has to be equal to 0."},{"Start":"02:57.370 ","End":"03:00.890","Text":"This will cross off and be very close to 0 because any number"},{"Start":"03:00.890 ","End":"03:04.715","Text":"divided by a huge number it\u0027s just going to approach 0."},{"Start":"03:04.715 ","End":"03:06.770","Text":"Then we see, therefore,"},{"Start":"03:06.770 ","End":"03:10.820","Text":"that our c is going to be equal to 0."},{"Start":"03:10.820 ","End":"03:12.679","Text":"That is perfect."},{"Start":"03:12.679 ","End":"03:14.225","Text":"Then I get my answer,"},{"Start":"03:14.225 ","End":"03:15.740","Text":"which is exactly what I wanted,"},{"Start":"03:15.740 ","End":"03:20.525","Text":"that my potential for a point charge is equal to kq"},{"Start":"03:20.525 ","End":"03:25.800","Text":"divided by r. This is what you have to understand, how to calculate,"},{"Start":"03:25.800 ","End":"03:27.525","Text":"or you have to remember the answer,"},{"Start":"03:27.525 ","End":"03:31.700","Text":"and that\u0027s the potential of a point charge is equal to kq"},{"Start":"03:31.700 ","End":"03:36.440","Text":"divided by r. That is assuming that the potential at infinity,"},{"Start":"03:36.440 ","End":"03:40.230","Text":"so when we\u0027re very, very far away from the point charge, is equal to 0."},{"Start":"03:40.230 ","End":"03:41.690","Text":"That, for the most part,"},{"Start":"03:41.690 ","End":"03:47.705","Text":"is going to be our assumption for almost all of the questions that we\u0027ll get."},{"Start":"03:47.705 ","End":"03:50.880","Text":"That\u0027s the end of this lesson."}],"ID":14201},{"Watched":false,"Name":"First Method - Potential Of A Finite Wire","Duration":"11m 30s","ChapterTopicVideoID":12149,"CourseChapterTopicPlaylistID":157337,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12149.jpeg","UploadDate":"2018-06-28T04:58:36.6870000","DurationForVideoObject":"PT11M30S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.950","Text":"Hello. In this lesson,"},{"Start":"00:01.950 ","End":"00:06.960","Text":"we\u0027re going to be seeing how to work out the potential of a charged finite wire."},{"Start":"00:06.960 ","End":"00:10.231","Text":"Now, if we had a charged infinite wire, then of course,"},{"Start":"00:10.231 ","End":"00:13.200","Text":"we would use the technique for finding the potential which we"},{"Start":"00:13.200 ","End":"00:16.800","Text":"use for our shapes that we can use Gauss\u0027s law on."},{"Start":"00:16.800 ","End":"00:18.945","Text":"However, here it\u0027s something different."},{"Start":"00:18.945 ","End":"00:21.795","Text":"We\u0027re going to be using the superposition principle."},{"Start":"00:21.795 ","End":"00:28.260","Text":"Let\u0027s imagine that our wire is of length L and that it has charge Q."},{"Start":"00:28.260 ","End":"00:32.895","Text":"Now, we\u0027re saying that our charge is uniformly distributed."},{"Start":"00:32.895 ","End":"00:42.260","Text":"That means that our charge distribution per unit length is equal to Q divided by L. Now,"},{"Start":"00:42.260 ","End":"00:47.330","Text":"we\u0027re being asked what is the potential at this point over here,"},{"Start":"00:47.330 ","End":"00:54.480","Text":"which is located a distance of y above our wire?"},{"Start":"00:55.060 ","End":"00:58.220","Text":"I\u0027ve now put an axis over here."},{"Start":"00:58.220 ","End":"01:03.035","Text":"Let\u0027s say this is the x-direction and this is the y-direction."},{"Start":"01:03.035 ","End":"01:08.460","Text":"What we\u0027re trying to do is we\u0027re trying to find the potential at this point."},{"Start":"01:08.810 ","End":"01:14.305","Text":"Let\u0027s do a brief reminder of how to solve this and then we\u0027ll solve this question."},{"Start":"01:14.305 ","End":"01:16.175","Text":"Because we have a finite wire,"},{"Start":"01:16.175 ","End":"01:18.775","Text":"we don\u0027t have a case of symmetry."},{"Start":"01:18.775 ","End":"01:21.020","Text":"That means that we\u0027re going to have to use"},{"Start":"01:21.020 ","End":"01:26.585","Text":"our superposition principle or using a technique similar to Coulomb\u0027s law."},{"Start":"01:26.585 ","End":"01:32.080","Text":"What we\u0027re going to do is we\u0027re going to split up our wire into little pieces."},{"Start":"01:32.080 ","End":"01:34.395","Text":"Let\u0027s symbolize it like that."},{"Start":"01:34.395 ","End":"01:38.840","Text":"Then we\u0027re going to say that the charge in each piece is some kind of dq."},{"Start":"01:38.840 ","End":"01:44.020","Text":"Then from that, we\u0027re going to find our d Phi."},{"Start":"01:44.020 ","End":"01:49.505","Text":"We know from a previous lesson that our potential for a point charge,"},{"Start":"01:49.505 ","End":"01:53.180","Text":"where we regard each of these little pieces as a point charge,"},{"Start":"01:53.180 ","End":"01:56.555","Text":"so our potential for each point charge is equal to"},{"Start":"01:56.555 ","End":"02:00.950","Text":"kq divided by r. Then we\u0027re going to find"},{"Start":"02:00.950 ","End":"02:04.805","Text":"the total potential of our finite wire by"},{"Start":"02:04.805 ","End":"02:12.170","Text":"integrating along all of these point charges for the entire length of the wire."},{"Start":"02:12.170 ","End":"02:15.095","Text":"Let\u0027s begin."},{"Start":"02:15.095 ","End":"02:18.770","Text":"I\u0027m going to cut up this wire into lots of little pieces."},{"Start":"02:18.770 ","End":"02:21.680","Text":"Let\u0027s just take a look at this little piece."},{"Start":"02:21.680 ","End":"02:28.650","Text":"It\u0027s located here and it\u0027s a distance x from our origin."},{"Start":"02:30.970 ","End":"02:35.635","Text":"This little piece has a charge of dq."},{"Start":"02:35.635 ","End":"02:37.740","Text":"Let\u0027s write this here dq,"},{"Start":"02:37.740 ","End":"02:41.570","Text":"and dq is always going to be equal to the charge density per"},{"Start":"02:41.570 ","End":"02:47.260","Text":"unit length multiplied by the little length, the unit of length."},{"Start":"02:47.260 ","End":"02:50.515","Text":"Here specifically, because we\u0027re going along the x-axis,"},{"Start":"02:50.515 ","End":"02:52.595","Text":"we\u0027ll call our dl,"},{"Start":"02:52.595 ","End":"02:57.180","Text":"dx because it\u0027s minor changes along the x-direction."},{"Start":"02:58.250 ","End":"03:02.222","Text":"Now, we want to know what the potential at this point is"},{"Start":"03:02.222 ","End":"03:06.230","Text":"due to this point charge over here, dq."},{"Start":"03:06.230 ","End":"03:10.940","Text":"We know that the potential due to a point charge,"},{"Start":"03:10.940 ","End":"03:12.425","Text":"so here it\u0027s d Phi,"},{"Start":"03:12.425 ","End":"03:21.945","Text":"is going to be equal to kdq divided by r. A quick little note,"},{"Start":"03:21.945 ","End":"03:27.365","Text":"this r isn\u0027t the point where our charge dq is,"},{"Start":"03:27.365 ","End":"03:30.485","Text":"but r is the distance from"},{"Start":"03:30.485 ","End":"03:35.270","Text":"our point charge to the point in space where we\u0027re measuring our potential."},{"Start":"03:35.270 ","End":"03:38.825","Text":"That\u0027s this diagonal distance."},{"Start":"03:38.825 ","End":"03:44.480","Text":"I wrote this down under here because it\u0027s really important speaking about this r,"},{"Start":"03:44.480 ","End":"03:46.475","Text":"lots of people get confused with this."},{"Start":"03:46.475 ","End":"03:48.660","Text":"Don\u0027t be one of them."},{"Start":"03:48.940 ","End":"03:56.010","Text":"In the diagram, this is our value for r. I\u0027ll draw this in blue as well."},{"Start":"03:57.710 ","End":"04:02.795","Text":"This whole thing maybe later will become a little bit more confusing."},{"Start":"04:02.795 ","End":"04:07.295","Text":"Just try and understand as much as you can now."},{"Start":"04:07.295 ","End":"04:12.710","Text":"The magnitude of our r is going to be equal to,"},{"Start":"04:12.710 ","End":"04:18.350","Text":"by Pythagoras\u0027s theorem, x^2 plus y^2,"},{"Start":"04:18.350 ","End":"04:20.960","Text":"and then the square root of all of that."},{"Start":"04:20.960 ","End":"04:27.105","Text":"Now, we have all of the information that we need in order to find our potential."},{"Start":"04:27.105 ","End":"04:35.190","Text":"We can write that our potential is therefore the integral on d Phi."},{"Start":"04:35.190 ","End":"04:40.240","Text":"That\u0027s going to be the integral on our potential for all of our point charges,"},{"Start":"04:40.240 ","End":"04:45.630","Text":"which is going to be equal to kdq divided by our r,"},{"Start":"04:45.630 ","End":"04:52.125","Text":"where our value for r is here the square root of x^2 plus y^2."},{"Start":"04:52.125 ","End":"04:54.695","Text":"Now instead of writing dq,"},{"Start":"04:54.695 ","End":"04:56.810","Text":"I\u0027m going to substitute in what I have for it."},{"Start":"04:56.810 ","End":"05:00.990","Text":"It\u0027s k Lambda dx. Now,"},{"Start":"05:00.990 ","End":"05:03.030","Text":"let\u0027s talk about our bounds."},{"Start":"05:03.030 ","End":"05:08.075","Text":"What I\u0027m doing here is I\u0027m summing up along all of my pieces,"},{"Start":"05:08.075 ","End":"05:10.940","Text":"all of the pieces that I\u0027ve split this finite wire into,"},{"Start":"05:10.940 ","End":"05:13.150","Text":"so I\u0027m summing up all of them."},{"Start":"05:13.150 ","End":"05:16.505","Text":"As we can see, I have my axis over here."},{"Start":"05:16.505 ","End":"05:20.780","Text":"I have my direction in this direction"},{"Start":"05:20.780 ","End":"05:25.335","Text":"and my opposite side of the wire are going over here."},{"Start":"05:25.335 ","End":"05:28.695","Text":"Each length is L divided by 2."},{"Start":"05:28.695 ","End":"05:32.965","Text":"I\u0027m summing up from negative L divided by 2 until they get here,"},{"Start":"05:32.965 ","End":"05:36.000","Text":"and then up until L divided by 2."},{"Start":"05:36.000 ","End":"05:40.350","Text":"Let\u0027s put that in negative L divided by 2,"},{"Start":"05:40.350 ","End":"05:43.630","Text":"up until L divided by 2."},{"Start":"05:44.480 ","End":"05:49.120","Text":"This format for the integration comes up a lot where we have"},{"Start":"05:49.120 ","End":"05:55.615","Text":"dx divided by the square root of x^2 plus some constant."},{"Start":"05:55.615 ","End":"05:57.725","Text":"y here is a constant."},{"Start":"05:57.725 ","End":"06:01.000","Text":"Instead of working out now the integration,"},{"Start":"06:01.000 ","End":"06:03.085","Text":"you should write this in your notes."},{"Start":"06:03.085 ","End":"06:07.765","Text":"It\u0027s important and it will speed up how you solve questions on the exam."},{"Start":"06:07.765 ","End":"06:15.470","Text":"The integral of dx divided by the square root of x^2 plus b,"},{"Start":"06:15.470 ","End":"06:17.060","Text":"where b is a constant,"},{"Start":"06:17.060 ","End":"06:18.611","Text":"just like y is."},{"Start":"06:18.611 ","End":"06:21.245","Text":"That is equal to ln,"},{"Start":"06:21.245 ","End":"06:29.250","Text":"the natural log of x plus the square root of x^2 plus b."},{"Start":"06:31.550 ","End":"06:35.470","Text":"Of course, plus c if we don\u0027t have bounds,"},{"Start":"06:35.470 ","End":"06:41.220","Text":"so plus c, where b cannot be equal to 0."},{"Start":"06:42.050 ","End":"06:45.045","Text":"This, you should write down."},{"Start":"06:45.045 ","End":"06:48.255","Text":"Let\u0027s carry on our integration."},{"Start":"06:48.255 ","End":"06:51.320","Text":"Then we\u0027ll get that our integration is"},{"Start":"06:51.320 ","End":"06:54.320","Text":"going to be equal to k Lambda because they are constant,"},{"Start":"06:54.320 ","End":"07:03.635","Text":"multiplied by ln of x plus the square root of x^2 plus y^2."},{"Start":"07:03.635 ","End":"07:07.235","Text":"Then we have to substitute in our bounds,"},{"Start":"07:07.235 ","End":"07:13.410","Text":"which is from negative L divided by 2 up until L divided by 2."},{"Start":"07:13.610 ","End":"07:16.550","Text":"Then that is going to be equal to,"},{"Start":"07:16.550 ","End":"07:19.115","Text":"once we substitute this in."},{"Start":"07:19.115 ","End":"07:21.365","Text":"Let\u0027s go down here."},{"Start":"07:21.365 ","End":"07:25.410","Text":"Our Phi r potential is equal to k Lambda ln."},{"Start":"07:26.270 ","End":"07:28.470","Text":"Then when we\u0027re dealing with ln\u0027s,"},{"Start":"07:28.470 ","End":"07:32.195","Text":"when we have ln of something minus ln of something else,"},{"Start":"07:32.195 ","End":"07:37.520","Text":"we can write it simply as ln of something divided by the something else."},{"Start":"07:37.520 ","End":"07:40.845","Text":"It\u0027s rule of logarithms."},{"Start":"07:40.845 ","End":"07:43.425","Text":"Let\u0027s just write it in that format."},{"Start":"07:43.425 ","End":"07:48.335","Text":"We\u0027re going to have ln and then of the absolute value of L"},{"Start":"07:48.335 ","End":"07:53.665","Text":"divided by 2 plus the square root of L divided by 2^2 plus"},{"Start":"07:53.665 ","End":"08:00.735","Text":"y^2 divided by negative L"},{"Start":"08:00.735 ","End":"08:04.530","Text":"over 2 plus the square root of negative L"},{"Start":"08:04.530 ","End":"08:08.373","Text":"over 2 squared is simply going to be L over 2 squared,"},{"Start":"08:08.373 ","End":"08:12.070","Text":"it\u0027s same thing, plus y^2."},{"Start":"08:14.180 ","End":"08:16.390","Text":"That is our answer."},{"Start":"08:16.390 ","End":"08:21.720","Text":"This is the electric potential due to a finite wire."},{"Start":"08:22.640 ","End":"08:25.145","Text":"Now, just an added bonus."},{"Start":"08:25.145 ","End":"08:27.320","Text":"In a lot of exams,"},{"Start":"08:27.320 ","End":"08:31.709","Text":"a follow-up question will be to work out,"},{"Start":"08:31.709 ","End":"08:33.815","Text":"first of all, what is the potential?"},{"Start":"08:33.815 ","End":"08:40.020","Text":"Then they\u0027ll ask you, what is the electric field once you\u0027ve worked out the potential?"},{"Start":"08:40.020 ","End":"08:45.515","Text":"As we can see, our potential is as a function of y."},{"Start":"08:45.515 ","End":"08:47.270","Text":"We had variables x,"},{"Start":"08:47.270 ","End":"08:50.645","Text":"but then once we summed up along our finite wire,"},{"Start":"08:50.645 ","End":"08:52.880","Text":"we substituted in our bounds."},{"Start":"08:52.880 ","End":"08:58.505","Text":"Then we were left with our potential being simply as a function of y."},{"Start":"08:58.505 ","End":"09:01.190","Text":"Also, if we go back to our diagram,"},{"Start":"09:01.190 ","End":"09:04.235","Text":"we can see that our potential"},{"Start":"09:04.235 ","End":"09:08.045","Text":"is only going to be in the y-direction because from symmetry,"},{"Start":"09:08.045 ","End":"09:13.700","Text":"the potential in the x-direction is going to cancel out from both sides."},{"Start":"09:14.120 ","End":"09:20.390","Text":"In that case, if we\u0027re being asked to find the electric field, first of all,"},{"Start":"09:20.390 ","End":"09:23.330","Text":"we know that the electric field is only going to have a y component because"},{"Start":"09:23.330 ","End":"09:26.433","Text":"our potential only has a y component."},{"Start":"09:26.433 ","End":"09:32.260","Text":"Then that means that our electric field is going to be negative d Phi,"},{"Start":"09:32.260 ","End":"09:35.110","Text":"where Phi is our potential."},{"Start":"09:35.540 ","End":"09:44.175","Text":"It\u0027s not our angle in cylindrical coordinates as our potential, divided by dy."},{"Start":"09:44.175 ","End":"09:49.790","Text":"That means that we\u0027re taking the negative derivative of our potential with respect to y."},{"Start":"09:49.790 ","End":"09:52.550","Text":"Then we\u0027ll get that our electric field is"},{"Start":"09:52.550 ","End":"09:55.565","Text":"equal to whatever it will be in the y-direction."},{"Start":"09:55.565 ","End":"10:01.715","Text":"Now, of course, if our electric potential also had x or z components,"},{"Start":"10:01.715 ","End":"10:03.710","Text":"then to find our electric field,"},{"Start":"10:03.710 ","End":"10:07.190","Text":"we would also have to take the derivative of"},{"Start":"10:07.190 ","End":"10:11.420","Text":"our potential with respect to x or z or whatever else it might be."},{"Start":"10:11.420 ","End":"10:14.630","Text":"This technique is also used, for instance,"},{"Start":"10:14.630 ","End":"10:20.645","Text":"when we want to find the potential of a charged ring, so something like that."},{"Start":"10:20.645 ","End":"10:23.840","Text":"Another quick tip is if, let\u0027s say,"},{"Start":"10:23.840 ","End":"10:26.510","Text":"I\u0027m given this shape,"},{"Start":"10:26.800 ","End":"10:30.640","Text":"here specifically, it\u0027s an octagon,"},{"Start":"10:30.640 ","End":"10:33.050","Text":"we could also have a hexagon, whatever it might be."},{"Start":"10:33.050 ","End":"10:39.845","Text":"I\u0027m asked to find the potential along the axis of symmetry of this octagon shape."},{"Start":"10:39.845 ","End":"10:41.820","Text":"I didn\u0027t draw this very well."},{"Start":"10:41.820 ","End":"10:49.265","Text":"What I can do is I can find the potential due to just 1 finite wire."},{"Start":"10:49.265 ","End":"10:51.440","Text":"Just this piece over here,"},{"Start":"10:51.440 ","End":"10:54.170","Text":"which is exactly what we did in this question,"},{"Start":"10:54.170 ","End":"10:58.430","Text":"where this distance over here is y,"},{"Start":"10:58.430 ","End":"11:00.530","Text":"just like what we had over here."},{"Start":"11:00.530 ","End":"11:03.663","Text":"Then I can just use superposition."},{"Start":"11:03.663 ","End":"11:06.845","Text":"Once I get the answer for this finite wire,"},{"Start":"11:06.845 ","End":"11:09.088","Text":"which is exactly what we worked out over here,"},{"Start":"11:09.088 ","End":"11:14.240","Text":"then I just multiply that answer by the amount of sides that my shape has."},{"Start":"11:14.240 ","End":"11:16.880","Text":"Here I have an octagon, so I\u0027ll multiply it by 8."},{"Start":"11:16.880 ","End":"11:20.540","Text":"If I had a hexagon, I\u0027ll multiply it by 6,"},{"Start":"11:20.540 ","End":"11:27.845","Text":"and so on, where this will also be y and so on all the way around."},{"Start":"11:27.845 ","End":"11:31.020","Text":"That\u0027s the end of this lesson."}],"ID":14202},{"Watched":false,"Name":"Potential Of a Ring","Duration":"4m 20s","ChapterTopicVideoID":12150,"CourseChapterTopicPlaylistID":157337,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12150.jpeg","UploadDate":"2018-06-28T04:59:53.9530000","DurationForVideoObject":"PT4M20S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.770","Text":"Hello. In this question,"},{"Start":"00:01.770 ","End":"00:07.335","Text":"we\u0027re being asked to calculate the potential along a ring\u0027s axis of symmetry."},{"Start":"00:07.335 ","End":"00:09.840","Text":"Now we\u0027re being told that the ring has a radius of"},{"Start":"00:09.840 ","End":"00:15.615","Text":"R and charge density per unit length of Lambda."},{"Start":"00:15.615 ","End":"00:21.210","Text":"The way that we\u0027re going to find our potential for this shape is we\u0027re going to use"},{"Start":"00:21.210 ","End":"00:27.300","Text":"the idea of breaking up our ring into lots of tiny little pieces."},{"Start":"00:27.300 ","End":"00:31.110","Text":"What we\u0027re going to do is we\u0027re going to break the ring into tiny little pieces,"},{"Start":"00:31.110 ","End":"00:33.285","Text":"consider each piece as a point charge,"},{"Start":"00:33.285 ","End":"00:35.940","Text":"and then sum up to find the total potential at"},{"Start":"00:35.940 ","End":"00:40.395","Text":"this point due to all of the point charges along the ring."},{"Start":"00:40.395 ","End":"00:44.045","Text":"Let\u0027s break up a tiny piece,"},{"Start":"00:44.045 ","End":"00:51.425","Text":"some arbitrary piece along a ring and the length of this piece is dl."},{"Start":"00:51.425 ","End":"00:55.230","Text":"Now let\u0027s write out our dq."},{"Start":"00:55.230 ","End":"01:01.110","Text":"This length dl has some charge which is dq and we can consider it a point charge."},{"Start":"01:01.110 ","End":"01:05.515","Text":"Dq is given as Lambda dl."},{"Start":"01:05.515 ","End":"01:09.885","Text":"Now because we\u0027re working with a circle over here,"},{"Start":"01:09.885 ","End":"01:13.820","Text":"so we\u0027re going to be using polar coordinates."},{"Start":"01:13.820 ","End":"01:20.955","Text":"This length dl in polar coordinates is going to be equal to Rd Theta."},{"Start":"01:20.955 ","End":"01:26.785","Text":"R because that\u0027s the radius of our circle and our r is not changing."},{"Start":"01:26.785 ","End":"01:31.220","Text":"Now we know that from previous lessons,"},{"Start":"01:31.220 ","End":"01:33.125","Text":"the potential of a point charge,"},{"Start":"01:33.125 ","End":"01:36.530","Text":"which is exactly what this little slice over here is,"},{"Start":"01:36.530 ","End":"01:39.260","Text":"is equal to kdq,"},{"Start":"01:39.260 ","End":"01:45.110","Text":"so the charge of the point charge divided by r,"},{"Start":"01:45.110 ","End":"01:46.610","Text":"where r is,"},{"Start":"01:46.610 ","End":"01:52.260","Text":"I\u0027m reminding you the distance from this piece on our ring,"},{"Start":"01:52.260 ","End":"01:58.130","Text":"our dl piece until this point over here where we\u0027re measuring our potential."},{"Start":"01:58.130 ","End":"02:03.105","Text":"Let\u0027s just draw this out so that no one will get confused."},{"Start":"02:03.105 ","End":"02:07.775","Text":"This blue arrow is what we\u0027re calling r. This is our potential."},{"Start":"02:07.775 ","End":"02:12.340","Text":"Now let\u0027s find out what our size r is equal to."},{"Start":"02:12.340 ","End":"02:15.740","Text":"Let\u0027s imagine that our axis of symmetry over here is"},{"Start":"02:15.740 ","End":"02:20.090","Text":"the z-axis and that we\u0027re measuring our potential at this point over here,"},{"Start":"02:20.090 ","End":"02:25.705","Text":"which is some arbitrary length z, up our z-axis."},{"Start":"02:25.705 ","End":"02:30.620","Text":"Now we can see through Pythagoras\u0027s theorem that our r over here,"},{"Start":"02:30.620 ","End":"02:40.050","Text":"this horizontal length is simply going to be the square root of R^2 plus z^2."},{"Start":"02:40.050 ","End":"02:42.060","Text":"That\u0027s great. Now we have our r,"},{"Start":"02:42.060 ","End":"02:48.365","Text":"so all we have to do is sum up all of our dls along the entire ring."},{"Start":"02:48.365 ","End":"02:53.615","Text":"Our total potential is going to be the integral of our dv,"},{"Start":"02:53.615 ","End":"02:57.320","Text":"which is simply going to be the integral of,"},{"Start":"02:57.320 ","End":"03:07.365","Text":"let\u0027s substitute in so k now our dq is Lambda Id Theta divided by r,"},{"Start":"03:07.365 ","End":"03:12.940","Text":"and our r is the square root of I^2 plus z^2."},{"Start":"03:12.940 ","End":"03:16.490","Text":"Now what we have to do is we have to set in our bounds,"},{"Start":"03:16.490 ","End":"03:19.235","Text":"so we\u0027re going from our Theta=0,"},{"Start":"03:19.235 ","End":"03:24.645","Text":"so let\u0027s say that we have an x-axis over here."},{"Start":"03:24.645 ","End":"03:28.825","Text":"From Theta=0, and then we\u0027re going full circle"},{"Start":"03:28.825 ","End":"03:33.550","Text":"around the ring and summing up everything until we get back to our x-axis."},{"Start":"03:33.550 ","End":"03:39.420","Text":"That means that we\u0027re summing from 0 until 2Pi because that\u0027s a full circle."},{"Start":"03:39.420 ","End":"03:40.960","Text":"Now in our integral,"},{"Start":"03:40.960 ","End":"03:43.750","Text":"we have a super easy integral because we see that we don\u0027t"},{"Start":"03:43.750 ","End":"03:47.470","Text":"have Theta lurking around in our equation,"},{"Start":"03:47.470 ","End":"03:53.845","Text":"so we can simply just multiply everything by 2Pi minus 0 multiplied by everything."},{"Start":"03:53.845 ","End":"03:58.975","Text":"We\u0027re simply going to get 2Pi k Lambda r"},{"Start":"03:58.975 ","End":"04:05.465","Text":"divided by the square root of I^2 plus z^2."},{"Start":"04:05.465 ","End":"04:10.040","Text":"The square root of something can also be written like this to the power of a half."},{"Start":"04:10.040 ","End":"04:17.435","Text":"This is our answer for the potential due to a uniformly charged ring,"},{"Start":"04:17.435 ","End":"04:21.000","Text":"and that is the end of our lesson."}],"ID":14203},{"Watched":false,"Name":"Conductors","Duration":"13m 16s","ChapterTopicVideoID":12137,"CourseChapterTopicPlaylistID":157337,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12137.jpeg","UploadDate":"2018-06-28T04:04:10.6400000","DurationForVideoObject":"PT13M16S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.785","Text":"Hello. In this lesson,"},{"Start":"00:01.785 ","End":"00:04.605","Text":"we\u0027re going to be speaking about conductors."},{"Start":"00:04.605 ","End":"00:07.680","Text":"Now 2 things that we\u0027re specifically going to be discussing,"},{"Start":"00:07.680 ","End":"00:12.120","Text":"and that is that the electric field inside a conductor is equal to 0"},{"Start":"00:12.120 ","End":"00:17.560","Text":"and that the surface of a conductor carries charge."},{"Start":"00:18.440 ","End":"00:20.880","Text":"Conductors are, generally,"},{"Start":"00:20.880 ","End":"00:23.325","Text":"made out of a conducting material,"},{"Start":"00:23.325 ","End":"00:26.280","Text":"which is, generally, a metal."},{"Start":"00:26.280 ","End":"00:29.235","Text":"In a conductor and in a metal,"},{"Start":"00:29.235 ","End":"00:34.605","Text":"there are lots of negative charges,"},{"Start":"00:34.605 ","End":"00:39.240","Text":"and there are positive charges also,"},{"Start":"00:39.240 ","End":"00:41.864","Text":"and these charges can move."},{"Start":"00:41.864 ","End":"00:46.567","Text":"So technically speaking, only the negative charges can move."},{"Start":"00:46.567 ","End":"00:47.720","Text":"But it doesn\u0027t matter."},{"Start":"00:47.720 ","End":"00:52.860","Text":"There\u0027s positive and negative charges that can move in the conductor."},{"Start":"00:53.450 ","End":"01:01.010","Text":"Let\u0027s say that there\u0027s an electric field over here in the conductor."},{"Start":"01:01.010 ","End":"01:07.130","Text":"This electric field is going to apply a force to all of the charges."},{"Start":"01:07.130 ","End":"01:15.260","Text":"As we know, that force is equal to q,"},{"Start":"01:15.260 ","End":"01:18.360","Text":"the charge, multiplied by the electric field."},{"Start":"01:18.650 ","End":"01:24.060","Text":"As we know, the force acts in the direction of the electric field."},{"Start":"01:24.060 ","End":"01:26.650","Text":"So if we have a positive charge,"},{"Start":"01:26.650 ","End":"01:31.780","Text":"so it\u0027s going to move in the exact direction of the electric field."},{"Start":"01:31.780 ","End":"01:38.445","Text":"In that case, all of the positive charges will move up here."},{"Start":"01:38.445 ","End":"01:40.465","Text":"If this is a negative charge,"},{"Start":"01:40.465 ","End":"01:44.090","Text":"it\u0027s going to move in opposite direction to the electric field,"},{"Start":"01:44.090 ","End":"01:46.195","Text":"so that means down here."},{"Start":"01:46.195 ","End":"01:55.310","Text":"So we\u0027ll have these negative charges like so and positive charges at the top."},{"Start":"01:56.040 ","End":"02:00.355","Text":"Then we get this type of charge distribution where,"},{"Start":"02:00.355 ","End":"02:04.565","Text":"at the top, we have positive charges and,"},{"Start":"02:04.565 ","End":"02:08.165","Text":"at the bottom, we have negative charges."},{"Start":"02:08.165 ","End":"02:12.995","Text":"Then by the fact that we have positive charges over here and negative charges over here,"},{"Start":"02:12.995 ","End":"02:21.590","Text":"a force or an electric field due to this charge distribution is formed."},{"Start":"02:21.590 ","End":"02:26.450","Text":"Of course, we know that the electric field goes from positive to negative."},{"Start":"02:26.450 ","End":"02:29.936","Text":"So here, we\u0027ll have a different E field,"},{"Start":"02:29.936 ","End":"02:32.360","Text":"and let\u0027s also put it in blue,"},{"Start":"02:32.360 ","End":"02:33.755","Text":"but also this wave,"},{"Start":"02:33.755 ","End":"02:36.590","Text":"to show that it\u0027s a different E field to this E field."},{"Start":"02:36.590 ","End":"02:38.780","Text":"This is an external E field over here,"},{"Start":"02:38.780 ","End":"02:44.060","Text":"and this is the E field created by this charge distribution."},{"Start":"02:44.870 ","End":"02:47.195","Text":"As time goes on,"},{"Start":"02:47.195 ","End":"02:53.360","Text":"this external E field is going to cause this charge distribution of"},{"Start":"02:53.360 ","End":"03:00.759","Text":"positive charges at the top and negative charges along the bottom."},{"Start":"03:00.759 ","End":"03:08.460","Text":"Then that means that we\u0027re going to get this E wave field."},{"Start":"03:08.460 ","End":"03:18.660","Text":"More of this E field is going to grow in strength until these 2 E fields are equal."},{"Start":"03:18.660 ","End":"03:23.735","Text":"What we\u0027ll get is that our E field in red is eventually"},{"Start":"03:23.735 ","End":"03:28.460","Text":"going to equal our E wave field that we drew in blue,"},{"Start":"03:28.460 ","End":"03:33.770","Text":"which is the E field due to the charged distribution,"},{"Start":"03:33.770 ","End":"03:38.970","Text":"and then we can see that they\u0027re equal and opposite."},{"Start":"03:42.560 ","End":"03:45.825","Text":"Let\u0027s say this is the z direction."},{"Start":"03:45.825 ","End":"03:48.590","Text":"So this is in the positive z direction,"},{"Start":"03:48.590 ","End":"03:52.770","Text":"and this is in the negative z direction."},{"Start":"03:54.470 ","End":"04:00.935","Text":"Therefore, we will see that the total E field"},{"Start":"04:00.935 ","End":"04:07.985","Text":"inside the conductor is therefore equal to 0 because they\u0027re equal and opposite,"},{"Start":"04:07.985 ","End":"04:10.025","Text":"and they\u0027ll balance each other out,"},{"Start":"04:10.025 ","End":"04:12.780","Text":"and it will be equal to 0."},{"Start":"04:14.540 ","End":"04:21.125","Text":"That is why we\u0027ll have an E field that is equal to 0 inside a conductor."},{"Start":"04:21.125 ","End":"04:24.710","Text":"What is important to note is that this only happens if"},{"Start":"04:24.710 ","End":"04:28.670","Text":"we\u0027re dealing with static electricity. What does that mean?"},{"Start":"04:28.670 ","End":"04:31.460","Text":"Obviously, we can see that from the time that"},{"Start":"04:31.460 ","End":"04:34.954","Text":"an external E field is applied to the conductor,"},{"Start":"04:34.954 ","End":"04:39.880","Text":"it\u0027s going to take a while for this charge distribution to build up."},{"Start":"04:39.880 ","End":"04:41.970","Text":"The charges are moving,"},{"Start":"04:41.970 ","End":"04:44.090","Text":"and then after a certain amount of time,"},{"Start":"04:44.090 ","End":"04:48.275","Text":"we\u0027ll get this charge distribution where all of the positive charges are at the top,"},{"Start":"04:48.275 ","End":"04:51.230","Text":"all of the negative charges are at the bottom, and therefore,"},{"Start":"04:51.230 ","End":"04:54.290","Text":"the 2 E fields balance each other out, and therefore,"},{"Start":"04:54.290 ","End":"04:59.315","Text":"we get a total E field of 0 inside."},{"Start":"04:59.315 ","End":"05:03.330","Text":"But for this to happen, it takes time."},{"Start":"05:03.940 ","End":"05:08.405","Text":"If we\u0027re dealing with a question that is dealing with static electricity,"},{"Start":"05:08.405 ","End":"05:10.010","Text":"that means that we\u0027ve assumed,"},{"Start":"05:10.010 ","End":"05:14.165","Text":"that from turning on this external E field,"},{"Start":"05:14.165 ","End":"05:17.510","Text":"that enough time has gone past,"},{"Start":"05:17.510 ","End":"05:23.089","Text":"such that our system is already stable and in equilibrium,"},{"Start":"05:23.089 ","End":"05:26.600","Text":"which means that this charge distribution has occurred,"},{"Start":"05:26.600 ","End":"05:29.760","Text":"and now the system is stable in this state."},{"Start":"05:29.760 ","End":"05:32.000","Text":"But we\u0027ll see later on that,"},{"Start":"05:32.000 ","End":"05:34.805","Text":"when we\u0027re dealing not with static electricity,"},{"Start":"05:34.805 ","End":"05:39.019","Text":"so when we\u0027re dealing from the moment that this external E field is applied,"},{"Start":"05:39.019 ","End":"05:42.275","Text":"so the particles are still moving, and slowly,"},{"Start":"05:42.275 ","End":"05:45.350","Text":"we\u0027ll get this charge distribution,"},{"Start":"05:45.350 ","End":"05:48.800","Text":"and we\u0027ll see how it changes within time."},{"Start":"05:48.800 ","End":"05:51.890","Text":"The point to take home for this,"},{"Start":"05:51.890 ","End":"05:53.330","Text":"for Number 1 over here,"},{"Start":"05:53.330 ","End":"05:57.205","Text":"is that the electric field inside a conductor is equal to 0,"},{"Start":"05:57.205 ","End":"05:59.990","Text":"but this is when we\u0027re already assuming"},{"Start":"05:59.990 ","End":"06:06.155","Text":"that enough time has passed by so that the system is stable."},{"Start":"06:06.155 ","End":"06:12.350","Text":"Another important point to note is that sometimes inside a conductor,"},{"Start":"06:12.350 ","End":"06:17.150","Text":"we can get an electric field which is not equal to 0."},{"Start":"06:17.150 ","End":"06:18.875","Text":"How can that be?"},{"Start":"06:18.875 ","End":"06:23.870","Text":"Let\u0027s imagine that we don\u0027t have an external electric field being applied,"},{"Start":"06:23.870 ","End":"06:27.304","Text":"but we have a force being applied to our conductor,"},{"Start":"06:27.304 ","End":"06:31.160","Text":"and this force is a magnetic field."},{"Start":"06:31.160 ","End":"06:35.885","Text":"So we\u0027ll have a magnetic field,"},{"Start":"06:35.885 ","End":"06:37.010","Text":"B, over here,"},{"Start":"06:37.010 ","End":"06:44.790","Text":"and then that means that we\u0027re going to have some force due to the B field."},{"Start":"06:44.790 ","End":"06:50.570","Text":"This B field causes the positive charges to"},{"Start":"06:50.570 ","End":"06:57.105","Text":"be at the bottom and the negative charges to move upwards."},{"Start":"06:57.105 ","End":"07:00.935","Text":"Then we can see, because of this charge distribution,"},{"Start":"07:00.935 ","End":"07:05.400","Text":"we\u0027re going to have in this direction an E field."},{"Start":"07:06.530 ","End":"07:15.330","Text":"Then we can see that we\u0027re going to have an electrical force for the E field."},{"Start":"07:15.330 ","End":"07:17.040","Text":"We can see that they\u0027re in"},{"Start":"07:17.040 ","End":"07:20.540","Text":"the opposite direction so the E field is in the positive z direction,"},{"Start":"07:20.540 ","End":"07:24.565","Text":"and the B field is in the negative direction."},{"Start":"07:24.565 ","End":"07:27.140","Text":"What we can see is that, eventually,"},{"Start":"07:27.140 ","End":"07:31.310","Text":"the size of these fields are going to be the same,"},{"Start":"07:31.310 ","End":"07:39.290","Text":"but they\u0027re going to be just in opposite directions but of equal magnitude."},{"Start":"07:39.290 ","End":"07:49.280","Text":"Therefore, what we\u0027ll get is that the total force within the conductor is equal to 0."},{"Start":"07:49.280 ","End":"07:54.955","Text":"However, we do have some E-field,"},{"Start":"07:54.955 ","End":"07:57.065","Text":"and the E field does not equal to 0."},{"Start":"07:57.065 ","End":"08:01.940","Text":"We have an E field due to this charge distribution."},{"Start":"08:01.940 ","End":"08:05.105","Text":"This is something important to note,"},{"Start":"08:05.105 ","End":"08:13.400","Text":"that the most important thing is that the forces will be balanced out."},{"Start":"08:13.400 ","End":"08:16.025","Text":"Generally, in most of the questions that we deal with,"},{"Start":"08:16.025 ","End":"08:19.560","Text":"we\u0027re dealing with an external electric field."},{"Start":"08:19.560 ","End":"08:22.055","Text":"When there is an external electric field,"},{"Start":"08:22.055 ","End":"08:25.189","Text":"it will cause this charge distribution,"},{"Start":"08:25.189 ","End":"08:29.555","Text":"and then the charge distribution will itself form"},{"Start":"08:29.555 ","End":"08:34.056","Text":"an electric field in the opposite direction but of equal magnitude,"},{"Start":"08:34.056 ","End":"08:37.310","Text":"and therefore the forces will balance out."},{"Start":"08:37.310 ","End":"08:41.420","Text":"But because we\u0027re dealing specifically with forces to do with an electric field,"},{"Start":"08:41.420 ","End":"08:46.195","Text":"so we\u0027ll get that the subsequent E field inside is equal to 0."},{"Start":"08:46.195 ","End":"08:51.320","Text":"The E field inside a conductor is only equal to"},{"Start":"08:51.320 ","End":"08:56.990","Text":"0 if the only force being applied is an electric field force."},{"Start":"08:56.990 ","End":"09:00.560","Text":"If, however, a different force is causing"},{"Start":"09:00.560 ","End":"09:05.660","Text":"this charge distribution such as we saw a magnetic field force,"},{"Start":"09:05.660 ","End":"09:09.320","Text":"or let\u0027s say due to gravity or any other force,"},{"Start":"09:09.320 ","End":"09:12.710","Text":"then we will get that the forces balance"},{"Start":"09:12.710 ","End":"09:17.195","Text":"out but that the electric field is equal to something."},{"Start":"09:17.195 ","End":"09:20.060","Text":"It isn\u0027t equal to 0."},{"Start":"09:21.050 ","End":"09:28.860","Text":"Now let\u0027s speak about why the surface of a conductor does carry charge."},{"Start":"09:30.260 ","End":"09:35.105","Text":"When we\u0027re dealing with the usual cases of a conductor"},{"Start":"09:35.105 ","End":"09:40.505","Text":"being placed in the center of some external electric field,"},{"Start":"09:40.505 ","End":"09:42.500","Text":"then what we\u0027ll get is that"},{"Start":"09:42.500 ","End":"09:48.410","Text":"the internal electric field inside the conductor is equal to 0."},{"Start":"09:48.410 ","End":"09:53.330","Text":"Now let\u0027s take this random point over here inside the conductor,"},{"Start":"09:53.330 ","End":"09:57.380","Text":"and let\u0027s see if there is some charge to this point."},{"Start":"09:57.380 ","End":"10:02.240","Text":"We\u0027ll do this with a Gaussian shell,"},{"Start":"10:02.240 ","End":"10:06.940","Text":"a very small one where the radius is approaching 0."},{"Start":"10:06.940 ","End":"10:10.375","Text":"Now let\u0027s work out the flux."},{"Start":"10:10.375 ","End":"10:17.035","Text":"The flux is equal to the integral on E.ds."},{"Start":"10:17.035 ","End":"10:20.944","Text":"We already know that E is equal to 0,"},{"Start":"10:20.944 ","End":"10:27.535","Text":"so that means that this integral is equal to 0, which means that Q_in,"},{"Start":"10:27.535 ","End":"10:30.940","Text":"so the charge inside the shell,"},{"Start":"10:30.940 ","End":"10:35.110","Text":"divided by Epsilon naught is also equal to 0."},{"Start":"10:35.110 ","End":"10:40.385","Text":"Therefore, that means that the total charge inside is equal to 0."},{"Start":"10:40.385 ","End":"10:42.415","Text":"Now just a note,"},{"Start":"10:42.415 ","End":"10:44.365","Text":"the fact that Q_in is equal to 0,"},{"Start":"10:44.365 ","End":"10:49.280","Text":"it doesn\u0027t mean that there are no charges inside of this shell."},{"Start":"10:49.280 ","End":"10:51.710","Text":"There can be many charges,"},{"Start":"10:51.710 ","End":"10:57.845","Text":"but the sum of all of the charges balances out such that the total charge is equal to 0."},{"Start":"10:57.845 ","End":"11:03.320","Text":"So there\u0027s lots of positive charges and an equal amount of negative charges,"},{"Start":"11:03.320 ","End":"11:04.550","Text":"so they cancel each other out,"},{"Start":"11:04.550 ","End":"11:06.220","Text":"so the total charge inside is 0,"},{"Start":"11:06.220 ","End":"11:12.620","Text":"or conversely, we could also have that there really are no charges inside,"},{"Start":"11:12.620 ","End":"11:14.970","Text":"and then we\u0027ll get the same thing."},{"Start":"11:15.950 ","End":"11:22.985","Text":"This is what happens when we\u0027re looking for a charge on the inside of a conductor."},{"Start":"11:22.985 ","End":"11:26.945","Text":"But what happens on the surface of a conductor?"},{"Start":"11:26.945 ","End":"11:30.725","Text":"So we choose this point over here on the surface of the conductor."},{"Start":"11:30.725 ","End":"11:36.815","Text":"Now again, we\u0027re going to do this Gaussian shell with the teeny-tiny radius,"},{"Start":"11:36.815 ","End":"11:39.540","Text":"which is approaching 0."},{"Start":"11:40.460 ","End":"11:43.785","Text":"So this is the flux for Number 1,"},{"Start":"11:43.785 ","End":"11:44.957","Text":"which is at the center,"},{"Start":"11:44.957 ","End":"11:48.895","Text":"and now let\u0027s see what the flux is for Number 2,"},{"Start":"11:48.895 ","End":"11:50.300","Text":"which is at the surface."},{"Start":"11:50.300 ","End":"11:54.435","Text":"Now, because we see that we are at the surface,"},{"Start":"11:54.435 ","End":"11:59.200","Text":"so we saw that because we had our external E field,"},{"Start":"11:59.200 ","End":"12:01.295","Text":"so let\u0027s do this over here,"},{"Start":"12:01.295 ","End":"12:04.385","Text":"E_External in this direction."},{"Start":"12:04.385 ","End":"12:15.570","Text":"That meant that we got another E field in this direction due to the charge distribution."},{"Start":"12:15.570 ","End":"12:23.880","Text":"That means that we can see that we have lots of negative charges over here."},{"Start":"12:24.290 ","End":"12:28.460","Text":"So then we can see that we have negative charges over here"},{"Start":"12:28.460 ","End":"12:32.405","Text":"and an uneven of positive charges to balance this out."},{"Start":"12:32.405 ","End":"12:37.160","Text":"That means that we are going to get some flux over"},{"Start":"12:37.160 ","End":"12:42.860","Text":"here in this section that comes out of the conductor surface."},{"Start":"12:42.860 ","End":"12:47.720","Text":"Therefore, we\u0027ll get E.ds."},{"Start":"12:47.720 ","End":"12:51.330","Text":"Here E is not equal to 0."},{"Start":"12:51.410 ","End":"13:00.290","Text":"Therefore, we\u0027ll get that we have some charge in over here that is not equal to 0."},{"Start":"13:00.290 ","End":"13:06.438","Text":"Therefore, we can tell that only the surface of a conductor carries charge,"},{"Start":"13:06.438 ","End":"13:13.180","Text":"however, the center or the inner part of the conductor does not carry any charge."},{"Start":"13:13.310 ","End":"13:16.990","Text":"That is the end of this lesson."}],"ID":14204},{"Watched":false,"Name":"Difference Between Conductors and Insulators","Duration":"3m 38s","ChapterTopicVideoID":12138,"CourseChapterTopicPlaylistID":157337,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12138.jpeg","UploadDate":"2018-06-28T04:06:18.2030000","DurationForVideoObject":"PT3M38S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.700","Text":"Hello. In this lesson we\u0027re going to be speaking about"},{"Start":"00:02.700 ","End":"00:05.760","Text":"the difference between conductors and insulators."},{"Start":"00:05.760 ","End":"00:09.150","Text":"A lot of people sometimes get a bit confused."},{"Start":"00:09.150 ","End":"00:14.190","Text":"In conductors, we will have some kind of material,"},{"Start":"00:14.190 ","End":"00:18.560","Text":"usually a metal, where charges can move."},{"Start":"00:18.560 ","End":"00:23.115","Text":"In insulators, we will have some kind of insulating material,"},{"Start":"00:23.115 ","End":"00:29.380","Text":"such as wood or a certain fabrics where charges cannot move."},{"Start":"00:30.470 ","End":"00:33.075","Text":"If charges can move,"},{"Start":"00:33.075 ","End":"00:37.230","Text":"so we saw in the previous video that that means that if we look at"},{"Start":"00:37.230 ","End":"00:41.985","Text":"somewhere in the center of the conductor,"},{"Start":"00:41.985 ","End":"00:46.955","Text":"so here we won\u0027t have any charge distribution."},{"Start":"00:46.955 ","End":"00:52.345","Text":"Whereas conversely, in an insulator we can have charges,"},{"Start":"00:52.345 ","End":"00:54.095","Text":"so if we\u0027re looking at this,"},{"Start":"00:54.095 ","End":"00:56.720","Text":"we can have charges,"},{"Start":"00:56.720 ","End":"00:59.630","Text":"charge distribution anywhere, also in the middle and"},{"Start":"00:59.630 ","End":"01:02.690","Text":"also on the surface of the insulator."},{"Start":"01:02.690 ","End":"01:08.880","Text":"Whereas on the conductor we can only have charge distribution on the surface."},{"Start":"01:10.580 ","End":"01:16.039","Text":"In a conductor, the charge distribution is only on the surface,"},{"Start":"01:16.039 ","End":"01:20.675","Text":"so we can only have a charge density per unit area Sigma,"},{"Start":"01:20.675 ","End":"01:23.630","Text":"whereas in an insulator we can have charged"},{"Start":"01:23.630 ","End":"01:27.530","Text":"distribution everywhere so we can have Sigma and Rho,"},{"Start":"01:27.530 ","End":"01:28.580","Text":"where Rho is of course,"},{"Start":"01:28.580 ","End":"01:31.190","Text":"the charge density per unit volume."},{"Start":"01:31.190 ","End":"01:35.804","Text":"In all the questions where we were dealing with some kind of"},{"Start":"01:35.804 ","End":"01:41.060","Text":"solid sphere where it had charge density per unit volume Rho,"},{"Start":"01:41.060 ","End":"01:43.820","Text":"so we were dealing with an insulator,"},{"Start":"01:43.820 ","End":"01:45.500","Text":"and then of course,"},{"Start":"01:45.500 ","End":"01:49.015","Text":"or if we had Rho as a function of r,"},{"Start":"01:49.015 ","End":"01:50.730","Text":"this was an insulator,"},{"Start":"01:50.730 ","End":"01:52.729","Text":"and then we\u0027ll have charges,"},{"Start":"01:52.729 ","End":"01:58.280","Text":"whatever it is throughout the entire sphere from the center and all the way out."},{"Start":"01:58.280 ","End":"02:01.480","Text":"However, if we\u0027re dealing with a conductor,"},{"Start":"02:01.480 ","End":"02:06.500","Text":"so all the questions where we\u0027re either told that we\u0027re using"},{"Start":"02:06.500 ","End":"02:12.663","Text":"a conducting sphere or a similar question."},{"Start":"02:12.663 ","End":"02:14.855","Text":"So mathematically and physically speaking,"},{"Start":"02:14.855 ","End":"02:18.470","Text":"it\u0027s the same if we have just a spherical shell."},{"Start":"02:18.470 ","End":"02:27.015","Text":"So we\u0027ll have charge density only along the surface so we\u0027ll only have the Sigma,"},{"Start":"02:27.015 ","End":"02:32.405","Text":"so this is an example of a conductor and where we can have charges"},{"Start":"02:32.405 ","End":"02:40.480","Text":"everywhere all over so this is definitely an insulator."},{"Start":"02:40.610 ","End":"02:43.480","Text":"When we\u0027re dealing with insulators,"},{"Start":"02:43.480 ","End":"02:48.925","Text":"we can have a solid sphere with charge distribution everywhere,"},{"Start":"02:48.925 ","End":"02:53.320","Text":"and we could also have a spherical shell with charge distribution just at"},{"Start":"02:53.320 ","End":"02:57.730","Text":"the edge is but both of those can be insulators,"},{"Start":"02:57.730 ","End":"03:00.190","Text":"so let\u0027s do that and we can also have it where we just have"},{"Start":"03:00.190 ","End":"03:04.225","Text":"the charge distribution here if it\u0027s a spherical shell."},{"Start":"03:04.225 ","End":"03:09.205","Text":"However, if it\u0027s a spherical shell or it\u0027s a conducting sphere,"},{"Start":"03:09.205 ","End":"03:11.770","Text":"then we can only have,"},{"Start":"03:11.770 ","End":"03:13.615","Text":"if we\u0027re dealing with a conductor,"},{"Start":"03:13.615 ","End":"03:17.157","Text":"charge distribution on the surface."},{"Start":"03:17.157 ","End":"03:20.335","Text":"Of course, the final thing which we spoke about in the previous lesson,"},{"Start":"03:20.335 ","End":"03:26.135","Text":"in conductors, the electric field inside the conductor is going to be equal to 0."},{"Start":"03:26.135 ","End":"03:33.470","Text":"Whereas in an insulator we will have an electric field inside."},{"Start":"03:33.470 ","End":"03:36.305","Text":"Those are the differences between conductors and insulators,"},{"Start":"03:36.305 ","End":"03:39.000","Text":"and that is the end of this lesson."}],"ID":14205},{"Watched":false,"Name":"Symmetry","Duration":"15m 53s","ChapterTopicVideoID":12139,"CourseChapterTopicPlaylistID":157337,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12139.jpeg","UploadDate":"2018-06-28T04:13:47.0300000","DurationForVideoObject":"PT15M53S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.695","Text":"Hello. In this lesson,"},{"Start":"00:01.695 ","End":"00:04.790","Text":"we\u0027re going to be speaking about symmetry,"},{"Start":"00:04.790 ","End":"00:11.475","Text":"so we\u0027re going to see what it means in a mathematical sense and how we can use it."},{"Start":"00:11.475 ","End":"00:16.275","Text":"Let\u0027s imagine that we have this infinite wire."},{"Start":"00:16.275 ","End":"00:17.970","Text":"Whether it\u0027s charged or not,"},{"Start":"00:17.970 ","End":"00:20.700","Text":"it doesn\u0027t matter, but it\u0027s just an infinite wire."},{"Start":"00:20.700 ","End":"00:23.760","Text":"Then let\u0027s say that over here,"},{"Start":"00:23.760 ","End":"00:25.275","Text":"we have the origin."},{"Start":"00:25.275 ","End":"00:27.165","Text":"So this is the y-axis,"},{"Start":"00:27.165 ","End":"00:28.830","Text":"and this is the x-axis,"},{"Start":"00:28.830 ","End":"00:33.450","Text":"and what we want to see is if we have symmetry for this problem."},{"Start":"00:33.450 ","End":"00:38.170","Text":"First, let\u0027s check if there\u0027s symmetry in the x direction."},{"Start":"00:38.300 ","End":"00:42.695","Text":"How are we going to check this?"},{"Start":"00:42.695 ","End":"00:46.430","Text":"So we\u0027re going to define a new value for x,"},{"Start":"00:46.430 ","End":"00:55.590","Text":"which we\u0027re going to say is equal to x plus some constant a."},{"Start":"00:55.590 ","End":"01:02.715","Text":"What that means is we\u0027re taking the origin of this axis that we set up,"},{"Start":"01:02.715 ","End":"01:04.270","Text":"we\u0027re taking the origin,"},{"Start":"01:04.270 ","End":"01:09.550","Text":"and we\u0027re moving it some distance a along the x-axis."},{"Start":"01:09.550 ","End":"01:13.505","Text":"If our origin started off here,"},{"Start":"01:13.505 ","End":"01:16.850","Text":"so then we\u0027ll move it a distance of,"},{"Start":"01:16.850 ","End":"01:19.220","Text":"let\u0027s say this is the distance, a,"},{"Start":"01:19.220 ","End":"01:26.180","Text":"so then we\u0027ll have something like so going like this."},{"Start":"01:26.180 ","End":"01:32.780","Text":"So this is the new x-axis and the new y-axis, the new position."},{"Start":"01:32.780 ","End":"01:34.875","Text":"If this was x,"},{"Start":"01:34.875 ","End":"01:37.690","Text":"so this is now x\u0027."},{"Start":"01:38.530 ","End":"01:42.380","Text":"Now in order to see if I have symmetry,"},{"Start":"01:42.380 ","End":"01:50.335","Text":"I\u0027m going to look at my new problem where I have my axis located like so,"},{"Start":"01:50.335 ","End":"01:54.754","Text":"and then I look and see if there\u0027s any difference."},{"Start":"01:54.754 ","End":"01:56.420","Text":"I see that there\u0027s no difference."},{"Start":"01:56.420 ","End":"01:59.833","Text":"My wire is still infinity."},{"Start":"01:59.833 ","End":"02:01.190","Text":"It\u0027s infinitely long."},{"Start":"02:01.190 ","End":"02:05.940","Text":"So I have an infinite length going in the rightwards direction,"},{"Start":"02:05.940 ","End":"02:10.670","Text":"and I still have an infinite length going in the leftwards direction,"},{"Start":"02:10.670 ","End":"02:14.120","Text":"so really there\u0027s no difference if I solve the question"},{"Start":"02:14.120 ","End":"02:18.395","Text":"with an axis over here or over here."},{"Start":"02:18.395 ","End":"02:24.030","Text":"So we can see that I have symmetry in the x direction."},{"Start":"02:24.650 ","End":"02:30.300","Text":"So what we can see is that we have symmetry if I can move my coordinate system,"},{"Start":"02:30.300 ","End":"02:33.105","Text":"and I have the exact same problem."},{"Start":"02:33.105 ","End":"02:36.555","Text":"Now let\u0027s check my symmetry in y."},{"Start":"02:36.555 ","End":"02:43.640","Text":"Here is my origin with respect to y."},{"Start":"02:43.640 ","End":"02:46.415","Text":"What I\u0027m going to do is the exact same thing."},{"Start":"02:46.415 ","End":"02:50.265","Text":"I\u0027m going to define a new position for my origin y\u0027,"},{"Start":"02:50.265 ","End":"02:53.445","Text":"and this is at y plus a."},{"Start":"02:53.445 ","End":"02:57.725","Text":"Now I\u0027m going to move this up to,"},{"Start":"02:57.725 ","End":"02:59.165","Text":"let\u0027s say, over here."},{"Start":"02:59.165 ","End":"03:02.065","Text":"This is my new y position,"},{"Start":"03:02.065 ","End":"03:05.445","Text":"and this is my new x position."},{"Start":"03:05.445 ","End":"03:10.535","Text":"Now if I look at my problem,"},{"Start":"03:10.535 ","End":"03:16.670","Text":"so originally I could see that my wire was exactly at the origin."},{"Start":"03:16.670 ","End":"03:20.705","Text":"It ran through the origin or at y is equal to 0."},{"Start":"03:20.705 ","End":"03:22.805","Text":"But now I can see,"},{"Start":"03:22.805 ","End":"03:29.130","Text":"this is y\u0027,"},{"Start":"03:29.130 ","End":"03:34.590","Text":"that there is this distance a between the wire and the origin."},{"Start":"03:34.590 ","End":"03:38.050","Text":"So I can see that the configuration of my question has"},{"Start":"03:38.050 ","End":"03:42.610","Text":"completely changed before my axis sat on my wire,"},{"Start":"03:42.610 ","End":"03:48.660","Text":"and now my wire is a distance of negative a away from the origin."},{"Start":"03:48.660 ","End":"03:54.165","Text":"My problem set has changed due to me moving my axis,"},{"Start":"03:54.165 ","End":"04:01.150","Text":"so therefore, we can see that I have no symmetry in y."},{"Start":"04:02.000 ","End":"04:07.210","Text":"Let\u0027s go back to x where we saw that we did have symmetry."},{"Start":"04:07.210 ","End":"04:12.380","Text":"What does it mean mathematically if we have symmetry in the x direction?"},{"Start":"04:12.380 ","End":"04:18.480","Text":"That means that if I take the derivative of something in the x direction,"},{"Start":"04:18.480 ","End":"04:20.570","Text":"so the derivative of anything,"},{"Start":"04:20.570 ","End":"04:24.320","Text":"if this is a scalar or a vector, it doesn\u0027t matter,"},{"Start":"04:24.320 ","End":"04:25.925","Text":"but if it\u0027s in the x direction,"},{"Start":"04:25.925 ","End":"04:31.095","Text":"the derivative of it with respect to x will be equal to 0."},{"Start":"04:31.095 ","End":"04:34.738","Text":"This was what defined symmetry."},{"Start":"04:34.738 ","End":"04:35.810","Text":"Let`s take a look."},{"Start":"04:35.810 ","End":"04:41.670","Text":"Let\u0027s say I have a temperature gradient all over here and,"},{"Start":"04:41.670 ","End":"04:43.745","Text":"at this height, over here, at this point,"},{"Start":"04:43.745 ","End":"04:47.180","Text":"I see that my temperature is equal to 30 degrees."},{"Start":"04:47.180 ","End":"04:53.225","Text":"That means that if I shift this point over here along the x-axis, in the x direction,"},{"Start":"04:53.225 ","End":"04:56.510","Text":"the temperature over here will also be equal to"},{"Start":"04:56.510 ","End":"05:00.425","Text":"30 degrees centigrade and so on and so forth."},{"Start":"05:00.425 ","End":"05:01.610","Text":"At all of these points,"},{"Start":"05:01.610 ","End":"05:05.210","Text":"the temperature will be 30 degrees centigrade."},{"Start":"05:05.210 ","End":"05:06.905","Text":"That means that if,"},{"Start":"05:06.905 ","End":"05:10.880","Text":"in this example, I took the derivative of temperature,"},{"Start":"05:10.880 ","End":"05:12.950","Text":"which is a function of its position,"},{"Start":"05:12.950 ","End":"05:16.500","Text":"wherever that might be with respect to x,"},{"Start":"05:16.500 ","End":"05:19.645","Text":"it will be equal to 0."},{"Start":"05:19.645 ","End":"05:23.946","Text":"Another example is if we have an E field."},{"Start":"05:23.946 ","End":"05:27.040","Text":"So if, at this position over here in y,"},{"Start":"05:27.040 ","End":"05:29.672","Text":"we have an E field in this direction,"},{"Start":"05:29.672 ","End":"05:32.355","Text":"so if we shift along the x-axis,"},{"Start":"05:32.355 ","End":"05:36.695","Text":"we\u0027re going to have the exact same E field pointing in the exact same direction."},{"Start":"05:36.695 ","End":"05:39.229","Text":"Now of course, if we change our y position,"},{"Start":"05:39.229 ","End":"05:44.532","Text":"it could be that the direction or the magnitude of our E field changes."},{"Start":"05:44.532 ","End":"05:46.455","Text":"But if y is constant,"},{"Start":"05:46.455 ","End":"05:48.480","Text":"and we just move along the x,"},{"Start":"05:48.480 ","End":"05:50.880","Text":"our E field is going to be the same."},{"Start":"05:50.880 ","End":"05:58.030","Text":"So then if we took the derivative of our E field with respect to x,"},{"Start":"05:58.030 ","End":"06:04.450","Text":"we\u0027ll also get that it is equal to 0 because we have symmetry in the x direction."},{"Start":"06:04.450 ","End":"06:08.445","Text":"So if we take our E field again,"},{"Start":"06:08.445 ","End":"06:13.681","Text":"and we look at it, and we want to split it up into its different components,"},{"Start":"06:13.681 ","End":"06:17.420","Text":"so here we have the y component of the E field, so E_y,"},{"Start":"06:17.420 ","End":"06:23.270","Text":"and here we have the x-component of the E field, so E_x."},{"Start":"06:23.270 ","End":"06:26.255","Text":"What we can see is, if we shift along the x-axis,"},{"Start":"06:26.255 ","End":"06:30.110","Text":"the y component of the electric field and"},{"Start":"06:30.110 ","End":"06:35.615","Text":"the x-component of the electric field don\u0027t change."},{"Start":"06:35.615 ","End":"06:41.720","Text":"So therefore, what we can see is if we look at this,"},{"Start":"06:41.720 ","End":"06:48.110","Text":"then this means that if we take the derivative of the different components of"},{"Start":"06:48.110 ","End":"06:55.740","Text":"whatever the scalar or vector equation or whatever it might be with respect to x,"},{"Start":"06:55.740 ","End":"06:57.550","Text":"so A_x with respect to x,"},{"Start":"06:57.550 ","End":"06:59.860","Text":"we\u0027re taking the derivative that will equal to 0,"},{"Start":"06:59.860 ","End":"07:05.818","Text":"and also if we take the derivative of the y component with respect to x,"},{"Start":"07:05.818 ","End":"07:12.318","Text":"so this will also still be equal to 0 because we can see it\u0027s not changing."},{"Start":"07:12.318 ","End":"07:15.150","Text":"Now let\u0027s look at a different type of symmetry,"},{"Start":"07:15.150 ","End":"07:18.675","Text":"and this is symmetry with respect to an angle."},{"Start":"07:18.675 ","End":"07:24.030","Text":"Let\u0027s imagine that we have some cylinder like so."},{"Start":"07:24.030 ","End":"07:28.935","Text":"Imagine that I\u0027ve drawn the cylinder properly,"},{"Start":"07:28.935 ","End":"07:34.365","Text":"and so here we have some axis going through it like this."},{"Start":"07:34.365 ","End":"07:39.270","Text":"I can have some symmetry with respect to an angle."},{"Start":"07:39.270 ","End":"07:43.380","Text":"That means that I have the same material over"},{"Start":"07:43.380 ","End":"07:49.980","Text":"here going all the way around 360 degrees around the cylinder."},{"Start":"07:49.980 ","End":"07:53.850","Text":"Let\u0027s say I have some charge density, Rho."},{"Start":"07:53.850 ","End":"07:58.065","Text":"It can be as a function of r,"},{"Start":"07:58.065 ","End":"08:04.365","Text":"where r is coming from the origin and out towards the edge of the cylinder."},{"Start":"08:04.365 ","End":"08:07.260","Text":"I can have a different density over here,"},{"Start":"08:07.260 ","End":"08:10.205","Text":"closer to the origin,"},{"Start":"08:10.205 ","End":"08:14.220","Text":"and then the density changes as we come closer to the edge of the cylinder."},{"Start":"08:14.220 ","End":"08:16.590","Text":"However, if our r is constant,"},{"Start":"08:16.590 ","End":"08:18.225","Text":"we\u0027re located on the same r,"},{"Start":"08:18.225 ","End":"08:22.543","Text":"but we move around with a different angle,"},{"Start":"08:22.543 ","End":"08:29.310","Text":"we\u0027re meant to get the same value for the problem,"},{"Start":"08:29.310 ","End":"08:33.150","Text":"but if we have symmetry with respect to an angle."},{"Start":"08:33.150 ","End":"08:39.000","Text":"Symmetry with respect to an angle is if I define a new angle, Theta,"},{"Start":"08:39.000 ","End":"08:42.090","Text":"so I define it as the original angle that we were"},{"Start":"08:42.090 ","End":"08:46.725","Text":"at plus some Theta naught, some constant."},{"Start":"08:46.725 ","End":"08:51.345","Text":"If I rotate the cylinder by this Theta naught,"},{"Start":"08:51.345 ","End":"08:53.565","Text":"so if I look back at my problem,"},{"Start":"08:53.565 ","End":"08:57.465","Text":"I meant to have the exact same question in front of me,"},{"Start":"08:57.465 ","End":"09:01.890","Text":"as if the question hasn\u0027t changed even though I rotated the cylinder."},{"Start":"09:01.890 ","End":"09:04.680","Text":"That means that I have symmetry with"},{"Start":"09:04.680 ","End":"09:08.085","Text":"respect to Theta or symmetry with respect to an angle."},{"Start":"09:08.085 ","End":"09:11.460","Text":"So as we\u0027ve seen, that means that if I take"},{"Start":"09:11.460 ","End":"09:16.695","Text":"the derivative of anything with respect to that angle,"},{"Start":"09:16.695 ","End":"09:22.335","Text":"Theta, so my derivative is meant to be equal to 0."},{"Start":"09:22.335 ","End":"09:25.335","Text":"If I have an electric field,"},{"Start":"09:25.335 ","End":"09:29.970","Text":"like so, and my electric field is in the radial direction,"},{"Start":"09:29.970 ","End":"09:36.600","Text":"so that means that it can also have a component with the r coordinate,"},{"Start":"09:36.600 ","End":"09:40.530","Text":"but it can also have a component in the Theta direction."},{"Start":"09:40.530 ","End":"09:44.350","Text":"This means, if it\u0027s in the radial direction,"},{"Start":"09:45.020 ","End":"09:52.425","Text":"as I go along the cylinder,"},{"Start":"09:52.425 ","End":"10:00.045","Text":"so I have the exact same E field in the radial direction,"},{"Start":"10:00.045 ","End":"10:03.340","Text":"and it\u0027s the same magnitude."},{"Start":"10:03.590 ","End":"10:06.300","Text":"As my angle Theta changes,"},{"Start":"10:06.300 ","End":"10:09.360","Text":"my electric field remains the same because"},{"Start":"10:09.360 ","End":"10:13.350","Text":"of this because I have symmetry in the Theta direction."},{"Start":"10:13.350 ","End":"10:22.215","Text":"Similarly, if I had a perpendicular component over here for some B field,"},{"Start":"10:22.215 ","End":"10:24.378","Text":"going like so,"},{"Start":"10:24.378 ","End":"10:29.580","Text":"the arrows in blue is this perpendicular B field."},{"Start":"10:29.580 ","End":"10:36.330","Text":"This is also symmetrical in the Theta direction because it"},{"Start":"10:36.330 ","End":"10:43.690","Text":"stays the same magnitude and direction as our Theta coordinate changes."},{"Start":"10:44.330 ","End":"10:49.320","Text":"This is an example of a finite cylinder."},{"Start":"10:49.320 ","End":"10:58.659","Text":"But what happens if this is the z-axis and our cylinder is infinite in the z direction?"},{"Start":"10:58.659 ","End":"11:01.537","Text":"So then it becomes like this infinite wire,"},{"Start":"11:01.537 ","End":"11:06.810","Text":"just the wire is slightly thicker because we\u0027ve made it into this cylinder."},{"Start":"11:06.810 ","End":"11:10.185","Text":"Now the cylinder is infinite in the z direction."},{"Start":"11:10.185 ","End":"11:14.505","Text":"That means that if we take some coordinate, z\u0027,"},{"Start":"11:14.505 ","End":"11:18.360","Text":"and we define it as z plus some constant,"},{"Start":"11:18.360 ","End":"11:20.450","Text":"that\u0027s again use a,"},{"Start":"11:20.450 ","End":"11:24.330","Text":"then we can see that if we look at"},{"Start":"11:24.330 ","End":"11:30.525","Text":"this section of the infinite cylinder or this section up top of the infinite cylinder,"},{"Start":"11:30.525 ","End":"11:34.170","Text":"we still have the exact same question in front of us,"},{"Start":"11:34.170 ","End":"11:36.870","Text":"which is just an infinite cylinder."},{"Start":"11:36.870 ","End":"11:42.163","Text":"Therefore, that means that we can say that we have symmetry in the z axis,"},{"Start":"11:42.163 ","End":"11:46.260","Text":"so then we can write something like so."},{"Start":"11:46.260 ","End":"11:48.390","Text":"If we take the derivative of a,"},{"Start":"11:48.390 ","End":"11:55.420","Text":"which is either a vector or a scalar equation or number, it doesn\u0027t matter."},{"Start":"11:55.420 ","End":"11:57.200","Text":"But we take the derivative with respect to"},{"Start":"11:57.200 ","End":"11:59.690","Text":"z because we have symmetry in the z direction,"},{"Start":"11:59.690 ","End":"12:02.820","Text":"this will also be equal to 0."},{"Start":"12:03.830 ","End":"12:06.390","Text":"As we saw over here,"},{"Start":"12:06.390 ","End":"12:08.040","Text":"we have this E field,"},{"Start":"12:08.040 ","End":"12:11.130","Text":"and we saw that it has symmetry in the Theta direction."},{"Start":"12:11.130 ","End":"12:13.350","Text":"As our Theta coordinate changes,"},{"Start":"12:13.350 ","End":"12:15.795","Text":"E, the electric field,"},{"Start":"12:15.795 ","End":"12:17.430","Text":"is exactly the same."},{"Start":"12:17.430 ","End":"12:22.191","Text":"Now we\u0027re being told that we also have the symmetry in the z direction."},{"Start":"12:22.191 ","End":"12:25.980","Text":"That means that if I go a little bit above here,"},{"Start":"12:25.980 ","End":"12:31.470","Text":"I have this exact same E field because we have"},{"Start":"12:31.470 ","End":"12:40.510","Text":"this symmetry in the z direction, something like so."},{"Start":"12:41.600 ","End":"12:46.540","Text":"That means that along this entire cylindrical shell,"},{"Start":"12:46.540 ","End":"12:51.045","Text":"the surface of the cylinder,"},{"Start":"12:51.045 ","End":"12:54.675","Text":"all the way around going up and down,"},{"Start":"12:54.675 ","End":"13:01.960","Text":"I\u0027m going to have this symmetry where I have this even E field throughout."},{"Start":"13:02.240 ","End":"13:06.780","Text":"Now let\u0027s speak about spherical symmetry."},{"Start":"13:06.780 ","End":"13:10.425","Text":"If I have a sphere, and it\u0027s solid,"},{"Start":"13:10.425 ","End":"13:15.600","Text":"and it has charge density per unit volume of Rho,"},{"Start":"13:15.600 ","End":"13:18.070","Text":"and let\u0027s say that Rho is dependent on r,"},{"Start":"13:18.070 ","End":"13:24.990","Text":"so it\u0027s dependent on the radius that we are at within the sphere."},{"Start":"13:24.990 ","End":"13:28.590","Text":"So as we move from the origin to the edge of the sphere,"},{"Start":"13:28.590 ","End":"13:32.370","Text":"our charge is changing because our charge density is changing."},{"Start":"13:32.370 ","End":"13:34.410","Text":"So in a sphere,"},{"Start":"13:34.410 ","End":"13:37.410","Text":"we can see that with respect to r,"},{"Start":"13:37.410 ","End":"13:39.240","Text":"we don\u0027t have symmetry because,"},{"Start":"13:39.240 ","End":"13:41.280","Text":"if we look at different values of r,"},{"Start":"13:41.280 ","End":"13:44.025","Text":"we have a different problem in front of us,"},{"Start":"13:44.025 ","End":"13:45.357","Text":"a different charge,"},{"Start":"13:45.357 ","End":"13:49.920","Text":"but we can see that this is only dependent on r. So if I"},{"Start":"13:49.920 ","End":"13:56.445","Text":"spin this sphere around like so in the Theta direction,"},{"Start":"13:56.445 ","End":"14:00.300","Text":"I can see that I\u0027m left with the exact same problem."},{"Start":"14:00.300 ","End":"14:02.865","Text":"I\u0027m looking at the exact same thing."},{"Start":"14:02.865 ","End":"14:08.175","Text":"So that means that if I take the derivative of something with respect to Theta,"},{"Start":"14:08.175 ","End":"14:10.230","Text":"because I have symmetry in the Theta direction,"},{"Start":"14:10.230 ","End":"14:12.270","Text":"this is going to be equal to 0."},{"Start":"14:12.270 ","End":"14:18.540","Text":"Similarly, if I spin this around in this direction,"},{"Start":"14:18.540 ","End":"14:25.237","Text":"like so, so it\u0027s doing roly-polies,"},{"Start":"14:25.237 ","End":"14:29.120","Text":"I can see that still I have the exact same problem ahead of me."},{"Start":"14:29.120 ","End":"14:34.970","Text":"It\u0027s like taking a football that doesn\u0027t have any markings on it and rolling it around."},{"Start":"14:34.970 ","End":"14:36.845","Text":"If I\u0027m looking at the surface,"},{"Start":"14:36.845 ","End":"14:41.000","Text":"the football is going to look exactly the same as before I rolled it."},{"Start":"14:41.000 ","End":"14:47.354","Text":"So then we can see that I have symmetry in the Phi direction."},{"Start":"14:47.354 ","End":"14:52.415","Text":"So that I can say that if I take the derivative with respect to Phi,"},{"Start":"14:52.415 ","End":"14:56.760","Text":"I\u0027m also going to get 0."},{"Start":"14:57.620 ","End":"15:01.595","Text":"What I have is a spherical shell,"},{"Start":"15:01.595 ","End":"15:04.445","Text":"where on the spherical shell,"},{"Start":"15:04.445 ","End":"15:10.115","Text":"if all of these dots represent E fields coming out in the radial direction,"},{"Start":"15:10.115 ","End":"15:13.629","Text":"like so all around,"},{"Start":"15:13.629 ","End":"15:18.440","Text":"so I just have a charged spherical shell that has an even E field."},{"Start":"15:18.440 ","End":"15:22.460","Text":"As I change the depth that I\u0027m at within the sphere,"},{"Start":"15:22.460 ","End":"15:24.094","Text":"so the E field again,"},{"Start":"15:24.094 ","End":"15:28.565","Text":"if I spin the sphere around in the Theta or the Phi direction,"},{"Start":"15:28.565 ","End":"15:32.600","Text":"I\u0027m going to have the same charge density"},{"Start":"15:32.600 ","End":"15:36.125","Text":"as long as I\u0027m located always at the same depth."},{"Start":"15:36.125 ","End":"15:38.600","Text":"If I look at a different radius,"},{"Start":"15:38.600 ","End":"15:40.227","Text":"so let\u0027s say over here,"},{"Start":"15:40.227 ","End":"15:44.245","Text":"so I\u0027ll have a different problem in front of me."},{"Start":"15:44.245 ","End":"15:47.035","Text":"But if I just change my Theta and Phi,"},{"Start":"15:47.035 ","End":"15:50.135","Text":"then I\u0027m looking at the same exact thing."},{"Start":"15:50.135 ","End":"15:54.030","Text":"That is the end of this lesson."}],"ID":14206},{"Watched":false,"Name":"Second Method - Gauss Law","Duration":"50m 45s","ChapterTopicVideoID":12140,"CourseChapterTopicPlaylistID":157337,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12140.jpeg","UploadDate":"2018-06-28T04:33:44.2700000","DurationForVideoObject":"PT50M45S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.354","Text":"Hello. In this lesson,"},{"Start":"00:02.354 ","End":"00:06.510","Text":"we\u0027re going to be speaking about the second method or the second technique"},{"Start":"00:06.510 ","End":"00:11.805","Text":"for solving questions where we\u0027re asked to find the potential."},{"Start":"00:11.805 ","End":"00:16.875","Text":"The second method deals with questions involving Gauss\u0027 law."},{"Start":"00:16.875 ","End":"00:20.475","Text":"If we\u0027re ever told to find the potential of an infinite wire,"},{"Start":"00:20.475 ","End":"00:21.735","Text":"or an infinite cylinder,"},{"Start":"00:21.735 ","End":"00:23.130","Text":"or an infinite plane,"},{"Start":"00:23.130 ","End":"00:24.840","Text":"or a spherical shell,"},{"Start":"00:24.840 ","End":"00:29.565","Text":"we can already see that we can use Gauss\u0027 Law to find the electric field,"},{"Start":"00:29.565 ","End":"00:34.139","Text":"and then we can find the potential by taking the negative"},{"Start":"00:34.139 ","End":"00:39.735","Text":"integral on this electric field that we found using Gauss\u0027 law."},{"Start":"00:39.735 ","End":"00:44.810","Text":"We\u0027re going to see how to solve this by using an example."},{"Start":"00:44.810 ","End":"00:48.230","Text":"We\u0027re being asked what is the potential of this"},{"Start":"00:48.230 ","End":"00:52.420","Text":"over here and what is the charge distribution?"},{"Start":"00:52.420 ","End":"00:54.780","Text":"Let\u0027s go over what we have here."},{"Start":"00:54.780 ","End":"01:02.495","Text":"Here in black, we have a solid conducting sphere of radius R and of charge Q."},{"Start":"01:02.495 ","End":"01:05.550","Text":"This is a solid sphere."},{"Start":"01:05.920 ","End":"01:11.345","Text":"Then after it, we have this blue circle over here."},{"Start":"01:11.345 ","End":"01:16.744","Text":"This blue circle we can see as of radius 2R and this is"},{"Start":"01:16.744 ","End":"01:26.164","Text":"a grounded conducting thin spherical shell and then over here in pink,"},{"Start":"01:26.164 ","End":"01:30.920","Text":"we have a conducting thick spherical shell."},{"Start":"01:30.920 ","End":"01:35.059","Text":"It has a width of R. We can see that"},{"Start":"01:35.059 ","End":"01:44.010","Text":"the inner side of this fixed spherical shell is at 3R and the outer edge is at 4R."},{"Start":"01:44.010 ","End":"01:53.170","Text":"The total width over here is R and we\u0027re being told that it has a charge of negative 2Q."},{"Start":"01:54.890 ","End":"01:58.999","Text":"We have a solid sphere,"},{"Start":"01:58.999 ","End":"02:01.715","Text":"a thin spherical shell,"},{"Start":"02:01.715 ","End":"02:05.670","Text":"and a thick spherical shell."},{"Start":"02:06.200 ","End":"02:09.169","Text":"First, let\u0027s calculate the potential."},{"Start":"02:09.169 ","End":"02:14.329","Text":"Now, we can see that we have spheres and spherical shells throughout this question,"},{"Start":"02:14.329 ","End":"02:18.769","Text":"which means that a very easy way to solve this potential is by using"},{"Start":"02:18.769 ","End":"02:23.665","Text":"Gauss\u0027 law in order to find the electric field and then just integrating along it,"},{"Start":"02:23.665 ","End":"02:26.080","Text":"and then we\u0027ll find the potential."},{"Start":"02:26.080 ","End":"02:28.760","Text":"The first step that we\u0027re going to have to do is to"},{"Start":"02:28.760 ","End":"02:32.310","Text":"find the electric field using Gauss\u0027 law."},{"Start":"02:34.370 ","End":"02:38.064","Text":"What we\u0027re going to do is we\u0027re going to split this up into"},{"Start":"02:38.064 ","End":"02:42.260","Text":"different sections and we\u0027ll find the electric field in each section."},{"Start":"02:42.780 ","End":"02:50.025","Text":"So small r is some radius that we\u0027re changing constantly."},{"Start":"02:50.025 ","End":"02:53.560","Text":"When this radius is smaller than R,"},{"Start":"02:53.560 ","End":"02:58.389","Text":"that means that we\u0027re located somewhere within the solid conducting sphere."},{"Start":"02:58.389 ","End":"03:02.245","Text":"Now, as we know, if we have some conductor,"},{"Start":"03:02.245 ","End":"03:08.010","Text":"that means that the electric field within the conductor is equal to 0."},{"Start":"03:08.010 ","End":"03:11.320","Text":"In most of the questions that we\u0027re going to be dealing with,"},{"Start":"03:11.320 ","End":"03:15.330","Text":"because we\u0027re dealing with an external electric field over here,"},{"Start":"03:15.330 ","End":"03:20.065","Text":"which causes the electric field within the conductor to be equal to 0."},{"Start":"03:20.065 ","End":"03:24.760","Text":"That is something good to remember."},{"Start":"03:26.600 ","End":"03:29.990","Text":"Now, what we want to do is we want to find"},{"Start":"03:29.990 ","End":"03:35.374","Text":"the electric field at someplace between this solid conducting sphere,"},{"Start":"03:35.374 ","End":"03:39.630","Text":"and this grounded conducting thin spherical shell."},{"Start":"03:39.630 ","End":"03:41.895","Text":"Let\u0027s choose a random point."},{"Start":"03:41.895 ","End":"03:46.310","Text":"We want to find the electrical field at this point over here."},{"Start":"03:46.310 ","End":"03:51.829","Text":"Now, we can say that this is our radius r"},{"Start":"03:51.829 ","End":"03:57.990","Text":"and we\u0027re trying to find the electrical field at this point over here."},{"Start":"03:58.220 ","End":"04:00.829","Text":"If you have a little bit of intuition,"},{"Start":"04:00.829 ","End":"04:06.289","Text":"you can already tell that we\u0027re going to get the electric field of a point charge,"},{"Start":"04:06.289 ","End":"04:10.864","Text":"because a point charge is just a conducting sphere,"},{"Start":"04:10.864 ","End":"04:12.885","Text":"which is very, very small."},{"Start":"04:12.885 ","End":"04:15.710","Text":"We can already see that that\u0027s going to be our electric field."},{"Start":"04:15.710 ","End":"04:17.795","Text":"But let\u0027s calculate it anyway."},{"Start":"04:17.795 ","End":"04:25.965","Text":"Now, where in the region where r is between R and 2R."},{"Start":"04:25.965 ","End":"04:30.490","Text":"We\u0027re between this radius and this radius."},{"Start":"04:31.700 ","End":"04:36.049","Text":"First of all, we know from the previous lesson where we learned about"},{"Start":"04:36.049 ","End":"04:40.850","Text":"symmetry that if we\u0027re dealing with some shape,"},{"Start":"04:40.850 ","End":"04:45.485","Text":"that if we spin the shape around or move it around,"},{"Start":"04:45.485 ","End":"04:47.810","Text":"we still have the exact same problem."},{"Start":"04:47.810 ","End":"04:51.709","Text":"We even spoke about spherical shape,"},{"Start":"04:51.709 ","End":"04:54.244","Text":"spherical shells, and solid spheres."},{"Start":"04:54.244 ","End":"04:59.300","Text":"If we spin this around in the Theta or Phi direction,"},{"Start":"04:59.300 ","End":"05:02.375","Text":"we\u0027re going to get the exact same problem in front of us."},{"Start":"05:02.375 ","End":"05:05.149","Text":"Because as far as we\u0027re concerned in this question,"},{"Start":"05:05.149 ","End":"05:07.099","Text":"because we haven\u0027t been told otherwise."},{"Start":"05:07.099 ","End":"05:11.699","Text":"This charge Q is evenly distributed throughout the sphere,"},{"Start":"05:12.220 ","End":"05:15.590","Text":"or rather because it\u0027s a conducting sphere,"},{"Start":"05:15.590 ","End":"05:20.170","Text":"so this charge q is evenly distributed along the surface of the sphere."},{"Start":"05:20.170 ","End":"05:23.784","Text":"It\u0027s like looking at a problem of a spherical shell"},{"Start":"05:23.784 ","End":"05:28.345","Text":"and we saw that spherical shells if we have even charged distribution."},{"Start":"05:28.345 ","End":"05:33.630","Text":"That means that we have this symmetry."},{"Start":"05:33.630 ","End":"05:38.330","Text":"Therefore, we know that the electric field at this point over here is"},{"Start":"05:38.330 ","End":"05:43.070","Text":"going to be the same as the electric field over here and here and here and here and here,"},{"Start":"05:43.070 ","End":"05:48.075","Text":"all the way along this circle that we\u0027re making here."},{"Start":"05:48.075 ","End":"05:52.969","Text":"What we can see is that this is like a Gaussian surface,"},{"Start":"05:52.969 ","End":"05:58.200","Text":"a spherical shell, Gaussian surface or Gaussian envelope."},{"Start":"05:58.610 ","End":"06:02.540","Text":"The electric field is going to be constant along"},{"Start":"06:02.540 ","End":"06:07.745","Text":"this red dotted line or along this Gaussian surface and so therefore,"},{"Start":"06:07.745 ","End":"06:16.735","Text":"we don\u0027t have to integrate along E d S and we can just say E.S,"},{"Start":"06:16.735 ","End":"06:19.850","Text":"where S is, of course, the surface area."},{"Start":"06:19.850 ","End":"06:25.220","Text":"We don\u0027t have to integrate because we know that this is a constant electrical fields."},{"Start":"06:25.220 ","End":"06:30.419","Text":"We\u0027re just going to multiply the electrical field by the surface area of this envelope"},{"Start":"06:30.419 ","End":"06:36.500","Text":"and we know from Gauss\u0027 law that this is equal to Q in divided by Epsilon naught."},{"Start":"06:36.500 ","End":"06:39.979","Text":"E is what we\u0027re trying to find, so E.S."},{"Start":"06:39.979 ","End":"06:47.332","Text":"So the surface area of a spherical shell is 4 Pi r^2 and this is equal to the Q inside,"},{"Start":"06:47.332 ","End":"06:48.410","Text":"so the charge inside,"},{"Start":"06:48.410 ","End":"06:49.699","Text":"which is just this over here,"},{"Start":"06:49.699 ","End":"06:53.060","Text":"Q divided by Epsilon naught."},{"Start":"06:53.060 ","End":"06:58.770","Text":"Then we\u0027re just going to isolate out our E, our electric field."},{"Start":"06:58.770 ","End":"07:06.350","Text":"We\u0027ll get that this is equal to Q divided by 4 Pi Epsilon naught r^2 and of course,"},{"Start":"07:06.350 ","End":"07:10.700","Text":"1 divided by 4 Pi Epsilon naught is just k. This is equal to"},{"Start":"07:10.700 ","End":"07:17.390","Text":"kQ divided by r^2 and then if we do this in a vector,"},{"Start":"07:17.390 ","End":"07:22.460","Text":"we know that it\u0027s in the radial direction and so this is the electric field that we get,"},{"Start":"07:22.460 ","End":"07:26.134","Text":"which is the exact same electric field of a point charge,"},{"Start":"07:26.134 ","End":"07:30.400","Text":"which is what we already expected to get."},{"Start":"07:31.760 ","End":"07:35.610","Text":"Now, let\u0027s look at the next area."},{"Start":"07:35.610 ","End":"07:38.675","Text":"Now, we\u0027re going to be dealing with this area over here."},{"Start":"07:38.675 ","End":"07:44.885","Text":"Let\u0027s say I want to find the electric field at some point between the radius 2R and 3R."},{"Start":"07:44.885 ","End":"07:50.240","Text":"Somewhere between this grounded conducting"},{"Start":"07:50.240 ","End":"07:57.290","Text":"thin spherical shell and the inner edge of my conducting thick spherical shell."},{"Start":"07:57.290 ","End":"08:01.440","Text":"Let\u0027s say I want to find the electric field at this point over here."},{"Start":"08:02.630 ","End":"08:04.964","Text":"This is the radius,"},{"Start":"08:04.964 ","End":"08:06.935","Text":"and now, just like before,"},{"Start":"08:06.935 ","End":"08:13.340","Text":"I\u0027m going to put on this Gaussian envelope or this Gaussian surface,"},{"Start":"08:13.340 ","End":"08:16.865","Text":"which is just the spherical shell going like so,"},{"Start":"08:16.865 ","End":"08:21.184","Text":"of radius r, and just like before,"},{"Start":"08:21.184 ","End":"08:25.130","Text":"I\u0027m assuming symmetry because I haven\u0027t been told otherwise."},{"Start":"08:25.130 ","End":"08:29.735","Text":"That means that the electric field all the way around"},{"Start":"08:29.735 ","End":"08:35.100","Text":"this spherical surface is going to be the same."},{"Start":"08:35.470 ","End":"08:41.675","Text":"Therefore, I can work it out just like I did for this section over here."},{"Start":"08:41.675 ","End":"08:46.380","Text":"Now, we\u0027re just dealing with this area over here."},{"Start":"08:48.500 ","End":"08:53.499","Text":"First of all, we\u0027re going to get an electric fields similar to that of a point charge."},{"Start":"08:53.499 ","End":"08:56.260","Text":"The only thing we have to take into account is this."},{"Start":"08:56.260 ","End":"08:59.649","Text":"One, we have our conducting sphere,"},{"Start":"08:59.649 ","End":"09:02.875","Text":"where we have this charge of capital Q."},{"Start":"09:02.875 ","End":"09:04.285","Text":"But also over here,"},{"Start":"09:04.285 ","End":"09:06.999","Text":"on this conducting thin spherical shell,"},{"Start":"09:06.999 ","End":"09:09.660","Text":"we also have some charge."},{"Start":"09:09.660 ","End":"09:13.839","Text":"The fact that it\u0027s grounded doesn\u0027t mean that we don\u0027t have a charge."},{"Start":"09:13.839 ","End":"09:19.815","Text":"The fact that it\u0027s grounded means that the sphere has a potential of 0."},{"Start":"09:19.815 ","End":"09:21.759","Text":"If something is grounded,"},{"Start":"09:21.759 ","End":"09:23.665","Text":"it has a potential of 0,"},{"Start":"09:23.665 ","End":"09:27.400","Text":"but it generally doesn\u0027t have a charge of 0."},{"Start":"09:27.400 ","End":"09:32.689","Text":"There will be some charge and when we say that something is grounded,"},{"Start":"09:32.689 ","End":"09:35.539","Text":"it means that if we do have a charge over here,"},{"Start":"09:35.539 ","End":"09:38.869","Text":"so more charges are traveling through"},{"Start":"09:38.869 ","End":"09:43.730","Text":"here in order to cancel out the overall charge over here."},{"Start":"09:43.730 ","End":"09:49.955","Text":"We definitely will have some charge on this grounded spherical shell."},{"Start":"09:49.955 ","End":"09:53.699","Text":"It has a charge different to 0,"},{"Start":"09:53.699 ","End":"09:55.680","Text":"but it has a 0 potential."},{"Start":"09:55.680 ","End":"09:57.900","Text":"I don\u0027t know what the charge is."},{"Start":"09:57.900 ","End":"10:01.075","Text":"In the meantime, I\u0027m just going to write that it\u0027s equal to q,"},{"Start":"10:01.075 ","End":"10:02.830","Text":"q is an unknown,"},{"Start":"10:02.830 ","End":"10:06.260","Text":"but I\u0027m going to work with it in the meantime and then later on,"},{"Start":"10:06.260 ","End":"10:08.480","Text":"we\u0027re going to work out what Q\u0027s."},{"Start":"10:08.480 ","End":"10:12.979","Text":"All we\u0027re going to have is that we have this electric field which is equal"},{"Start":"10:12.979 ","End":"10:17.525","Text":"to k and then we\u0027re just going to superimpose these 2 charges."},{"Start":"10:17.525 ","End":"10:20.494","Text":"We\u0027re just going to add them up, Q plus q,"},{"Start":"10:20.494 ","End":"10:27.645","Text":"where this one is of course unknown divided by r^2 in the r direction."},{"Start":"10:27.645 ","End":"10:32.680","Text":"It\u0027s still the electric field similar to a point,"},{"Start":"10:32.680 ","End":"10:34.279","Text":"of a point charge,"},{"Start":"10:34.279 ","End":"10:37.055","Text":"just like we saw in the previous section, how we get it."},{"Start":"10:37.055 ","End":"10:42.425","Text":"But we\u0027re adding in the charge of this extra spherical shell over here."},{"Start":"10:42.425 ","End":"10:47.073","Text":"Of course, we don\u0027t know it will work it out later."},{"Start":"10:47.073 ","End":"10:51.369","Text":"Now, let\u0027s go into the next section where we\u0027re located"},{"Start":"10:51.369 ","End":"10:56.230","Text":"in here between the radius of 3R and 4R."},{"Start":"10:56.230 ","End":"11:00.910","Text":"We\u0027re located inside the conducting thick spherical shell."},{"Start":"11:00.910 ","End":"11:07.580","Text":"Now, we\u0027re located between 3R and 4R."},{"Start":"11:07.710 ","End":"11:12.309","Text":"Again, we\u0027re located in a conducting spherical shell,"},{"Start":"11:12.309 ","End":"11:18.052","Text":"which means that straight away we can say that the electric field in there is equal to 0."},{"Start":"11:18.052 ","End":"11:21.565","Text":"Now, our final region."},{"Start":"11:21.565 ","End":"11:27.010","Text":"Let\u0027s take a look over here where we\u0027re just outside this whole system over here."},{"Start":"11:27.010 ","End":"11:33.350","Text":"We\u0027re located at a radius which is greater than 4R."},{"Start":"11:34.020 ","End":"11:41.260","Text":"We\u0027re going to have this electric field that goes all the way around."},{"Start":"11:41.260 ","End":"11:48.690","Text":"We\u0027re using a Gaussian surface over here as well."},{"Start":"11:48.690 ","End":"11:53.890","Text":"We have no reason to suspect that the charge distribution isn\u0027t even over here."},{"Start":"11:53.890 ","End":"11:56.770","Text":"We can say that the electric field at"},{"Start":"11:56.770 ","End":"12:00.745","Text":"every single point on this Gaussian surface is exactly the same."},{"Start":"12:00.745 ","End":"12:07.225","Text":"Therefore, we can again consider this as a point charge."},{"Start":"12:07.225 ","End":"12:12.160","Text":"Then we\u0027re finding the electric field outside of a point charge."},{"Start":"12:12.160 ","End":"12:15.219","Text":"That we already know what it is, but again,"},{"Start":"12:15.219 ","End":"12:18.700","Text":"we need to take into account all of the charges inside."},{"Start":"12:18.700 ","End":"12:22.030","Text":"First of all, we have this Q for the inner sphere."},{"Start":"12:22.030 ","End":"12:27.580","Text":"Then we have this unknown q over here that we\u0027re going to calculate later."},{"Start":"12:27.580 ","End":"12:31.374","Text":"Then we also have the charge of this outer spherical shell,"},{"Start":"12:31.374 ","End":"12:33.940","Text":"which has a charge of negative 2Q."},{"Start":"12:33.940 ","End":"12:37.735","Text":"Our electrical field is going to be equal to"},{"Start":"12:37.735 ","End":"12:40.780","Text":"k. Then we\u0027re superimposing all of the charges,"},{"Start":"12:40.780 ","End":"12:46.615","Text":"so we have Q negative 2Q plus this lowercase q,"},{"Start":"12:46.615 ","End":"12:48.639","Text":"which is unknown in the meantime,"},{"Start":"12:48.639 ","End":"12:53.780","Text":"divided by r^2 in the r direction."},{"Start":"12:53.820 ","End":"13:00.010","Text":"Now, we\u0027ve found the electric field in all of the regions."},{"Start":"13:00.010 ","End":"13:05.300","Text":"Now, all we have to do is we have to find the potential."},{"Start":"13:05.850 ","End":"13:10.360","Text":"Now, we\u0027re going to integrate along everything."},{"Start":"13:10.360 ","End":"13:15.280","Text":"Let\u0027s scroll a little bit to the side to give us some space."},{"Start":"13:15.280 ","End":"13:17.739","Text":"There are two ways of integrating."},{"Start":"13:17.739 ","End":"13:23.662","Text":"We can integrate with bounds or without bounds,"},{"Start":"13:23.662 ","End":"13:24.869","Text":"or in other words,"},{"Start":"13:24.869 ","End":"13:28.035","Text":"have a definite or an indefinite integral."},{"Start":"13:28.035 ","End":"13:33.190","Text":"Right now, I\u0027m going to use the idea of an indefinite integral."},{"Start":"13:33.330 ","End":"13:35.934","Text":"That\u0027s how we\u0027re going to do this."},{"Start":"13:35.934 ","End":"13:37.660","Text":"Let\u0027s do that now."},{"Start":"13:37.660 ","End":"13:40.824","Text":"First of all, we\u0027re going to start in the region"},{"Start":"13:40.824 ","End":"13:45.400","Text":"where r is smaller than capital R, the radius."},{"Start":"13:45.400 ","End":"13:51.610","Text":"We\u0027re inside the conducting sphere where we saw that our electric field is equal to 0."},{"Start":"13:51.610 ","End":"13:52.900","Text":"Just to remind you,"},{"Start":"13:52.900 ","End":"13:58.850","Text":"the potential is equal to the negative integral on E.dr."},{"Start":"13:59.580 ","End":"14:02.320","Text":"So because E is equal to 0,"},{"Start":"14:02.320 ","End":"14:03.639","Text":"so we\u0027re integrating on 0,"},{"Start":"14:03.639 ","End":"14:05.169","Text":"which is equal to 0."},{"Start":"14:05.169 ","End":"14:09.370","Text":"But remember whenever we have an indefinite integral,"},{"Start":"14:09.370 ","End":"14:12.370","Text":"we always have to add some constant."},{"Start":"14:12.370 ","End":"14:15.775","Text":"Over here, we\u0027re going to add a constant,"},{"Start":"14:15.775 ","End":"14:19.040","Text":"C_1, the first constant."},{"Start":"14:19.590 ","End":"14:26.005","Text":"Now the next region is between R and 2R."},{"Start":"14:26.005 ","End":"14:31.570","Text":"Then our potential over here between R and"},{"Start":"14:31.570 ","End":"14:37.945","Text":"2R is going to be equal to the negative integral of E,"},{"Start":"14:37.945 ","End":"14:43.644","Text":"which is kQ divided by r^2 dr,"},{"Start":"14:43.644 ","End":"14:53.170","Text":"and then this is simply going to be equal to kQ divided by r. Of course,"},{"Start":"14:53.170 ","End":"14:55.420","Text":"because it\u0027s an indefinite integral,"},{"Start":"14:55.420 ","End":"14:59.570","Text":"we\u0027re going to add another constant, so C_2."},{"Start":"15:00.840 ","End":"15:04.975","Text":"Then in the region between 2R and 3R,"},{"Start":"15:04.975 ","End":"15:08.919","Text":"what we\u0027re going to get is the same as what we got before,"},{"Start":"15:08.919 ","End":"15:11.005","Text":"just taking into account this q,"},{"Start":"15:11.005 ","End":"15:13.300","Text":"which is unknown, the lowercase q."},{"Start":"15:13.300 ","End":"15:20.925","Text":"This is just going to be equal to k Q of plus unknown q divided by r,"},{"Start":"15:20.925 ","End":"15:26.030","Text":"plus of course, because it\u0027s an indefinite integral, some constant C_3."},{"Start":"15:26.030 ","End":"15:33.670","Text":"Then we\u0027re going into the region between 3R and 4R,"},{"Start":"15:33.670 ","End":"15:35.319","Text":"where are we got an electric field of,"},{"Start":"15:35.319 ","End":"15:38.050","Text":"0 and of course,"},{"Start":"15:38.050 ","End":"15:40.629","Text":"because we\u0027re doing an indefinite integral,"},{"Start":"15:40.629 ","End":"15:43.015","Text":"we have to add some constant."},{"Start":"15:43.015 ","End":"15:44.665","Text":"It\u0027s equal to C_4."},{"Start":"15:44.665 ","End":"15:49.465","Text":"Then in the region outside of the whole system,"},{"Start":"15:49.465 ","End":"15:52.390","Text":"so when r is bigger than 4R,"},{"Start":"15:52.390 ","End":"15:54.925","Text":"we\u0027re going to use this format."},{"Start":"15:54.925 ","End":"15:59.499","Text":"it\u0027s just equal to k and then Q minus 2Q"},{"Start":"15:59.499 ","End":"16:05.669","Text":"plus q divided by r and then of course,"},{"Start":"16:05.669 ","End":"16:07.379","Text":"because it\u0027s an indefinite integral,"},{"Start":"16:07.379 ","End":"16:11.740","Text":"we have to add another constant, C_5."},{"Start":"16:12.830 ","End":"16:19.290","Text":"The next thing that we have to do is we have to calibrate our answers."},{"Start":"16:19.290 ","End":"16:22.860","Text":"That means that we\u0027re going to be finding all of these constants,"},{"Start":"16:22.860 ","End":"16:26.110","Text":"C_1 through to C_5."},{"Start":"16:26.610 ","End":"16:31.045","Text":"We\u0027re going to be using the idea of continuity."},{"Start":"16:31.045 ","End":"16:37.330","Text":"The potential has to somewhere have a value equal to 0."},{"Start":"16:37.330 ","End":"16:42.819","Text":"Just like when we have the potential due to gravity, so mgh."},{"Start":"16:42.819 ","End":"16:46.134","Text":"We defined somewhere where h is considered 0"},{"Start":"16:46.134 ","End":"16:50.455","Text":"and that therefore there the potential energy is equal to 0."},{"Start":"16:50.455 ","End":"16:53.095","Text":"Similarly in electric potential,"},{"Start":"16:53.095 ","End":"16:57.265","Text":"we have to define an area where the potential is equal to 0."},{"Start":"16:57.265 ","End":"17:01.608","Text":"Usually we define that area at infinity."},{"Start":"17:01.608 ","End":"17:04.150","Text":"The potential at infinity,"},{"Start":"17:04.150 ","End":"17:06.789","Text":"when we\u0027re infinitely far from the system,"},{"Start":"17:06.789 ","End":"17:10.280","Text":"we define as being equal to 0."},{"Start":"17:10.710 ","End":"17:13.795","Text":"This will give us our calibration."},{"Start":"17:13.795 ","End":"17:19.870","Text":"Now what we\u0027re going to do is we\u0027re going to substitute in a radius of infinity."},{"Start":"17:19.870 ","End":"17:23.080","Text":"We can see that that\u0027s going to be in this region."},{"Start":"17:23.080 ","End":"17:28.555","Text":"Infinity is in the region which is bigger than 4R."},{"Start":"17:28.555 ","End":"17:31.789","Text":"If I plug this in over here,"},{"Start":"17:31.800 ","End":"17:39.370","Text":"I\u0027ll see that Phi of infinity in"},{"Start":"17:39.370 ","End":"17:46.120","Text":"this region is equal to k unknown q minus Q divided by r,"},{"Start":"17:46.120 ","End":"17:49.840","Text":"which we\u0027re saying is infinitely big, plus C_5."},{"Start":"17:49.840 ","End":"17:51.490","Text":"We know from this,"},{"Start":"17:51.490 ","End":"17:55.255","Text":"which we can take as a given is going to be equal to 0."},{"Start":"17:55.255 ","End":"17:58.000","Text":"First of all if the denominator is so big,"},{"Start":"17:58.000 ","End":"18:01.735","Text":"we can say that this whole fraction is approaching 0."},{"Start":"18:01.735 ","End":"18:03.609","Text":"Then we have this,"},{"Start":"18:03.609 ","End":"18:07.750","Text":"which we can say is 0 plus C_5 has to equal 0."},{"Start":"18:07.750 ","End":"18:13.480","Text":"Therefore we can say that C_5 equals 0."},{"Start":"18:13.480 ","End":"18:19.180","Text":"Therefore, we can cross this out and say it\u0027s equal to 0."},{"Start":"18:19.180 ","End":"18:24.830","Text":"We found one constant and now we have to find the others."},{"Start":"18:25.260 ","End":"18:30.820","Text":"We already use calibration and now we\u0027re going to use continuity."},{"Start":"18:30.820 ","End":"18:36.565","Text":"What\u0027s important to remember is that the potential is a continuous function."},{"Start":"18:36.565 ","End":"18:41.005","Text":"If I draw a graph of the potential,"},{"Start":"18:41.005 ","End":"18:42.789","Text":"knowing it\u0027s a continuous function."},{"Start":"18:42.789 ","End":"18:44.889","Text":"Here on the y-axis I have potential."},{"Start":"18:44.889 ","End":"18:46.810","Text":"Here on the x-axis I have r,"},{"Start":"18:46.810 ","End":"18:48.850","Text":"so that I have my different regions."},{"Start":"18:48.850 ","End":"18:50.830","Text":"R, 2R,"},{"Start":"18:50.830 ","End":"18:54.729","Text":"3R, 4R and whatever."},{"Start":"18:54.729 ","End":"18:56.920","Text":"I can see that up until R,"},{"Start":"18:56.920 ","End":"19:01.374","Text":"I have a potential which is constant, it\u0027s just C_1."},{"Start":"19:01.374 ","End":"19:04.225","Text":"Then \u0027til 2R,"},{"Start":"19:04.225 ","End":"19:07.840","Text":"I have this potential which is reducing as"},{"Start":"19:07.840 ","End":"19:12.064","Text":"a function of 1 divided by r plus some constant."},{"Start":"19:12.064 ","End":"19:15.055","Text":"Then again between 2R and 3R,"},{"Start":"19:15.055 ","End":"19:24.715","Text":"I also have this potential which is also being reduced as a function of 1 divided by r,"},{"Start":"19:24.715 ","End":"19:27.700","Text":"and so on and so forth for the other two regions."},{"Start":"19:27.700 ","End":"19:29.349","Text":"Then between 3R and 4R,"},{"Start":"19:29.349 ","End":"19:32.725","Text":"I have another constant function."},{"Start":"19:32.725 ","End":"19:43.405","Text":"What we can see is that I have areas of discontinuity over here as I\u0027ve drawn it."},{"Start":"19:43.405 ","End":"19:47.710","Text":"The potential can\u0027t just behave like this."},{"Start":"19:47.710 ","End":"19:53.845","Text":"These two lines, and these lines have to meet. They have to touch."},{"Start":"19:53.845 ","End":"19:57.580","Text":"That means that the function has to be continuous."},{"Start":"19:57.580 ","End":"20:06.915","Text":"I have to somehow join these up so that we have this continuous function."},{"Start":"20:06.915 ","End":"20:10.559","Text":"What we\u0027re going to do is we\u0027re going to see that at"},{"Start":"20:10.559 ","End":"20:16.539","Text":"the crossover point here between this region and this region,"},{"Start":"20:16.539 ","End":"20:19.795","Text":"we have a mutual point of capital R,"},{"Start":"20:19.795 ","End":"20:22.044","Text":"between this region and this region,"},{"Start":"20:22.044 ","End":"20:25.270","Text":"we have a mutual point of 2R,"},{"Start":"20:25.270 ","End":"20:27.880","Text":"then of 3R, and then of 4R."},{"Start":"20:27.880 ","End":"20:33.279","Text":"What we\u0027re going to show is that the potential over here when we substitute in"},{"Start":"20:33.279 ","End":"20:38.575","Text":"capital R. The potential over here when we substitute in this mutual point,"},{"Start":"20:38.575 ","End":"20:43.854","Text":"capital R has to be equal to the same number and this way,"},{"Start":"20:43.854 ","End":"20:45.100","Text":"or the same value."},{"Start":"20:45.100 ","End":"20:52.679","Text":"This way, we\u0027re going to get that our potential function is continuous."},{"Start":"20:52.679 ","End":"20:54.760","Text":"Let us do this."},{"Start":"20:54.760 ","End":"20:56.650","Text":"If my explanation was unclear,"},{"Start":"20:56.650 ","End":"20:58.525","Text":"it will be very clear now."},{"Start":"20:58.525 ","End":"21:06.895","Text":"What we\u0027re going to do is we\u0027re going to equate the potential in this region of here,"},{"Start":"21:06.895 ","End":"21:11.305","Text":"where we have 3R, r, 4R."},{"Start":"21:11.305 ","End":"21:14.125","Text":"In this region between 3R and 4R,"},{"Start":"21:14.125 ","End":"21:20.080","Text":"at the point where we have this over here,"},{"Start":"21:20.080 ","End":"21:21.850","Text":"4R and 4R,"},{"Start":"21:21.850 ","End":"21:24.809","Text":"at this point 4R."},{"Start":"21:24.809 ","End":"21:30.745","Text":"We\u0027re going to say that that has to be equal to the potential in the region where"},{"Start":"21:30.745 ","End":"21:38.074","Text":"r is greater than 4R at the same point 4R."},{"Start":"21:38.074 ","End":"21:42.350","Text":"At the mutual point over here."},{"Start":"21:42.720 ","End":"21:46.510","Text":"If we substitute in 4R over here,"},{"Start":"21:46.510 ","End":"21:49.090","Text":"we just see that it\u0027s equal to C_4."},{"Start":"21:49.090 ","End":"21:52.248","Text":"There\u0027s nowhere to plug in 4R,"},{"Start":"21:52.248 ","End":"21:53.770","Text":"and that is equal to, we are ready."},{"Start":"21:53.770 ","End":"21:55.495","Text":"C_5 is equal to 0,"},{"Start":"21:55.495 ","End":"21:57.249","Text":"that is equal to k,"},{"Start":"21:57.249 ","End":"22:01.770","Text":"q unknown minus Q divided by,"},{"Start":"22:01.770 ","End":"22:05.115","Text":"and then we substitute in here 4R."},{"Start":"22:05.115 ","End":"22:10.059","Text":"Then we can see that C_4 is equal to this."},{"Start":"22:10.350 ","End":"22:19.315","Text":"Now let\u0027s do the next area whether a potential in the region between 2R and 3R over here,"},{"Start":"22:19.315 ","End":"22:27.070","Text":"has to be equal to the potential in the region of 3R and 4R where of course,"},{"Start":"22:27.070 ","End":"22:30.235","Text":"the mutual point is 3R."},{"Start":"22:30.235 ","End":"22:34.021","Text":"We\u0027re calculating it this."},{"Start":"22:34.021 ","End":"22:37.765","Text":"In this region we\u0027re going to have k,"},{"Start":"22:37.765 ","End":"22:43.689","Text":"Q plus q unknown divided by r,"},{"Start":"22:43.689 ","End":"22:48.715","Text":"this point 3R plus this C_3,"},{"Start":"22:48.715 ","End":"22:55.255","Text":"and this is equal to the potential in the region between 3R and 4R."},{"Start":"22:55.255 ","End":"22:57.700","Text":"At this point 3R,"},{"Start":"22:57.700 ","End":"22:59.695","Text":"which we saw is equal to C_4,"},{"Start":"22:59.695 ","End":"23:00.879","Text":"which we just worked out."},{"Start":"23:00.879 ","End":"23:02.909","Text":"C_4 is equal to k,"},{"Start":"23:02.909 ","End":"23:07.790","Text":"q unknown minus Q divided by 4R."},{"Start":"23:09.690 ","End":"23:13.795","Text":"Then we just isolate out our C_3,"},{"Start":"23:13.795 ","End":"23:17.545","Text":"and then we get what our constant C_3 is equal to."},{"Start":"23:17.545 ","End":"23:21.115","Text":"Now, let\u0027s do the next region."},{"Start":"23:21.115 ","End":"23:26.634","Text":"We\u0027re taking the potential in this region"},{"Start":"23:26.634 ","End":"23:32.350","Text":"between R and 2R,"},{"Start":"23:32.350 ","End":"23:35.919","Text":"we can see the mutual point is this 2R over here,"},{"Start":"23:35.919 ","End":"23:43.000","Text":"at the point 2R and this has to be equal to the potential in the next region,"},{"Start":"23:43.000 ","End":"23:50.540","Text":"that\u0027s between 2R and 3R and at the same mutual point, which is 2R."},{"Start":"23:50.820 ","End":"23:57.258","Text":"Then that\u0027s equal to kQ divided by 2R,"},{"Start":"23:57.258 ","End":"24:00.985","Text":"and of course, plus C_2,"},{"Start":"24:00.985 ","End":"24:05.579","Text":"and this has to be equal to k,"},{"Start":"24:05.579 ","End":"24:14.655","Text":"Q plus q unknown divided by 2R plus C_3,"},{"Start":"24:14.655 ","End":"24:16.629","Text":"where C_3 we worked out over here."},{"Start":"24:16.629 ","End":"24:19.525","Text":"It\u0027s a little bit long, I\u0027m just going to write C_3."},{"Start":"24:19.525 ","End":"24:24.190","Text":"But it\u0027s this K, q minus Q over 4R minus k,"},{"Start":"24:24.190 ","End":"24:28.090","Text":"Q plus q over 3R."},{"Start":"24:28.090 ","End":"24:33.579","Text":"Then the final region that we\u0027re going to do is we\u0027re going to find"},{"Start":"24:33.579 ","End":"24:39.955","Text":"the potential in the region where R is smaller than r at the points."},{"Start":"24:39.955 ","End":"24:49.194","Text":"Here we can see that the mutual region or the mutual point is R. At this point R,"},{"Start":"24:49.194 ","End":"24:52.944","Text":"and this of course has to be equal to the potential in the region"},{"Start":"24:52.944 ","End":"24:58.780","Text":"between r and 2R at the same mutual point."},{"Start":"24:58.780 ","End":"25:04.150","Text":"Then we can see that C_1 is going to be equal to"},{"Start":"25:04.150 ","End":"25:09.999","Text":"kQ divided by the mutual point R plus C_2,"},{"Start":"25:09.999 ","End":"25:19.565","Text":"where C_2 we found by just isolating out over here and plugging in what we got for C_3."},{"Start":"25:19.565 ","End":"25:22.149","Text":"Now we\u0027ve found all of"},{"Start":"25:22.149 ","End":"25:26.020","Text":"our constants and all that\u0027s left is to just plug in the different values,"},{"Start":"25:26.020 ","End":"25:27.955","Text":"which I\u0027m not going to do because it\u0027s tedious."},{"Start":"25:27.955 ","End":"25:31.765","Text":"But there\u0027s one thing that is still left to do,"},{"Start":"25:31.765 ","End":"25:37.840","Text":"and that is that we still don\u0027t know what this q is equal to."},{"Start":"25:37.840 ","End":"25:41.905","Text":"Let\u0027s go back 1 second to our diagram."},{"Start":"25:41.905 ","End":"25:45.279","Text":"Here, we got this q, which we don\u0027t know,"},{"Start":"25:45.279 ","End":"25:51.060","Text":"and we got this due to the fact that this spherical shell is grounded."},{"Start":"25:51.060 ","End":"25:55.934","Text":"What did I say that I hopped on that when something is grounded,"},{"Start":"25:55.934 ","End":"25:58.919","Text":"that means that the potential at that exact point,"},{"Start":"25:58.919 ","End":"26:03.755","Text":"the potential at 2R is equal to 0."},{"Start":"26:03.755 ","End":"26:07.490","Text":"The potential in this spherical shell is equal to 0."},{"Start":"26:07.500 ","End":"26:13.343","Text":"Then what we can do is we can go back over here,"},{"Start":"26:13.343 ","End":"26:20.635","Text":"and then we can see that 2R appears in this region over here,"},{"Start":"26:20.635 ","End":"26:22.548","Text":"and also in this region over here,"},{"Start":"26:22.548 ","End":"26:26.950","Text":"and we just saw at this point 2R these 2 equations are equal."},{"Start":"26:26.950 ","End":"26:30.355","Text":"We can either use this equation or this equation."},{"Start":"26:30.355 ","End":"26:32.215","Text":"Let\u0027s use this equation."},{"Start":"26:32.215 ","End":"26:39.399","Text":"We know that the potential at 2R has to be equal to 0 by definition,"},{"Start":"26:39.399 ","End":"26:41.649","Text":"because it\u0027s a grounded spherical shell,"},{"Start":"26:41.649 ","End":"26:45.040","Text":"and this is equal to k,"},{"Start":"26:45.040 ","End":"26:54.685","Text":"Q plus q unknown divided by 2R plus C_3,"},{"Start":"26:54.685 ","End":"26:57.070","Text":"where we\u0027ve already found with C_3 is."},{"Start":"26:57.070 ","End":"26:59.470","Text":"It\u0027s this over here."},{"Start":"26:59.470 ","End":"27:01.615","Text":"Then we just plug this in."},{"Start":"27:01.615 ","End":"27:06.985","Text":"We work out the algebra to isolate out this q, and that\u0027s it."},{"Start":"27:06.985 ","End":"27:12.385","Text":"We\u0027ve found the q and then we\u0027ve solved the entire question."},{"Start":"27:12.385 ","End":"27:20.335","Text":"Now, we\u0027ve worked out what the potential is at every single point."},{"Start":"27:20.335 ","End":"27:23.050","Text":"Let\u0027s move on to the second part of the question,"},{"Start":"27:23.050 ","End":"27:26.230","Text":"which is, what is the charge distribution?"},{"Start":"27:26.230 ","End":"27:29.470","Text":"We\u0027re trying to find the charge distribution"},{"Start":"27:29.470 ","End":"27:34.400","Text":"for every single point in our system over here."},{"Start":"27:34.440 ","End":"27:37.420","Text":"Let\u0027s start with a solid sphere."},{"Start":"27:37.420 ","End":"27:41.185","Text":"We know that it\u0027s a conducting sphere, it\u0027s a conductor,"},{"Start":"27:41.185 ","End":"27:47.980","Text":"which means that all of the charge is going to be distributed only on the surface,"},{"Start":"27:47.980 ","End":"27:53.425","Text":"because that\u0027s where the charge is when dealing with a conductor."},{"Start":"27:53.425 ","End":"27:59.215","Text":"We see that the charge is only on the surface and we have a total charge of Q."},{"Start":"27:59.215 ","End":"28:08.380","Text":"That means that the charge distribution Sigma is simply the charge divided by the area."},{"Start":"28:08.380 ","End":"28:12.429","Text":"We have the charge which is Q divided by the total area,"},{"Start":"28:12.429 ","End":"28:14.979","Text":"which is just the surface area of a sphere."},{"Start":"28:14.979 ","End":"28:22.160","Text":"That is 4Pi and then of radius R^2."},{"Start":"28:23.130 ","End":"28:31.600","Text":"Then next we have here the charge for the grounded conducting thin sphere."},{"Start":"28:31.600 ","End":"28:33.730","Text":"Here again it\u0027s a thin sphere,"},{"Start":"28:33.730 ","End":"28:35.575","Text":"it\u0027s a surface area."},{"Start":"28:35.575 ","End":"28:37.600","Text":"Again, we\u0027re going to have,"},{"Start":"28:37.600 ","End":"28:38.755","Text":"let\u0027s have this one,"},{"Start":"28:38.755 ","End":"28:40.600","Text":"here we have 2."},{"Start":"28:40.600 ","End":"28:42.414","Text":"This is equal to,"},{"Start":"28:42.414 ","End":"28:43.509","Text":"again, the same thing,"},{"Start":"28:43.509 ","End":"28:45.849","Text":"the charge on the thin sphere,"},{"Start":"28:45.849 ","End":"28:48.655","Text":"which is Q divided by the surface area,"},{"Start":"28:48.655 ","End":"28:55.940","Text":"that\u0027s 4Pi and then we have the radius over here is 2R, 2R^2."},{"Start":"28:56.940 ","End":"29:04.400","Text":"Of course, q we already know because we found it when we were looking for the potential."},{"Start":"29:05.700 ","End":"29:09.879","Text":"Now, we get to this pink over here,"},{"Start":"29:09.879 ","End":"29:13.795","Text":"which is a conducting thick spherical shell."},{"Start":"29:13.795 ","End":"29:17.058","Text":"This is a spherical shell that has some width,"},{"Start":"29:17.058 ","End":"29:19.315","Text":"and of course, it\u0027s a conductor as well."},{"Start":"29:19.315 ","End":"29:23.304","Text":"First of all, we know that there\u0027s going to be no charge in the middle,"},{"Start":"29:23.304 ","End":"29:26.754","Text":"but there is going to be some charge distribution"},{"Start":"29:26.754 ","End":"29:32.260","Text":"along the inner edge and along the outer edge of the spherical shell."},{"Start":"29:32.260 ","End":"29:36.940","Text":"Let\u0027s say that along the inner edge we have another unknown charge."},{"Start":"29:36.940 ","End":"29:38.484","Text":"Let\u0027s draw it in red."},{"Start":"29:38.484 ","End":"29:43.389","Text":"We have q_3 along the inner edge,"},{"Start":"29:43.389 ","End":"29:47.454","Text":"and we have some charge q_4 along the outer edge."},{"Start":"29:47.454 ","End":"29:53.620","Text":"We know that the total charge that we have on this thick spherical shell is negative 2Q,"},{"Start":"29:53.620 ","End":"29:57.670","Text":"but we don\u0027t know how it is distributed."},{"Start":"29:57.670 ","End":"29:59.845","Text":"All we know is because it\u0027s a conductor,"},{"Start":"29:59.845 ","End":"30:06.290","Text":"the charge is somehow distributed along the inner and outer edges."},{"Start":"30:08.310 ","End":"30:13.359","Text":"Let\u0027s see. What we can know is that the total charge,"},{"Start":"30:13.359 ","End":"30:19.390","Text":"that is q_3 plus q_4 is equal to what we were given here,"},{"Start":"30:19.390 ","End":"30:22.075","Text":"which is equal to negative 2Q."},{"Start":"30:22.075 ","End":"30:24.400","Text":"Now, in some questions we\u0027ll get that"},{"Start":"30:24.400 ","End":"30:28.989","Text":"this conducting thick spherical shell has a total charge of 0."},{"Start":"30:28.989 ","End":"30:33.159","Text":"In that case, q_3 plus q_4 will equal 0,"},{"Start":"30:33.159 ","End":"30:37.869","Text":"and then we\u0027ll just get that q_3 is equal to negative q_4."},{"Start":"30:37.869 ","End":"30:39.460","Text":"But in our case over here,"},{"Start":"30:39.460 ","End":"30:42.950","Text":"we get this equation."},{"Start":"30:42.950 ","End":"30:46.975","Text":"First of all, we can see that we have 1 equation,"},{"Start":"30:46.975 ","End":"30:48.384","Text":"but we have 2 unknowns."},{"Start":"30:48.384 ","End":"30:50.800","Text":"We have q_3 and q_4."},{"Start":"30:50.800 ","End":"30:56.965","Text":"What I\u0027m going to do is I\u0027m going to use this gray pen,"},{"Start":"30:56.965 ","End":"30:59.380","Text":"and what I\u0027m going to do is a trick."},{"Start":"30:59.380 ","End":"31:03.370","Text":"I\u0027m going to draw a Gaussian surface right in"},{"Start":"31:03.370 ","End":"31:08.450","Text":"the middle of this conducting thick spherical shell, like so."},{"Start":"31:12.690 ","End":"31:17.920","Text":"Here we have our Gaussian surface."},{"Start":"31:17.920 ","End":"31:20.545","Text":"First of all, let\u0027s see,"},{"Start":"31:20.545 ","End":"31:24.295","Text":"because we know that this is a conductor,"},{"Start":"31:24.295 ","End":"31:30.205","Text":"so we know that the electric field within the Gaussian surface is equal to 0."},{"Start":"31:30.205 ","End":"31:31.600","Text":"That we already know."},{"Start":"31:31.600 ","End":"31:35.990","Text":"We saw that also when we were finding the potential inside here."},{"Start":"31:37.500 ","End":"31:41.605","Text":"Therefore, that means that the flux,"},{"Start":"31:41.605 ","End":"31:46.225","Text":"which is equal to E.S,"},{"Start":"31:46.225 ","End":"31:50.350","Text":"is also going to be equal to 0 because the electric field is equal to 0,"},{"Start":"31:50.350 ","End":"31:53.350","Text":"so 0 times the area is 0."},{"Start":"31:53.350 ","End":"32:00.630","Text":"Now, we also know that the flux is equal to Qin divided by Epsilon naught."},{"Start":"32:00.630 ","End":"32:02.595","Text":"Flux is equal to E.S,"},{"Start":"32:02.595 ","End":"32:06.640","Text":"which is equal to Qin divided by Epsilon naught."},{"Start":"32:06.640 ","End":"32:10.375","Text":"Qin, what do we know?"},{"Start":"32:10.375 ","End":"32:13.285","Text":"From the previous section when we were working out the potential,"},{"Start":"32:13.285 ","End":"32:17.200","Text":"we saw that the total charge encompassed by"},{"Start":"32:17.200 ","End":"32:23.380","Text":"this Gaussian surface is the charge from the conducting sphere, which is Q."},{"Start":"32:23.380 ","End":"32:27.699","Text":"It\u0027s the charge from this grounded spherical shell,"},{"Start":"32:27.699 ","End":"32:29.529","Text":"which is the small q,"},{"Start":"32:29.529 ","End":"32:32.409","Text":"which we also worked out what that is equal to,"},{"Start":"32:32.409 ","End":"32:39.535","Text":"and another charge that is encompassed within this Gaussian surface is this charge q_3,"},{"Start":"32:39.535 ","End":"32:42.649","Text":"which we don\u0027t know what it is yet."},{"Start":"32:42.660 ","End":"32:47.079","Text":"But it\u0027s also inside this Gaussian surface, and of course,"},{"Start":"32:47.079 ","End":"32:49.239","Text":"this is divided by Epsilon naught,"},{"Start":"32:49.239 ","End":"32:51.715","Text":"and we know that this is equal to 0."},{"Start":"32:51.715 ","End":"32:54.670","Text":"Well, we\u0027ve already written that it\u0027s equal to 0."},{"Start":"32:54.670 ","End":"32:57.055","Text":"Therefore, let\u0027s write it up here."},{"Start":"32:57.055 ","End":"33:07.690","Text":"Our next equation is that Q plus q plus q_3 is equal to 0."},{"Start":"33:07.690 ","End":"33:11.949","Text":"Now, we have 2 equations and 2 unknowns,"},{"Start":"33:11.949 ","End":"33:17.120","Text":"and so therefore we can solve what q_3 is and therefore what q_4 is."},{"Start":"33:18.440 ","End":"33:23.310","Text":"Then all we would do is we would substitute in what q is here,"},{"Start":"33:23.310 ","End":"33:25.230","Text":"and then we\u0027ll isolate out q_3,"},{"Start":"33:25.230 ","End":"33:27.825","Text":"and then we know what q_3 is."},{"Start":"33:27.825 ","End":"33:32.370","Text":"Then we\u0027ll have the charge distribution over here."},{"Start":"33:32.370 ","End":"33:36.720","Text":"We\u0027ll have q_3 divided by the surface area of the inner shell,"},{"Start":"33:36.720 ","End":"33:43.900","Text":"which is just going to be divided by 4 Pi 3R^2,"},{"Start":"33:43.900 ","End":"33:50.530","Text":"and then we\u0027ll plug that into this equation over here for q_3,"},{"Start":"33:50.530 ","End":"33:52.479","Text":"and then we\u0027ll find what q_4 is,"},{"Start":"33:52.479 ","End":"33:54.789","Text":"and then the charge distribution along"},{"Start":"33:54.789 ","End":"33:57.940","Text":"this edge will be q_4 divided by this surface area,"},{"Start":"33:57.940 ","End":"34:03.679","Text":"which will just be 4Pi multiplied by 4R^2."},{"Start":"34:04.020 ","End":"34:10.220","Text":"Then we\u0027ll have found the charge distribution for every surface here."},{"Start":"34:11.760 ","End":"34:14.905","Text":"Now, let\u0027s answer question 2."},{"Start":"34:14.905 ","End":"34:23.110","Text":"How much work is needed to bring a charge of 5 coulombs from infinity to 1.5R?"},{"Start":"34:23.110 ","End":"34:29.590","Text":"Let\u0027s call this charge of 5 coulombs, let\u0027s call it q_5,"},{"Start":"34:29.590 ","End":"34:31.734","Text":"because it has a charge of 5 Coulombs,"},{"Start":"34:31.734 ","End":"34:33.520","Text":"and we\u0027re bringing it from infinity,"},{"Start":"34:33.520 ","End":"34:36.700","Text":"so from very far away, until 1.5R,"},{"Start":"34:36.700 ","End":"34:41.125","Text":"so until around about this point over here."},{"Start":"34:41.125 ","End":"34:45.640","Text":"Let\u0027s say that our charge q_5 is located over here."},{"Start":"34:45.640 ","End":"34:49.390","Text":"Let\u0027s say that this is an infinite distance away from here and this"},{"Start":"34:49.390 ","End":"34:53.635","Text":"is at a location of 1.5R."},{"Start":"34:53.635 ","End":"34:57.399","Text":"We\u0027re asking for how much work is needed in order to do that,"},{"Start":"34:57.399 ","End":"35:03.459","Text":"and we know that an equation is equal to the negative change in energy."},{"Start":"35:03.459 ","End":"35:05.930","Text":"This is work."},{"Start":"35:07.110 ","End":"35:10.674","Text":"This is the potential energy of course,"},{"Start":"35:10.674 ","End":"35:15.625","Text":"and this equation is speaking about that the electrical work"},{"Start":"35:15.625 ","End":"35:22.060","Text":"done is equal to the negative change in the potential energy."},{"Start":"35:22.060 ","End":"35:26.334","Text":"But I know that I,"},{"Start":"35:26.334 ","End":"35:29.799","Text":"in order to bring this charge to this point,"},{"Start":"35:29.799 ","End":"35:33.384","Text":"I have to put in work,"},{"Start":"35:33.384 ","End":"35:40.420","Text":"I have to put it in mechanical work or some force in order to move this charge over here."},{"Start":"35:40.420 ","End":"35:45.800","Text":"That\u0027s going to be equal and opposite to this electrical work."},{"Start":"35:46.080 ","End":"35:48.565","Text":"We can write it like that."},{"Start":"35:48.565 ","End":"35:52.510","Text":"Therefore, I\u0027ll get that the work that I need to put in,"},{"Start":"35:52.510 ","End":"35:56.950","Text":"the work needed to bring this charge over to here is simply going"},{"Start":"35:56.950 ","End":"36:02.210","Text":"to be equal to the change in the potential energy."},{"Start":"36:02.670 ","End":"36:06.235","Text":"So what is our potential energy?"},{"Start":"36:06.235 ","End":"36:09.340","Text":"Potential energy uses potential."},{"Start":"36:09.340 ","End":"36:13.330","Text":"That\u0027s the whole point of finding the potential of something."},{"Start":"36:13.330 ","End":"36:16.030","Text":"It\u0027s so that we can find the potential energy of it."},{"Start":"36:16.030 ","End":"36:19.240","Text":"The potential energy is simply equal to the charge,"},{"Start":"36:19.240 ","End":"36:20.740","Text":"which here\u0027s q_5,"},{"Start":"36:20.740 ","End":"36:24.085","Text":"multiplied by the potential."},{"Start":"36:24.085 ","End":"36:26.529","Text":"In that case we can say,"},{"Start":"36:26.529 ","End":"36:31.914","Text":"that our work to bring this particle q_5 from infinity up until"},{"Start":"36:31.914 ","End":"36:37.675","Text":"1.5R is equal to the change in the potential energy."},{"Start":"36:37.675 ","End":"36:41.230","Text":"We\u0027ll take the potential energy at infinity."},{"Start":"36:41.230 ","End":"36:46.180","Text":"That\u0027s going to be q_5 multiplied by the potential at infinity,"},{"Start":"36:46.180 ","End":"36:48.820","Text":"which by definition we know is equal to 0,"},{"Start":"36:48.820 ","End":"36:50.889","Text":"so all of this is equal to 0,"},{"Start":"36:50.889 ","End":"36:55.675","Text":"minus the potential energy at 1.5R."},{"Start":"36:55.675 ","End":"37:02.230","Text":"That is q_5 multiplied by the potential at 1.5R."},{"Start":"37:02.230 ","End":"37:04.825","Text":"In the previous section,"},{"Start":"37:04.825 ","End":"37:07.750","Text":"in the first half of the video,"},{"Start":"37:07.750 ","End":"37:11.365","Text":"we found the potential at this point."},{"Start":"37:11.365 ","End":"37:17.489","Text":"It was the potential in the region between R and 2R,"},{"Start":"37:17.489 ","End":"37:19.570","Text":"so that\u0027s right over here,"},{"Start":"37:19.570 ","End":"37:25.585","Text":"and we found that this was equal to KQ divided by r plus C_2,"},{"Start":"37:25.585 ","End":"37:30.500","Text":"where before we found out what C_2 was equal to."},{"Start":"37:30.960 ","End":"37:34.375","Text":"Then we\u0027ll just substitute this in here."},{"Start":"37:34.375 ","End":"37:36.339","Text":"We\u0027re multiplying by k,"},{"Start":"37:36.339 ","End":"37:38.455","Text":"Q divided by the radius."},{"Start":"37:38.455 ","End":"37:46.270","Text":"Right now, we\u0027re located at 1.5R plus C_2,"},{"Start":"37:46.270 ","End":"37:47.470","Text":"and there we go."},{"Start":"37:47.470 ","End":"37:49.940","Text":"This is the work."},{"Start":"37:50.430 ","End":"37:53.395","Text":"Now, this is the answer."},{"Start":"37:53.395 ","End":"37:59.920","Text":"Another way to solve it is to say that a particle q_5 has some kind of velocity."},{"Start":"37:59.920 ","End":"38:04.780","Text":"We\u0027ve defined some initial velocity V_0 over here,"},{"Start":"38:04.780 ","End":"38:09.490","Text":"and then we have some final velocity V_f over here."},{"Start":"38:09.490 ","End":"38:15.050","Text":"Then we can just solve this question by using conservation of energy."},{"Start":"38:15.570 ","End":"38:18.369","Text":"If we\u0027re using conservation of energy,"},{"Start":"38:18.369 ","End":"38:22.015","Text":"we know that the initial energy is equal to the final energy."},{"Start":"38:22.015 ","End":"38:27.640","Text":"The initial energy comes from the energy from velocity or the kinetic energy."},{"Start":"38:27.640 ","End":"38:31.780","Text":"That\u0027s going to be half mv0^2,"},{"Start":"38:31.780 ","End":"38:33.670","Text":"where of course we have to know the mass."},{"Start":"38:33.670 ","End":"38:36.939","Text":"Therefore, of this particle plus its potential energy,"},{"Start":"38:36.939 ","End":"38:40.524","Text":"which we just saw initially at infinity,"},{"Start":"38:40.524 ","End":"38:44.575","Text":"is q_5 multiplied by the potential at infinity,"},{"Start":"38:44.575 ","End":"38:47.064","Text":"which we know that this is equal to 0,"},{"Start":"38:47.064 ","End":"38:51.535","Text":"and this has to be equal to the final energy right over here,"},{"Start":"38:51.535 ","End":"38:54.009","Text":"which also has kinetic energy,"},{"Start":"38:54.009 ","End":"39:02.034","Text":"so half mvf^2 plus the potential energy over here,"},{"Start":"39:02.034 ","End":"39:10.375","Text":"which is q_5 multiplied by the potential at 1.5R."},{"Start":"39:10.375 ","End":"39:14.155","Text":"This question is a classic question that comes up in exams."},{"Start":"39:14.155 ","End":"39:20.300","Text":"Version also with the first question, finding the potential."},{"Start":"39:20.310 ","End":"39:22.690","Text":"It takes a long time,"},{"Start":"39:22.690 ","End":"39:24.534","Text":"but it\u0027s worth learning how to do this."},{"Start":"39:24.534 ","End":"39:29.935","Text":"Now, we\u0027ve already spoken on how to solve question 1 and 2,"},{"Start":"39:29.935 ","End":"39:34.404","Text":"given that we\u0027re using indefinite integrals,"},{"Start":"39:34.404 ","End":"39:36.400","Text":"so an integral without bounds."},{"Start":"39:36.400 ","End":"39:38.349","Text":"Now, in the next half of the video,"},{"Start":"39:38.349 ","End":"39:42.910","Text":"I\u0027m going to show how to solve this question using a definite integral."},{"Start":"39:42.910 ","End":"39:46.645","Text":"That means using bounds in my integration."},{"Start":"39:46.645 ","End":"39:50.170","Text":"Here we have our system and I just pulled up"},{"Start":"39:50.170 ","End":"39:55.179","Text":"the electric fields that we found at the beginning of the lesson using Gauss\u0027 law,"},{"Start":"39:55.179 ","End":"39:58.960","Text":"and now we\u0027re going to use that to find the potential."},{"Start":"39:58.960 ","End":"40:05.575","Text":"We know that the potential is the negative integral of E.dr."},{"Start":"40:05.575 ","End":"40:09.309","Text":"Now, what we\u0027re going to do is we\u0027re going to find the potential in"},{"Start":"40:09.309 ","End":"40:15.085","Text":"the region where we\u0027re located bigger than 4R."},{"Start":"40:15.085 ","End":"40:20.755","Text":"We\u0027re located outside of our whole system over here."},{"Start":"40:20.755 ","End":"40:25.780","Text":"We can see that this is the electric field over here."},{"Start":"40:25.780 ","End":"40:29.515","Text":"What we\u0027re going to do is we\u0027re going to do an integral on that."},{"Start":"40:29.515 ","End":"40:36.595","Text":"We have that our E field is equal to k, Q minus 2_q."},{"Start":"40:36.595 ","End":"40:47.480","Text":"Let\u0027s just do this as negative Q plus this q over here divided by r squared,"},{"Start":"40:47.730 ","End":"40:52.554","Text":"and now, we\u0027re going to do this integral,"},{"Start":"40:52.554 ","End":"40:56.665","Text":"and we\u0027re going to put in the bounds in a second."},{"Start":"40:56.665 ","End":"41:00.834","Text":"We\u0027re going to have over here our E.dr,"},{"Start":"41:00.834 ","End":"41:03.909","Text":"and what do we usually do is we integrate"},{"Start":"41:03.909 ","End":"41:07.585","Text":"along bounds where we know what the potential is going to be."},{"Start":"41:07.585 ","End":"41:09.850","Text":"For instance, we know,"},{"Start":"41:09.850 ","End":"41:11.080","Text":"and it\u0027s generally defined,"},{"Start":"41:11.080 ","End":"41:15.670","Text":"that the potential at infinity is equal to 0."},{"Start":"41:15.670 ","End":"41:20.109","Text":"Therefore, we\u0027re going to integrate from infinity up"},{"Start":"41:20.109 ","End":"41:25.179","Text":"until we reach the area where we\u0027re trying to find the potential for."},{"Start":"41:25.179 ","End":"41:30.625","Text":"Over here, it will be up until r where our r is located in this region."},{"Start":"41:30.625 ","End":"41:36.560","Text":"We\u0027re still located outside of our spherical shells."},{"Start":"41:37.770 ","End":"41:40.855","Text":"As we know, what does this mean?"},{"Start":"41:40.855 ","End":"41:44.905","Text":"When we\u0027re integrating from infinity up until r,"},{"Start":"41:44.905 ","End":"41:47.919","Text":"so that we can write this as taking,"},{"Start":"41:47.919 ","End":"41:49.179","Text":"and because there\u0027s a minus over here,"},{"Start":"41:49.179 ","End":"41:56.320","Text":"so we\u0027re taking the potential at r and subtracting the potential at infinity."},{"Start":"41:56.320 ","End":"42:00.685","Text":"Because we know that the potential at infinity is equal to 0,"},{"Start":"42:00.685 ","End":"42:04.330","Text":"so then we\u0027ll have the potential at r minus 0,"},{"Start":"42:04.330 ","End":"42:07.719","Text":"and so then we\u0027re just left with the potential at r,"},{"Start":"42:07.719 ","End":"42:09.144","Text":"which is what we\u0027re trying to find,"},{"Start":"42:09.144 ","End":"42:15.220","Text":"which is equal to the negative integral from infinity up until r of E,"},{"Start":"42:15.220 ","End":"42:25.059","Text":"which is k multiplied by negative Q plus q divided by r squared,"},{"Start":"42:25.059 ","End":"42:28.720","Text":"and then the E field is in the radial direction,"},{"Start":"42:28.720 ","End":"42:31.990","Text":"and this dr vector is also in the radial direction."},{"Start":"42:31.990 ","End":"42:34.434","Text":"When we multiply those together,"},{"Start":"42:34.434 ","End":"42:44.575","Text":"we\u0027ll just be left with a scalar and we\u0027ll have this dr. Let\u0027s do the integral."},{"Start":"42:44.575 ","End":"42:50.920","Text":"What we\u0027ll get is we\u0027ll get k(q-Q) and"},{"Start":"42:50.920 ","End":"42:57.040","Text":"then we\u0027ll have multiplied by 1 divided by r. Then we\u0027ll substitute in our bounds,"},{"Start":"42:57.040 ","End":"42:59.470","Text":"which is from infinity,"},{"Start":"42:59.470 ","End":"43:07.015","Text":"and up until r. When we put infinity over here in the denominator we\u0027ll obviously get 0."},{"Start":"43:07.015 ","End":"43:15.534","Text":"What we\u0027ll in fact get is that the potential at r is simply equal to k(q-Q)"},{"Start":"43:15.534 ","End":"43:25.585","Text":"divided by r. This is the potential that we have in the region outside over here."},{"Start":"43:25.585 ","End":"43:30.530","Text":"This region where R is bigger than 4R."},{"Start":"43:31.290 ","End":"43:35.350","Text":"Now let\u0027s go to the next smaller region."},{"Start":"43:35.350 ","End":"43:38.019","Text":"This is where r is located,"},{"Start":"43:38.019 ","End":"43:41.950","Text":"between 3R and 4R."},{"Start":"43:41.950 ","End":"43:46.660","Text":"We\u0027re located within this conducting thick spherical shell."},{"Start":"43:46.660 ","End":"43:51.460","Text":"We can see that our electric field over here is equal to 0."},{"Start":"43:51.460 ","End":"43:56.110","Text":"Then when we do our integral like over here,"},{"Start":"43:56.110 ","End":"44:00.490","Text":"so we have to put in our integral bounds."},{"Start":"44:00.490 ","End":"44:06.879","Text":"We can do like before where we integrated from infinity until our region r. However,"},{"Start":"44:06.879 ","End":"44:12.669","Text":"we know that the potential up until 4R is equal to this over here,"},{"Start":"44:12.669 ","End":"44:14.095","Text":"we just worked it out."},{"Start":"44:14.095 ","End":"44:20.030","Text":"What we can do is we can integrate from 4R like so"},{"Start":"44:21.870 ","End":"44:26.620","Text":"until we get over to somewhere within"},{"Start":"44:26.620 ","End":"44:31.794","Text":"this region at some radius r within this region over here,"},{"Start":"44:31.794 ","End":"44:35.709","Text":"because we know what the potential is it 4R and"},{"Start":"44:35.709 ","End":"44:41.420","Text":"then we\u0027re going to do the integration of E.dr."},{"Start":"44:41.850 ","End":"44:45.205","Text":"Now again, when we integrate,"},{"Start":"44:45.205 ","End":"44:46.795","Text":"we\u0027re just summing the bounds."},{"Start":"44:46.795 ","End":"44:51.715","Text":"We\u0027re taking over here the potential at r"},{"Start":"44:51.715 ","End":"44:57.430","Text":"minus the potential at 4R."},{"Start":"44:57.430 ","End":"44:59.530","Text":"That\u0027s what this integral means."},{"Start":"44:59.530 ","End":"45:03.760","Text":"Because we know the electric field in this region is equal to 0,"},{"Start":"45:03.760 ","End":"45:07.210","Text":"so then this whole equation over here is equal to 0."},{"Start":"45:07.210 ","End":"45:12.984","Text":"Then we\u0027ll get that the potential at r minus the potential at"},{"Start":"45:12.984 ","End":"45:20.170","Text":"4R is equal to 0 and therefore will get that the potential at r,"},{"Start":"45:20.170 ","End":"45:21.744","Text":"when we just isolate this out,"},{"Start":"45:21.744 ","End":"45:27.019","Text":"is simply equal to the potential at 4R,"},{"Start":"45:28.380 ","End":"45:31.510","Text":"and of course, that\u0027s what we just got here."},{"Start":"45:31.510 ","End":"45:36.535","Text":"That\u0027s k negative q"},{"Start":"45:36.535 ","End":"45:42.685","Text":"plus q over here divided by r. Of course,"},{"Start":"45:42.685 ","End":"45:48.115","Text":"our r over here is 4R."},{"Start":"45:48.115 ","End":"45:52.190","Text":"This is our potential in this region over here."},{"Start":"45:52.980 ","End":"45:56.215","Text":"Now let\u0027s do the next region,"},{"Start":"45:56.215 ","End":"46:01.900","Text":"which is where r is located between 2R and 3R."},{"Start":"46:01.900 ","End":"46:06.910","Text":"Then again, we\u0027ll get that the potential at r minus,"},{"Start":"46:06.910 ","End":"46:13.435","Text":"so the next potential down that we\u0027ve already found is in this region over here,"},{"Start":"46:13.435 ","End":"46:16.970","Text":"so minus the potential at 3R,"},{"Start":"46:18.810 ","End":"46:26.635","Text":"and this is equal to the negative integral on the electric field in this region."},{"Start":"46:26.635 ","End":"46:32.485","Text":"That is at k(Q+q) divided by r"},{"Start":"46:32.485 ","End":"46:38.380","Text":"squared and then we can just do dr and because we\u0027ve done the dot multiplication,"},{"Start":"46:38.380 ","End":"46:42.850","Text":"the vector signs cancel out and our bounds,"},{"Start":"46:42.850 ","End":"46:47.860","Text":"so they\u0027re just going to be from 3R until"},{"Start":"46:47.860 ","End":"46:54.835","Text":"r. Then this is going to be equal to so,"},{"Start":"46:54.835 ","End":"46:57.625","Text":"first of all, let\u0027s just write this like so."},{"Start":"46:57.625 ","End":"46:59.410","Text":"We have Phi at r,"},{"Start":"46:59.410 ","End":"47:02.350","Text":"the potential at r minus the potential at 3R,"},{"Start":"47:02.350 ","End":"47:07.480","Text":"which is this k(q-Q) divided"},{"Start":"47:07.480 ","End":"47:14.289","Text":"by 4R and this is equal to the negative integral of this."},{"Start":"47:14.289 ","End":"47:16.705","Text":"That\u0027s just going to be"},{"Start":"47:16.705 ","End":"47:25.135","Text":"k(Q+q) multiplied by 1 divided by r,"},{"Start":"47:25.135 ","End":"47:33.879","Text":"with the bounds of 3R until r. Then we\u0027ll get,"},{"Start":"47:33.879 ","End":"47:35.275","Text":"let\u0027s just cut to the chase."},{"Start":"47:35.275 ","End":"47:38.889","Text":"We\u0027ll get that the potential over here at r is going to be"},{"Start":"47:38.889 ","End":"47:45.820","Text":"equal to k(Q+q) multiplied by 1"},{"Start":"47:45.820 ","End":"47:50.980","Text":"divided by r minus 1 divided by 3R and then"},{"Start":"47:50.980 ","End":"47:55.800","Text":"plus k(q-Q)"},{"Start":"47:55.800 ","End":"48:01.230","Text":"divided by 4R."},{"Start":"48:01.230 ","End":"48:04.690","Text":"This will be the potential in this region."},{"Start":"48:05.550 ","End":"48:08.859","Text":"At this stage, I\u0027m guessing that you already get"},{"Start":"48:08.859 ","End":"48:11.709","Text":"the idea of how we\u0027re doing the integrals."},{"Start":"48:11.709 ","End":"48:13.270","Text":"I\u0027m going to stop here."},{"Start":"48:13.270 ","End":"48:16.884","Text":"But the only thing that we still need to calculate,"},{"Start":"48:16.884 ","End":"48:19.119","Text":"because we\u0027re pretending like we didn\u0027t solve"},{"Start":"48:19.119 ","End":"48:22.240","Text":"this question with a different method earlier on,"},{"Start":"48:22.240 ","End":"48:25.629","Text":"is that we still have to find what this q is over here."},{"Start":"48:25.629 ","End":"48:27.009","Text":"Because this is an unknown."},{"Start":"48:27.009 ","End":"48:32.664","Text":"This is something that we just defined in order to solve this,"},{"Start":"48:32.664 ","End":"48:35.169","Text":"to get to this stage, but we still have to figure"},{"Start":"48:35.169 ","End":"48:39.170","Text":"out what this lower case q is actually equal to."},{"Start":"48:39.510 ","End":"48:41.815","Text":"How are we going to find this?"},{"Start":"48:41.815 ","End":"48:43.869","Text":"Let\u0027s go back to the diagram."},{"Start":"48:43.869 ","End":"48:46.825","Text":"We can see that the q,"},{"Start":"48:46.825 ","End":"48:48.490","Text":"we defined it to be over here,"},{"Start":"48:48.490 ","End":"48:53.470","Text":"which is the charge of this grounded conducting thin spherical shell."},{"Start":"48:53.470 ","End":"48:56.559","Text":"We know that the definition of"},{"Start":"48:56.559 ","End":"49:00.590","Text":"a grounded thin spherical shell"},{"Start":"49:01.020 ","End":"49:07.239","Text":"is that the potential at this point is equal to 0."},{"Start":"49:07.239 ","End":"49:10.285","Text":"That\u0027s what it means if something is grounded."},{"Start":"49:10.285 ","End":"49:13.510","Text":"That means that we know that at 2R,"},{"Start":"49:13.510 ","End":"49:17.589","Text":"which is where the grounded spherical shell is,"},{"Start":"49:17.589 ","End":"49:19.120","Text":"at a radius of 2R,"},{"Start":"49:19.120 ","End":"49:21.745","Text":"our potential is equal to 0."},{"Start":"49:21.745 ","End":"49:23.365","Text":"Let\u0027s come back here."},{"Start":"49:23.365 ","End":"49:30.565","Text":"Then I can just say that my potential at 2R is equal to 0,"},{"Start":"49:30.565 ","End":"49:33.745","Text":"but also I know that my potential at 2R,"},{"Start":"49:33.745 ","End":"49:37.885","Text":"because of course the function for the potential is continuous."},{"Start":"49:37.885 ","End":"49:45.354","Text":"I can say that my potential over here if I substitute in 2R is going to be equal to 0."},{"Start":"49:45.354 ","End":"49:50.199","Text":"I can say that this is equal to k(Q+q),"},{"Start":"49:50.199 ","End":"49:51.655","Text":"which is what I\u0027m trying to find,"},{"Start":"49:51.655 ","End":"49:57.969","Text":"multiplied by 1 divided by 2R minus 1 divided by"},{"Start":"49:57.969 ","End":"50:07.060","Text":"3R plus k(q-Q) divided by 4R."},{"Start":"50:07.060 ","End":"50:13.509","Text":"Then all I need to do is I have to isolate out my q and there I have it."},{"Start":"50:13.509 ","End":"50:15.710","Text":"I\u0027ll know what this is."},{"Start":"50:16.920 ","End":"50:20.619","Text":"Once we\u0027ve isolated out our lower case q,"},{"Start":"50:20.619 ","End":"50:23.560","Text":"we can substitute it into"},{"Start":"50:23.560 ","End":"50:28.600","Text":"the equations that we\u0027ve calculated for the potential in the different regions."},{"Start":"50:28.600 ","End":"50:31.929","Text":"Of course, you still have to complete the other 2 regions that"},{"Start":"50:31.929 ","End":"50:35.365","Text":"we still didn\u0027t calculate the potential for,"},{"Start":"50:35.365 ","End":"50:38.680","Text":"but I\u0027m just going to save time and once you\u0027ve done that,"},{"Start":"50:38.680 ","End":"50:42.400","Text":"then you\u0027ve finished this question."},{"Start":"50:42.400 ","End":"50:46.400","Text":"That is the end of this lesson."}],"ID":14207},{"Watched":false,"Name":"Exercise 1","Duration":"12m 53s","ChapterTopicVideoID":12141,"CourseChapterTopicPlaylistID":157337,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12141.jpeg","UploadDate":"2018-06-28T04:37:40.9530000","DurationForVideoObject":"PT12M53S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.295","Text":"Hello. In this lesson,"},{"Start":"00:02.295 ","End":"00:04.695","Text":"we\u0027re going to be answering the following question,"},{"Start":"00:04.695 ","End":"00:09.210","Text":"which is to find the potential of the spherical shell of radius R,"},{"Start":"00:09.210 ","End":"00:11.475","Text":"which has a charge Q."},{"Start":"00:11.475 ","End":"00:15.420","Text":"We\u0027re being told that the charge is uniformly distributed throughout."},{"Start":"00:15.420 ","End":"00:17.580","Text":"If you remember a few lessons ago,"},{"Start":"00:17.580 ","End":"00:21.735","Text":"we learned of 3 possible ways to calculate the potential,"},{"Start":"00:21.735 ","End":"00:24.915","Text":"and the second way was using Gauss\u0027s law,"},{"Start":"00:24.915 ","End":"00:29.700","Text":"where we said that this was useful way of calculating"},{"Start":"00:29.700 ","End":"00:31.650","Text":"the potential when we have"},{"Start":"00:31.650 ","End":"00:36.090","Text":"the infinite title or an infinite cylinder or an infinite plane,"},{"Start":"00:36.090 ","End":"00:38.960","Text":"or some kind of spherical shape,"},{"Start":"00:38.960 ","End":"00:42.065","Text":"whether it\u0027s a solid sphere or a spherical shell."},{"Start":"00:42.065 ","End":"00:43.940","Text":"Here we\u0027re dealing with a spherical shell,"},{"Start":"00:43.940 ","End":"00:48.235","Text":"so we know that we\u0027re going to be using this method."},{"Start":"00:48.235 ","End":"00:52.760","Text":"First of all, we remember that the equation for potential is"},{"Start":"00:52.760 ","End":"00:57.080","Text":"equal to the negative integral on our electric field,"},{"Start":"00:57.080 ","End":"01:01.850","Text":"E. dr. Then what we can"},{"Start":"01:01.850 ","End":"01:06.920","Text":"do is we can use Gauss\u0027s law in order to find the electric field,"},{"Start":"01:06.920 ","End":"01:08.195","Text":"and then we integrate."},{"Start":"01:08.195 ","End":"01:14.560","Text":"We can integrate with an indefinite integral or with a definite integral."},{"Start":"01:14.560 ","End":"01:18.670","Text":"Then we\u0027re going to find the potential."},{"Start":"01:19.250 ","End":"01:22.665","Text":"Let\u0027s first find the electric field."},{"Start":"01:22.665 ","End":"01:29.690","Text":"We\u0027re going to put a Gaussian surface around our shape over here,"},{"Start":"01:29.690 ","End":"01:37.680","Text":"so we have some kind of sphere and its radius r. Now,"},{"Start":"01:37.680 ","End":"01:39.345","Text":"we\u0027re going to integrate."},{"Start":"01:39.345 ","End":"01:44.330","Text":"We have this closed integral of E. ds,"},{"Start":"01:44.330 ","End":"01:50.830","Text":"so we\u0027re integrating along the surface area of a spherical shell of radius r,"},{"Start":"01:50.830 ","End":"01:57.305","Text":"and we know that this is equal to the charge enclosed divided by Epsilon_0."},{"Start":"01:57.305 ","End":"02:06.970","Text":"We know that our electric field is like so in the radial direction."},{"Start":"02:06.980 ","End":"02:12.935","Text":"Because we\u0027re being told in the question that the charge is uniformly distributed,"},{"Start":"02:12.935 ","End":"02:18.185","Text":"then we know that instead of having to integrate along the area,"},{"Start":"02:18.185 ","End":"02:23.660","Text":"we can just say that this is equal to simply the electric field in"},{"Start":"02:23.660 ","End":"02:30.455","Text":"the radial direction multiplied by the total surface area of the spherical shell."},{"Start":"02:30.455 ","End":"02:34.805","Text":"Then we\u0027ll get that our electric field multiplied by"},{"Start":"02:34.805 ","End":"02:40.483","Text":"the surface area of a Gaussian shell or a Gaussian surface,"},{"Start":"02:40.483 ","End":"02:42.470","Text":"so that is of a spherical shells,"},{"Start":"02:42.470 ","End":"02:49.470","Text":"so that\u0027s 4pi r^2 and this is equal to this over here,"},{"Start":"02:49.470 ","End":"02:51.725","Text":"so the enclosed charge, which is,"},{"Start":"02:51.725 ","End":"02:55.610","Text":"of course, Q divided by Epsilon_0."},{"Start":"02:55.610 ","End":"03:01.670","Text":"Therefore, we can isolate out our E and we\u0027ll get that our electric field is equal to"},{"Start":"03:01.670 ","End":"03:07.935","Text":"Q divided by 4pi Epsilon_0 r^2."},{"Start":"03:07.935 ","End":"03:14.570","Text":"Of course, we know that 1 over 4pi Epsilon_0 is equal to k. Therefore,"},{"Start":"03:14.570 ","End":"03:18.920","Text":"we can say that our electric field is equal to kQ"},{"Start":"03:18.920 ","End":"03:25.350","Text":"divided by r^2 and we know that it\u0027s in the radial direction."},{"Start":"03:25.580 ","End":"03:32.330","Text":"This is, of course, in the region where our r is larger than R,"},{"Start":"03:32.330 ","End":"03:36.050","Text":"so where we\u0027re located outside of the spherical shell."},{"Start":"03:36.050 ","End":"03:40.675","Text":"If, however, we\u0027re located within the spherical shell,"},{"Start":"03:40.675 ","End":"03:43.145","Text":"so we\u0027ll have something like this,"},{"Start":"03:43.145 ","End":"03:45.830","Text":"where of course,"},{"Start":"03:45.830 ","End":"03:50.030","Text":"we have this type of radio E field over"},{"Start":"03:50.030 ","End":"03:55.339","Text":"here and because our charge is uniformly distributed,"},{"Start":"03:55.339 ","End":"03:58.175","Text":"we can still use this E.S,"},{"Start":"03:58.175 ","End":"04:00.680","Text":"which is still equal to E,"},{"Start":"04:00.680 ","End":"04:07.925","Text":"and the surface area is still equal to 4pi r^2 squared, just this time,"},{"Start":"04:07.925 ","End":"04:11.220","Text":"r is smaller, r,"},{"Start":"04:11.220 ","End":"04:15.185","Text":"and this is equal to the charge enclosed, which we can see is 0,"},{"Start":"04:15.185 ","End":"04:19.595","Text":"the charge is only on this spherical shell."},{"Start":"04:19.595 ","End":"04:24.175","Text":"That is equal to 0 divided by Epsilon_0, so that\u0027s equal to 0,"},{"Start":"04:24.175 ","End":"04:29.660","Text":"and therefore, we get that our electric field in this region is equal to 0."},{"Start":"04:30.560 ","End":"04:39.670","Text":"What we can do is we can write that E field is simply equal to 0 in the region where"},{"Start":"04:39.670 ","End":"04:48.850","Text":"r is smaller than R and it\u0027s equal to kQ divided by r^2 in the radial direction,"},{"Start":"04:48.850 ","End":"04:54.640","Text":"when r is larger than R. Now,"},{"Start":"04:54.640 ","End":"04:57.445","Text":"we\u0027re going to deal with the next stage where we\u0027re going to integrate"},{"Start":"04:57.445 ","End":"05:01.030","Text":"along this electric field in order to find the potential."},{"Start":"05:01.030 ","End":"05:02.425","Text":"As we said before,"},{"Start":"05:02.425 ","End":"05:07.930","Text":"we can either do an indefinite integral where we don\u0027t set any bounds,"},{"Start":"05:07.930 ","End":"05:11.430","Text":"or we can do the definite integral where we do set bounds."},{"Start":"05:11.430 ","End":"05:16.610","Text":"Now, we\u0027re going to go for the indefinite integral because I prefer this way,"},{"Start":"05:16.610 ","End":"05:18.155","Text":"I think it\u0027s a bit easier."},{"Start":"05:18.155 ","End":"05:22.730","Text":"What we have to remember to do is when we take the indefinite integral,"},{"Start":"05:22.730 ","End":"05:27.950","Text":"we always have to add on constants for integration."},{"Start":"05:27.950 ","End":"05:32.450","Text":"Then afterwards, we\u0027re going to use the idea that our function"},{"Start":"05:32.450 ","End":"05:37.600","Text":"for a potential has to be a constant function."},{"Start":"05:38.180 ","End":"05:40.355","Text":"Then by using that,"},{"Start":"05:40.355 ","End":"05:44.540","Text":"we\u0027re going to find our integrating constants."},{"Start":"05:44.540 ","End":"05:47.580","Text":"Let\u0027s go ahead and do this."},{"Start":"05:48.620 ","End":"05:53.385","Text":"Let\u0027s calculate our potential function."},{"Start":"05:53.385 ","End":"05:57.705","Text":"We\u0027re going to have similar setup like here."},{"Start":"05:57.705 ","End":"06:00.870","Text":"Now, let\u0027s integrate along this region."},{"Start":"06:00.870 ","End":"06:07.265","Text":"We have the negative integral on 0. dr,"},{"Start":"06:07.265 ","End":"06:10.340","Text":"where we have an integral of 0,"},{"Start":"06:10.340 ","End":"06:12.470","Text":"it would obviously equals to 0,"},{"Start":"06:12.470 ","End":"06:16.520","Text":"so we have 0 plus are integrating constant C_1."},{"Start":"06:16.520 ","End":"06:24.615","Text":"And this is in the region where r is smaller than R. Then let\u0027s integrate over here."},{"Start":"06:24.615 ","End":"06:28.280","Text":"We\u0027re going to have the negative integral on E,"},{"Start":"06:28.280 ","End":"06:33.960","Text":"which is KQ divided by r^2 dr."},{"Start":"06:35.570 ","End":"06:42.360","Text":"The dot product between r hat and the dr vector just becomes 1."},{"Start":"06:42.360 ","End":"06:44.880","Text":"This is going to be equal to,"},{"Start":"06:44.880 ","End":"06:51.240","Text":"so the integral of 1 divided by r^2 is negative 1 divided by r,"},{"Start":"06:51.240 ","End":"06:54.340","Text":"and then we have another negative over here, so it cancels out,"},{"Start":"06:54.340 ","End":"06:58.310","Text":"so then we\u0027ll get kQ divided by r. Again,"},{"Start":"06:58.310 ","End":"06:59.900","Text":"because it\u0027s an indefinite integral,"},{"Start":"06:59.900 ","End":"07:03.320","Text":"we add on integrating constant C_2."},{"Start":"07:03.320 ","End":"07:12.240","Text":"This is in the region where r is larger than R. This is our potential function,"},{"Start":"07:12.240 ","End":"07:16.325","Text":"but we can see that we don\u0027t know what C_1 and C_2 is equal to."},{"Start":"07:16.325 ","End":"07:19.115","Text":"What do we have to do is we have to use idea"},{"Start":"07:19.115 ","End":"07:22.400","Text":"that the potential is a continuous function,"},{"Start":"07:22.400 ","End":"07:25.940","Text":"and we have to use the idea of calibration."},{"Start":"07:25.940 ","End":"07:30.110","Text":"What we mean by calibration is we take a point where we"},{"Start":"07:30.110 ","End":"07:34.815","Text":"know what the value of our potential function is going to be equal to."},{"Start":"07:34.815 ","End":"07:41.310","Text":"Generally, what we do is we say that the potential at infinity,"},{"Start":"07:41.660 ","End":"07:48.485","Text":"so that means when we\u0027re infinitely far away from our charge sphere is equal to 0."},{"Start":"07:48.485 ","End":"07:50.630","Text":"This is a definition."},{"Start":"07:50.630 ","End":"07:53.945","Text":"Really, really far away from this charge sphere,"},{"Start":"07:53.945 ","End":"07:58.265","Text":"we won\u0027t really feel the electric field or the potential."},{"Start":"07:58.265 ","End":"08:00.590","Text":"Therefore, if we know that,"},{"Start":"08:00.590 ","End":"08:03.635","Text":"then what we can do is we can substitute"},{"Start":"08:03.635 ","End":"08:08.990","Text":"into this equation over here in order to find C_2."},{"Start":"08:08.990 ","End":"08:11.945","Text":"Let\u0027s find the potential,"},{"Start":"08:11.945 ","End":"08:14.880","Text":"therefore, at infinity."},{"Start":"08:14.880 ","End":"08:18.315","Text":"That means that r is equal to infinity."},{"Start":"08:18.315 ","End":"08:21.746","Text":"That\u0027s in the region where r is greater than R,"},{"Start":"08:21.746 ","End":"08:24.660","Text":"we\u0027re located outside of the sphere,"},{"Start":"08:24.660 ","End":"08:28.495","Text":"so that\u0027s why we\u0027re using this equation and not this equation."},{"Start":"08:28.495 ","End":"08:32.555","Text":"That\u0027s equal to kQ divided by r,"},{"Start":"08:32.555 ","End":"08:35.885","Text":"which is infinity plus C_2,"},{"Start":"08:35.885 ","End":"08:39.060","Text":"and this is equal to 0."},{"Start":"08:39.100 ","End":"08:46.505","Text":"First of all, if we have a fraction where the denominator is a very, very large number,"},{"Start":"08:46.505 ","End":"08:49.339","Text":"especially relative to the numerator,"},{"Start":"08:49.339 ","End":"08:54.570","Text":"so that means that this whole fraction is going to be approaching 0."},{"Start":"08:55.760 ","End":"08:58.010","Text":"This is approaching 0,"},{"Start":"08:58.010 ","End":"09:01.705","Text":"which means that we have 0 plus C_2 is equal to 0."},{"Start":"09:01.705 ","End":"09:06.830","Text":"Therefore, we can say that C_2 is equal to 0."},{"Start":"09:08.000 ","End":"09:12.375","Text":"Now, we can rub this out over here."},{"Start":"09:12.375 ","End":"09:16.160","Text":"Now, what we\u0027ve just done is calibration;"},{"Start":"09:16.160 ","End":"09:20.465","Text":"we calibrated that we know the potential at infinity is equal to 0,"},{"Start":"09:20.465 ","End":"09:25.909","Text":"and then we used that calibration to substitute in this radius of infinity"},{"Start":"09:25.909 ","End":"09:32.380","Text":"into the equation that is relevant to where infinity is which is outside the sphere,"},{"Start":"09:32.380 ","End":"09:34.800","Text":"and then we got our constant."},{"Start":"09:34.800 ","End":"09:37.790","Text":"Now, we\u0027re going to be using the idea that"},{"Start":"09:37.790 ","End":"09:41.540","Text":"the potential function is continuous. What does that mean?"},{"Start":"09:41.540 ","End":"09:46.120","Text":"That means that at the mutual points in our region,"},{"Start":"09:46.120 ","End":"09:48.375","Text":"so that means at R,"},{"Start":"09:48.375 ","End":"09:51.080","Text":"so that means on the surface of the spherical shell,"},{"Start":"09:51.080 ","End":"09:57.680","Text":"we can see that R can fall into this region over here and into this region over here,"},{"Start":"09:57.680 ","End":"10:00.720","Text":"we have an overlap of this point."},{"Start":"10:01.370 ","End":"10:03.855","Text":"How can we use that?"},{"Start":"10:03.855 ","End":"10:10.235","Text":"That means that we know that the potential C_1 when r is equal to"},{"Start":"10:10.235 ","End":"10:14.870","Text":"R has to be equal to"},{"Start":"10:14.870 ","End":"10:22.405","Text":"the potential function where r over here is equal to R in this region."},{"Start":"10:22.405 ","End":"10:26.060","Text":"These 2 equations have to be the same"},{"Start":"10:26.060 ","End":"10:30.780","Text":"at the point R at the surface of the spherical shell."},{"Start":"10:31.310 ","End":"10:34.605","Text":"Let\u0027s substitute that point ends."},{"Start":"10:34.605 ","End":"10:39.095","Text":"We have C_1 and we have to substitute in R to this equation,"},{"Start":"10:39.095 ","End":"10:40.775","Text":"we see we don\u0027t have a place,"},{"Start":"10:40.775 ","End":"10:43.060","Text":"so we\u0027re just going to write out C_1."},{"Start":"10:43.060 ","End":"10:49.265","Text":"This has to be equal to this equation when we substitute in R. We\u0027re ready,"},{"Start":"10:49.265 ","End":"10:50.780","Text":"we can write this out."},{"Start":"10:50.780 ","End":"10:53.780","Text":"We have kQ divided by R,"},{"Start":"10:53.780 ","End":"10:57.215","Text":"where R this time is R. Now,"},{"Start":"10:57.215 ","End":"11:00.110","Text":"we just got our answer for C_1,"},{"Start":"11:00.110 ","End":"11:03.875","Text":"and we can just substitute this right in there."},{"Start":"11:03.875 ","End":"11:13.895","Text":"Let\u0027s rub that out. C_1 we just saw is equal to kQ divided by R. All right,"},{"Start":"11:13.895 ","End":"11:17.080","Text":"so this is the answer for our potential function."},{"Start":"11:17.080 ","End":"11:21.490","Text":"Let\u0027s just draw out the graph to see what we get."},{"Start":"11:21.490 ","End":"11:24.635","Text":"Let\u0027s say that this is R,"},{"Start":"11:24.635 ","End":"11:29.536","Text":"the radius of the spherical shell,"},{"Start":"11:29.536 ","End":"11:32.050","Text":"so what we can see is that up until R,"},{"Start":"11:32.050 ","End":"11:36.340","Text":"we have a constant potential which is located here at"},{"Start":"11:36.340 ","End":"11:42.495","Text":"kQ divided by R. This is our potential and this is our radius."},{"Start":"11:42.495 ","End":"11:45.970","Text":"We can see that up until this point,"},{"Start":"11:45.970 ","End":"11:49.404","Text":"our function is constant,"},{"Start":"11:49.404 ","End":"11:52.630","Text":"and then at this point R,"},{"Start":"11:52.630 ","End":"11:57.145","Text":"our function starts behaving like 1 divided by R,"},{"Start":"11:57.145 ","End":"11:59.120","Text":"or 1 divided by X."},{"Start":"11:59.120 ","End":"12:06.530","Text":"And we can see that it goes down as a function of R. We can see that"},{"Start":"12:06.530 ","End":"12:14.435","Text":"our function is continuous and we can see that at this point R over here,"},{"Start":"12:14.435 ","End":"12:20.120","Text":"we have the same value for the potential,"},{"Start":"12:20.120 ","End":"12:23.965","Text":"which is how we managed to work out the function."},{"Start":"12:23.965 ","End":"12:27.800","Text":"This was a very important lesson and knowing how to"},{"Start":"12:27.800 ","End":"12:31.400","Text":"calculate the potential of a spherical shell,"},{"Start":"12:31.400 ","End":"12:35.570","Text":"and also, even knowing this equation off by heart is super,"},{"Start":"12:35.570 ","End":"12:37.400","Text":"super, super important,"},{"Start":"12:37.400 ","End":"12:39.470","Text":"also for later work."},{"Start":"12:39.470 ","End":"12:41.960","Text":"Please go over all of the steps,"},{"Start":"12:41.960 ","End":"12:43.190","Text":"see how I did it,"},{"Start":"12:43.190 ","End":"12:44.930","Text":"make sure that you understand it,"},{"Start":"12:44.930 ","End":"12:46.445","Text":"memorize how to do it,"},{"Start":"12:46.445 ","End":"12:49.705","Text":"this is a very important lesson."},{"Start":"12:49.705 ","End":"12:54.060","Text":"This is the end of this important lesson."}],"ID":14208},{"Watched":false,"Name":"Exercise 2","Duration":"41m 14s","ChapterTopicVideoID":12142,"CourseChapterTopicPlaylistID":157337,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12142.jpeg","UploadDate":"2018-06-28T04:44:55.3730000","DurationForVideoObject":"PT41M14S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.845","Text":"Hello, in this lesson,"},{"Start":"00:01.845 ","End":"00:05.030","Text":"we\u0027re going to be answering the following question."},{"Start":"00:05.030 ","End":"00:10.470","Text":"Here we have a conductive cylinder in the center over here,"},{"Start":"00:10.470 ","End":"00:13.950","Text":"of radius R and length L,"},{"Start":"00:13.950 ","End":"00:16.905","Text":"and it has a charge of negative Q."},{"Start":"00:16.905 ","End":"00:22.815","Text":"What we can see is that we have a whole system of cylinders and cylindrical shells,"},{"Start":"00:22.815 ","End":"00:25.500","Text":"and we\u0027re looking at a slice of the whole system,"},{"Start":"00:25.500 ","End":"00:31.200","Text":"so we can\u0027t see the length L. Around this cylinder over here,"},{"Start":"00:31.200 ","End":"00:36.705","Text":"in dark gray, we have a thick conducting shell surrounding it."},{"Start":"00:36.705 ","End":"00:40.129","Text":"We have a cylindrical shell over here,"},{"Start":"00:40.129 ","End":"00:44.415","Text":"which is this ring in dark gray,"},{"Start":"00:44.415 ","End":"00:54.015","Text":"and the inner radius of this conducting shell is at 2R and the outer radius is at 3R."},{"Start":"00:54.015 ","End":"00:58.355","Text":"The length of this shell is also L. We\u0027re being"},{"Start":"00:58.355 ","End":"01:04.140","Text":"told that this cylindrical shell has a charge of negative 4Q."},{"Start":"01:05.180 ","End":"01:08.525","Text":"Then, around the thick cylindrical shell,"},{"Start":"01:08.525 ","End":"01:13.069","Text":"we have a grounded conducting shell,"},{"Start":"01:13.069 ","End":"01:17.420","Text":"which is thin, and that\u0027s this gray circle going along here."},{"Start":"01:17.420 ","End":"01:23.360","Text":"This is also a cylindrical shell and it is of radius 4R and also of"},{"Start":"01:23.360 ","End":"01:29.660","Text":"length L. We\u0027re not told the charge on this thin cylindrical shell,"},{"Start":"01:29.660 ","End":"01:34.460","Text":"but what we are told is that L is much larger than R,"},{"Start":"01:34.460 ","End":"01:36.989","Text":"the radius of the cylinder."},{"Start":"01:37.640 ","End":"01:41.029","Text":"Our first 2 questions is to find"},{"Start":"01:41.029 ","End":"01:45.199","Text":"the charge distribution of the system and to find the potential."},{"Start":"01:45.199 ","End":"01:46.850","Text":"Now a few lessons ago,"},{"Start":"01:46.850 ","End":"01:54.340","Text":"we did an example using spherical shells and solid spheres that were charged."},{"Start":"01:54.340 ","End":"01:56.325","Text":"What we did there was,"},{"Start":"01:56.325 ","End":"01:58.575","Text":"first we found the potential,"},{"Start":"01:58.575 ","End":"02:01.755","Text":"and then afterwards we found the charge distribution."},{"Start":"02:01.755 ","End":"02:05.700","Text":"You could answer this question also like that by first finding the potential,"},{"Start":"02:05.700 ","End":"02:09.010","Text":"so first answering question 2 and then answering question 1."},{"Start":"02:09.010 ","End":"02:13.100","Text":"However, I\u0027m going to show you another way of solving this type of question."},{"Start":"02:13.100 ","End":"02:17.734","Text":"We\u0027re first going to answer question 1 and then question 2 but of course,"},{"Start":"02:17.734 ","End":"02:20.434","Text":"you can do it the other way round."},{"Start":"02:20.434 ","End":"02:22.564","Text":"Now, something to note,"},{"Start":"02:22.564 ","End":"02:26.660","Text":"not always can we first find the charge distribution"},{"Start":"02:26.660 ","End":"02:31.980","Text":"and then the potential so sometimes this method doesn\u0027t work."},{"Start":"02:33.770 ","End":"02:37.280","Text":"Let\u0027s begin. First of all,"},{"Start":"02:37.280 ","End":"02:40.970","Text":"we have this inner cylinder and it\u0027s not a cylindrical shell,"},{"Start":"02:40.970 ","End":"02:42.440","Text":"it\u0027s a solid cylinder."},{"Start":"02:42.440 ","End":"02:45.155","Text":"Now, we\u0027re being told that it\u0027s a conductive cylinder,"},{"Start":"02:45.155 ","End":"02:47.509","Text":"so the cylinder is a conductor,"},{"Start":"02:47.509 ","End":"02:54.935","Text":"which means that the charge inside the cylinder has to be equal to 0,"},{"Start":"02:54.935 ","End":"02:57.740","Text":"and we know that the charge is therefore only"},{"Start":"02:57.740 ","End":"03:03.390","Text":"distributed along the surface of the cylinder."},{"Start":"03:04.220 ","End":"03:10.970","Text":"Therefore, we can say that the charge at radius equal to R,"},{"Start":"03:10.970 ","End":"03:15.634","Text":"so that\u0027s the surface area of this inner conducting cylinder,"},{"Start":"03:15.634 ","End":"03:20.805","Text":"has to be equal to the charge of the cylinder itself, negative Q."},{"Start":"03:20.805 ","End":"03:23.114","Text":"Because we know that this charge"},{"Start":"03:23.114 ","End":"03:27.470","Text":"negative Q is only on the surface of the cylinder because it\u0027s a conductor."},{"Start":"03:27.470 ","End":"03:34.860","Text":"Therefore, we can say that the charge density per unit area Sigma,"},{"Start":"03:34.860 ","End":"03:38.839","Text":"at radius R is equal to the charge,"},{"Start":"03:38.839 ","End":"03:40.265","Text":"which is negative Q,"},{"Start":"03:40.265 ","End":"03:42.724","Text":"divided by the surface area."},{"Start":"03:42.724 ","End":"03:46.950","Text":"The surface area of a cylinder is equal to 2PiRL."},{"Start":"03:50.750 ","End":"03:56.340","Text":"We can also say that we have a charge per unit length,"},{"Start":"03:56.340 ","End":"04:01.075","Text":"so that\u0027s how much charge as we go down the length of the cylinder."},{"Start":"04:01.075 ","End":"04:03.250","Text":"That\u0027s simply going to be the charge,"},{"Start":"04:03.250 ","End":"04:06.609","Text":"so that\u0027s negative Q divided by the length of the cylinder,"},{"Start":"04:06.609 ","End":"04:10.450","Text":"which is L. We\u0027re going to mainly be working with the Sigma,"},{"Start":"04:10.450 ","End":"04:12.010","Text":"but it\u0027s useful to also know this."},{"Start":"04:12.010 ","End":"04:17.739","Text":"Then, the next thing that we have is this next thing in gray,"},{"Start":"04:17.739 ","End":"04:22.410","Text":"which is a thick conducting cylindrical shell."},{"Start":"04:22.410 ","End":"04:26.274","Text":"Again, this is a conducting cylindrical shell,"},{"Start":"04:26.274 ","End":"04:28.555","Text":"which means that it\u0027s a conductor,"},{"Start":"04:28.555 ","End":"04:36.064","Text":"which means that there\u0027s going to be no charge in the center of this thick shell,"},{"Start":"04:36.064 ","End":"04:42.575","Text":"which means that the charge is going to be distributed along the inner and outer edges,"},{"Start":"04:42.575 ","End":"04:44.990","Text":"so let\u0027s call the inner edge,"},{"Start":"04:44.990 ","End":"04:49.520","Text":"let\u0027s say that it has a charge of q_2 and the outer edge,"},{"Start":"04:49.520 ","End":"04:53.760","Text":"let\u0027s say that it has a charge of q_3."},{"Start":"04:53.930 ","End":"05:00.160","Text":"Then we can say that the charge at R=2R=q_2,"},{"Start":"05:02.240 ","End":"05:05.205","Text":"of course, we don\u0027t yet know what this is,"},{"Start":"05:05.205 ","End":"05:08.290","Text":"and the charge at R=3R,"},{"Start":"05:08.290 ","End":"05:10.490","Text":"which is the outer edge."},{"Start":"05:10.490 ","End":"05:12.680","Text":"We\u0027re going to call this q_3,"},{"Start":"05:12.680 ","End":"05:14.104","Text":"and this, we also don\u0027t know."},{"Start":"05:14.104 ","End":"05:19.490","Text":"What we do know is that the total charge of this cylindrical shell is negative 4Q,"},{"Start":"05:19.490 ","End":"05:21.589","Text":"which has to be distributed along"},{"Start":"05:21.589 ","End":"05:26.705","Text":"the inner and outer edges and nowhere in-between because it\u0027s a conductor."},{"Start":"05:26.705 ","End":"05:33.840","Text":"Therefore, we can say that q_2+q_3= negative 4Q."},{"Start":"05:37.700 ","End":"05:41.210","Text":"Now what we\u0027re going to do is we\u0027re going to try and"},{"Start":"05:41.210 ","End":"05:44.515","Text":"calculate what q_2 and q_3 are equal to."},{"Start":"05:44.515 ","End":"05:52.590","Text":"We have a trick and that is to form a Gaussian surface somewhere in here."},{"Start":"05:52.820 ","End":"06:02.400","Text":"That means inside this conducting thick cylindrical shell."},{"Start":"06:02.570 ","End":"06:07.280","Text":"We know that because the charge is evenly distributed or"},{"Start":"06:07.280 ","End":"06:11.285","Text":"we haven\u0027t been told that it isn\u0027t evenly distributed,"},{"Start":"06:11.285 ","End":"06:14.150","Text":"we can assume that it is evenly distributed."},{"Start":"06:14.150 ","End":"06:17.420","Text":"We know that the electric field at each point is going out,"},{"Start":"06:17.420 ","End":"06:19.355","Text":"first of all in a radial direction,"},{"Start":"06:19.355 ","End":"06:22.265","Text":"and that we have a uniform electric field."},{"Start":"06:22.265 ","End":"06:25.660","Text":"Therefore, we can use our equation,"},{"Start":"06:25.960 ","End":"06:32.485","Text":"E.S=Qin divided by Epsilon naught."},{"Start":"06:32.485 ","End":"06:35.869","Text":"Now we know that the electric field within a conductor,"},{"Start":"06:35.869 ","End":"06:38.735","Text":"which is what this thick gray line is,"},{"Start":"06:38.735 ","End":"06:41.134","Text":"is always equal to 0."},{"Start":"06:41.134 ","End":"06:44.150","Text":"The electric field here is equal to 0."},{"Start":"06:44.150 ","End":"06:46.714","Text":"In order for this equation to balance out,"},{"Start":"06:46.714 ","End":"06:49.740","Text":"that must mean that Qin=0."},{"Start":"06:51.110 ","End":"06:56.405","Text":"We know that the charge enclosed within this red dotted line is equal to 0,"},{"Start":"06:56.405 ","End":"07:00.424","Text":"but we also know that we have 2 charges enclosed."},{"Start":"07:00.424 ","End":"07:02.570","Text":"Within this red dotted line,"},{"Start":"07:02.570 ","End":"07:04.535","Text":"we have this q_2,"},{"Start":"07:04.535 ","End":"07:07.070","Text":"which we still don\u0027t know what it\u0027s equal to,"},{"Start":"07:07.070 ","End":"07:10.165","Text":"and we have over here negative Q,"},{"Start":"07:10.165 ","End":"07:13.350","Text":"so plus negative Q."},{"Start":"07:13.350 ","End":"07:16.970","Text":"This has to be equal to 0, therefore,"},{"Start":"07:16.970 ","End":"07:19.673","Text":"we can isolate this out, and we get that q_2=Q."},{"Start":"07:19.673 ","End":"07:26.954","Text":"If q_2 = 2,"},{"Start":"07:26.954 ","End":"07:28.620","Text":"we can plug that in here,"},{"Start":"07:28.620 ","End":"07:32.235","Text":"so we have Q, for q_2,"},{"Start":"07:32.235 ","End":"07:38.160","Text":"+ q_3 has to be equal to negative 4Q."},{"Start":"07:38.160 ","End":"07:40.200","Text":"We\u0027re using this equation over here."},{"Start":"07:40.200 ","End":"07:49.810","Text":"Therefore, we get that q_3=negative 5Q."},{"Start":"07:49.920 ","End":"07:54.385","Text":"Now let\u0027s find our charge distributions."},{"Start":"07:54.385 ","End":"08:00.130","Text":"We have Sigma at a radius of 2R."},{"Start":"08:00.130 ","End":"08:03.910","Text":"That is simply going to be the charge at 2R,"},{"Start":"08:03.910 ","End":"08:05.200","Text":"which is q_2,"},{"Start":"08:05.200 ","End":"08:10.120","Text":"which we found is equal to Q divided by the surface area."},{"Start":"08:10.120 ","End":"08:14.650","Text":"It\u0027s the inner edge of this thick cylindrical shell that has"},{"Start":"08:14.650 ","End":"08:19.239","Text":"an area of 2Pi multiplied by the radius,"},{"Start":"08:19.239 ","End":"08:23.620","Text":"which here is 2R multiplied by L,"},{"Start":"08:23.620 ","End":"08:29.320","Text":"then the charge density per unit area for the outer edge."},{"Start":"08:29.320 ","End":"08:32.080","Text":"At 3R is equal to q_3,"},{"Start":"08:32.080 ","End":"08:38.380","Text":"which is negative 5Q divided by the surface area of the outer shell,"},{"Start":"08:38.380 ","End":"08:43.880","Text":"which is 2Pi multiplied by 3RL."},{"Start":"08:46.710 ","End":"08:53.514","Text":"Now let\u0027s talk about this thin grounded cylindrical shell."},{"Start":"08:53.514 ","End":"08:56.319","Text":"First of all, if it\u0027s grounded,"},{"Start":"08:56.319 ","End":"08:59.139","Text":"the first thing that has to come to your mind is that"},{"Start":"08:59.139 ","End":"09:05.065","Text":"the potential along this cylindrical shell has to be equal to 0."},{"Start":"09:05.065 ","End":"09:09.025","Text":"That\u0027s the definition of if something is grounded."},{"Start":"09:09.025 ","End":"09:11.814","Text":"If the potential is equal to 0,"},{"Start":"09:11.814 ","End":"09:17.995","Text":"because we know that the electric field is the derivative of the potential."},{"Start":"09:17.995 ","End":"09:22.494","Text":"If we take the derivative of the potential,"},{"Start":"09:22.494 ","End":"09:29.350","Text":"D Phi by over here will be by DR. Of course, the negative derivative."},{"Start":"09:29.350 ","End":"09:32.230","Text":"If the potential is equal to 0,"},{"Start":"09:32.230 ","End":"09:38.920","Text":"that means that the negative derivative of the potential is also going to be equal to 0,"},{"Start":"09:38.920 ","End":"09:42.534","Text":"and of course this is equal to the electric field."},{"Start":"09:42.534 ","End":"09:45.860","Text":"The electric field is a negative derivative."},{"Start":"09:47.880 ","End":"09:51.369","Text":"We got an E field of 0,"},{"Start":"09:51.369 ","End":"09:52.839","Text":"and just as before,"},{"Start":"09:52.839 ","End":"09:55.600","Text":"we have E.S,"},{"Start":"09:55.600 ","End":"10:01.690","Text":"the flux of the field is equal to Q_in the enclosed charge divided by Epsilon Naught."},{"Start":"10:01.690 ","End":"10:08.214","Text":"We just saw that the E field coming out through this section over here,"},{"Start":"10:08.214 ","End":"10:10.105","Text":"so let\u0027s draw it in blue."},{"Start":"10:10.105 ","End":"10:13.940","Text":"Around over here."},{"Start":"10:16.110 ","End":"10:22.450","Text":"We can also say if we take a Gaussian surface even bigger than this,"},{"Start":"10:22.450 ","End":"10:24.399","Text":"from what we just saw over here."},{"Start":"10:24.399 ","End":"10:31.225","Text":"We get that our E field over here is going to be equal to 0. We just saw that."},{"Start":"10:31.225 ","End":"10:34.825","Text":"In that case, we know that Q_in"},{"Start":"10:34.825 ","End":"10:39.145","Text":"also has to be equal to 0 in order for this equation to work."},{"Start":"10:39.145 ","End":"10:45.950","Text":"That means that the total charge enclosed over here has to be equal to 0."},{"Start":"10:46.590 ","End":"10:50.155","Text":"Let\u0027s scroll down for a little bit of space."},{"Start":"10:50.155 ","End":"10:52.705","Text":"The total charge here is equal to 0."},{"Start":"10:52.705 ","End":"10:59.830","Text":"We can say that the charge at a radius of 4R."},{"Start":"10:59.830 ","End":"11:01.959","Text":"That is over here at"},{"Start":"11:01.959 ","End":"11:09.130","Text":"this thin cylindrical grounded shell is equal to 0, what we just saw."},{"Start":"11:09.130 ","End":"11:14.035","Text":"But we also know that it\u0027s equal to all the charges enclosed,"},{"Start":"11:14.035 ","End":"11:18.200","Text":"so It\u0027s equal to q_4,"},{"Start":"11:19.110 ","End":"11:30.830","Text":"which is some charge over here along this shell,"},{"Start":"11:30.830 ","End":"11:36.340","Text":"q_4 plus the charge of this thick shell,"},{"Start":"11:36.340 ","End":"11:38.830","Text":"which it\u0027s q_2 plus q_3,"},{"Start":"11:38.830 ","End":"11:46.134","Text":"or just we saw it\u0027s equal to negative 4Q plus this charge over here,"},{"Start":"11:46.134 ","End":"11:49.300","Text":"which is equal to negative Q."},{"Start":"11:49.300 ","End":"11:52.419","Text":"What\u0027s important to note is that Q_in represents"},{"Start":"11:52.419 ","End":"11:58.059","Text":"the total enclosed charge within these Gaussian envelopes within,"},{"Start":"11:58.059 ","End":"12:00.339","Text":"over here, or over here,"},{"Start":"12:00.339 ","End":"12:02.540","Text":"a little bit outside."},{"Start":"12:02.760 ","End":"12:05.529","Text":"Now the reason I can take"},{"Start":"12:05.529 ","End":"12:09.235","Text":"the Gaussian surface either along this thin cylindrical, shallower,"},{"Start":"12:09.235 ","End":"12:10.614","Text":"just a bit outside,"},{"Start":"12:10.614 ","End":"12:18.350","Text":"is because we know that the potential at infinity is defined as being equal to 0."},{"Start":"12:19.400 ","End":"12:25.575","Text":"We can see that the potential along this outer shell is also equal to 0."},{"Start":"12:25.575 ","End":"12:30.345","Text":"The potential along the outer shell and the potential at infinity are the same,"},{"Start":"12:30.345 ","End":"12:35.455","Text":"which means that there\u0027s nothing in-between these 2 points."},{"Start":"12:35.455 ","End":"12:39.714","Text":"That means that if we take the Gaussian surface over here,"},{"Start":"12:39.714 ","End":"12:43.600","Text":"or we take it at a slightly larger radius over here."},{"Start":"12:43.600 ","End":"12:49.629","Text":"We can consider the electric field at any point from this outer surface and until"},{"Start":"12:49.629 ","End":"12:55.925","Text":"infinity as an area with a 0 E field or with a potential equal to 0."},{"Start":"12:55.925 ","End":"12:59.360","Text":"That\u0027s important to note."},{"Start":"12:59.360 ","End":"13:02.560","Text":"In actual fact, what we have is as if"},{"Start":"13:02.560 ","End":"13:11.810","Text":"this thin grounded cylindrical shell is an extremely thick conducting cylindrical shell."},{"Start":"13:11.850 ","End":"13:17.995","Text":"That means that the E field throughout from this point up until infinity is equal to 0."},{"Start":"13:17.995 ","End":"13:22.000","Text":"Which means that the charge enclosed between this point"},{"Start":"13:22.000 ","End":"13:27.324","Text":"over here at 4R and up until infinity is equal to 0."},{"Start":"13:27.324 ","End":"13:34.644","Text":"But of course we know that with the conductor we have charge located along,"},{"Start":"13:34.644 ","End":"13:36.580","Text":"if it\u0027s a thick cylindrical shell,"},{"Start":"13:36.580 ","End":"13:39.609","Text":"let\u0027s say along the inner surface like q_2,"},{"Start":"13:39.609 ","End":"13:43.220","Text":"and the outer surface is at q_3."},{"Start":"13:43.260 ","End":"13:46.329","Text":"We saw like here,"},{"Start":"13:46.329 ","End":"13:50.544","Text":"when we find the charge density or the charge distribution,"},{"Start":"13:50.544 ","End":"13:51.849","Text":"we take the charge,"},{"Start":"13:51.849 ","End":"13:55.044","Text":"the total charge, and divide it by the surface area."},{"Start":"13:55.044 ","End":"13:59.920","Text":"The surface area at infinity is going to have a radius of infinity."},{"Start":"13:59.920 ","End":"14:04.269","Text":"The denominator will approach infinity."},{"Start":"14:04.269 ","End":"14:08.065","Text":"The denominator will yes approach affinity or be infinity."},{"Start":"14:08.065 ","End":"14:14.485","Text":"Therefore, this fraction as a whole will be equal to 0 or will be approaching 0."},{"Start":"14:14.485 ","End":"14:21.324","Text":"Which means that we\u0027ll get that we have a 0 charge distribution along the outer edge,"},{"Start":"14:21.324 ","End":"14:25.930","Text":"or at infinity where the potential is equal to 0."},{"Start":"14:25.930 ","End":"14:31.210","Text":"Of course the potential is equal to 0 because there\u0027s no charge there."},{"Start":"14:31.210 ","End":"14:33.430","Text":"These 2 things work."},{"Start":"14:33.430 ","End":"14:36.370","Text":"It\u0027s important to note that the charge here,"},{"Start":"14:36.370 ","End":"14:38.574","Text":"the overall charge is equal to 0."},{"Start":"14:38.574 ","End":"14:43.464","Text":"But that doesn\u0027t mean that there\u0027s no charge q_4 over here along the,"},{"Start":"14:43.464 ","End":"14:48.799","Text":"quote unquote, inner edge of this infinite conductor."},{"Start":"14:49.880 ","End":"14:53.790","Text":"That just means that our q_4 over here has to"},{"Start":"14:53.790 ","End":"14:57.329","Text":"balance out the other 2 charges that we have within."},{"Start":"14:57.329 ","End":"15:03.705","Text":"If we just solve this and we isolate out our q_4,"},{"Start":"15:03.705 ","End":"15:08.590","Text":"we\u0027ll get that q_4 has to be equal to 5q."},{"Start":"15:09.710 ","End":"15:13.215","Text":"That means that our Sigma over here,"},{"Start":"15:13.215 ","End":"15:15.659","Text":"I charged density per unit area,"},{"Start":"15:15.659 ","End":"15:25.664","Text":"for 4R is going to be the charge which is 5Q divided by the surface area,"},{"Start":"15:25.664 ","End":"15:28.755","Text":"which is the surface area of a cylinder."},{"Start":"15:28.755 ","End":"15:31.439","Text":"That is 2Pi multiplied by the radius,"},{"Start":"15:31.439 ","End":"15:39.210","Text":"which is 4R multiplied by the length L. It\u0027s"},{"Start":"15:39.210 ","End":"15:42.735","Text":"important to note that the only way we could do this is"},{"Start":"15:42.735 ","End":"15:47.190","Text":"because the outer cylinder is the one that is grounded,"},{"Start":"15:47.190 ","End":"15:50.969","Text":"which means that it has the same potential as that infinity,"},{"Start":"15:50.969 ","End":"15:57.015","Text":"which is why the electric field from there onwards up until infinity is equal to 0."},{"Start":"15:57.015 ","End":"16:00.315","Text":"That\u0027s why the total charge over here is equal to 0."},{"Start":"16:00.315 ","End":"16:02.474","Text":"Then we could do this calculation."},{"Start":"16:02.474 ","End":"16:04.590","Text":"That is why when we were dealing with"},{"Start":"16:04.590 ","End":"16:08.834","Text":"the previous question with the spherical conductors,"},{"Start":"16:08.834 ","End":"16:13.619","Text":"we had our grounded surface not at the edge of the system,"},{"Start":"16:13.619 ","End":"16:16.065","Text":"but rather somewhere in the middle."},{"Start":"16:16.065 ","End":"16:19.409","Text":"That is why we wouldn\u0027t have been able to have"},{"Start":"16:19.409 ","End":"16:23.505","Text":"done this calculation and that\u0027s why we would have gotten stuck."},{"Start":"16:23.505 ","End":"16:30.630","Text":"You can only do this when the grounded surface is at the exterior."},{"Start":"16:30.630 ","End":"16:35.130","Text":"Then we have the same potential as we have at infinity."},{"Start":"16:35.130 ","End":"16:37.859","Text":"That\u0027s how we can get all of these steps,"},{"Start":"16:37.859 ","End":"16:42.610","Text":"that the E field is equal to 0 and therefore the Q_in=0."},{"Start":"16:43.970 ","End":"16:47.655","Text":"This is the answer to question number 1."},{"Start":"16:47.655 ","End":"16:49.815","Text":"Now let\u0027s take a look at question number 2."},{"Start":"16:49.815 ","End":"16:52.599","Text":"What is the potential?"},{"Start":"16:53.810 ","End":"16:56.669","Text":"In order to find the potential,"},{"Start":"16:56.669 ","End":"17:01.920","Text":"we can see that we\u0027re dealing with something that looks like an infinite cylinder."},{"Start":"17:01.920 ","End":"17:04.560","Text":"How can we consider it an infinite cylinder?"},{"Start":"17:04.560 ","End":"17:07.635","Text":"We\u0027re told that the cylinders all have a length L,"},{"Start":"17:07.635 ","End":"17:11.550","Text":"and we\u0027re told that L is significantly larger than R, the radius."},{"Start":"17:11.550 ","End":"17:13.829","Text":"Whenever we see something like this,"},{"Start":"17:13.829 ","End":"17:17.820","Text":"we can consider that the cylinder is infinitely long."},{"Start":"17:17.820 ","End":"17:21.210","Text":"Then if we\u0027re dealing with an infinitely long cylinder"},{"Start":"17:21.210 ","End":"17:25.049","Text":"so when a question such as what is the potential comes up,"},{"Start":"17:25.049 ","End":"17:27.780","Text":"we know how to use Gauss\u0027s law."},{"Start":"17:27.780 ","End":"17:31.799","Text":"What we\u0027re going to do is first we\u0027re going to find the E field at"},{"Start":"17:31.799 ","End":"17:36.075","Text":"every single point using Gauss\u0027s law."},{"Start":"17:36.075 ","End":"17:39.119","Text":"Then what we\u0027re going to do is we\u0027re going to take the negative"},{"Start":"17:39.119 ","End":"17:43.274","Text":"integral of the E field in order to find the potential."},{"Start":"17:43.274 ","End":"17:47.580","Text":"Then what we\u0027re going to do is we\u0027re going to"},{"Start":"17:47.580 ","End":"17:52.605","Text":"use because we\u0027re going to take the negative integral by using an indefinite integral."},{"Start":"17:52.605 ","End":"17:57.434","Text":"That means that we\u0027re going to add on integrating constants."},{"Start":"17:57.434 ","End":"18:00.629","Text":"Then what we\u0027re going to do is we\u0027re going to use the idea"},{"Start":"18:00.629 ","End":"18:03.630","Text":"of calibration and the fact that"},{"Start":"18:03.630 ","End":"18:10.035","Text":"the potential function has to be a continuous function in order to find the potential."},{"Start":"18:10.035 ","End":"18:13.930","Text":"Let begin."},{"Start":"18:15.890 ","End":"18:21.000","Text":"Let\u0027s take first where we\u0027re inside the inner cylinder,"},{"Start":"18:21.000 ","End":"18:22.830","Text":"which as we know, is a solid cylinder,"},{"Start":"18:22.830 ","End":"18:25.560","Text":"so R is less than I."},{"Start":"18:25.560 ","End":"18:30.030","Text":"Now because we know that this solid cylinder is a conductor,"},{"Start":"18:30.030 ","End":"18:33.975","Text":"so that means that the E field over here is equal to 0."},{"Start":"18:33.975 ","End":"18:41.050","Text":"Simple. Now, let\u0027s take the E field when we\u0027re between R and 2R."},{"Start":"18:45.380 ","End":"18:50.025","Text":"Here we\u0027re located within this white empty space."},{"Start":"18:50.025 ","End":"18:54.569","Text":"What we know is that our electric field is uniform."},{"Start":"18:54.569 ","End":"18:56.774","Text":"We have no reason to assume otherwise."},{"Start":"18:56.774 ","End":"19:04.540","Text":"We can say that E.S=Q_in divided by Epsilon naught."},{"Start":"19:04.910 ","End":"19:11.820","Text":"We get that the electric field multiplied by the surface area of a cylinder,"},{"Start":"19:11.820 ","End":"19:19.190","Text":"which is equal to 2PirL=Q_in."},{"Start":"19:19.190 ","End":"19:26.730","Text":"Our Q_in our charging closes this negative Q divided by Epsilon naught."},{"Start":"19:26.730 ","End":"19:29.850","Text":"If we isolate out our E,"},{"Start":"19:29.850 ","End":"19:35.870","Text":"we get that our electric field is equal to negative Q divided"},{"Start":"19:35.870 ","End":"19:43.710","Text":"by 2PirL Epsilon naught."},{"Start":"19:44.500 ","End":"19:48.109","Text":"This is the electric fields and we can make it"},{"Start":"19:48.109 ","End":"19:51.710","Text":"into a vector because of course it\u0027s in the radial direction."},{"Start":"19:51.710 ","End":"19:57.630","Text":"Now let\u0027s look in the region between 2R and 3R,"},{"Start":"19:57.630 ","End":"20:03.660","Text":"so now we\u0027re located within this thick conducting cylindrical shell."},{"Start":"20:03.660 ","End":"20:06.974","Text":"Again, this is a conductor,"},{"Start":"20:06.974 ","End":"20:11.400","Text":"which means that the electric field over here is equal to 0."},{"Start":"20:11.400 ","End":"20:14.579","Text":"Now what we\u0027re going to do is we\u0027re going to"},{"Start":"20:14.579 ","End":"20:18.340","Text":"look in the region between 3R and 4R, so like so."},{"Start":"20:20.780 ","End":"20:25.770","Text":"In this white space between the thick conducting"},{"Start":"20:25.770 ","End":"20:31.290","Text":"shell and this thin conducting grounded shell."},{"Start":"20:31.290 ","End":"20:36.659","Text":"Here again, we have a uniform electric field in the radial direction."},{"Start":"20:36.659 ","End":"20:38.745","Text":"We have no reason to think that we don\u0027t."},{"Start":"20:38.745 ","End":"20:42.465","Text":"We have the electric field multiplied by the surface area,"},{"Start":"20:42.465 ","End":"20:44.620","Text":"which is 2PirL,"},{"Start":"20:45.080 ","End":"20:48.570","Text":"and this is equal to Q_in."},{"Start":"20:48.570 ","End":"20:50.685","Text":"So our Q_in,"},{"Start":"20:50.685 ","End":"20:55.350","Text":"at this point is equal to our charge over here,"},{"Start":"20:55.350 ","End":"21:00.329","Text":"which is negative 4Q plus our charge over here,"},{"Start":"21:00.329 ","End":"21:06.539","Text":"which is negative Q divided by Epsilon naught."},{"Start":"21:06.539 ","End":"21:13.020","Text":"Then we will get that this is equal to negative 5Q divided by Epsilon naught."},{"Start":"21:13.020 ","End":"21:19.034","Text":"Then we can isolate out our E so we\u0027ll get that this is equal to negative 5Q"},{"Start":"21:19.034 ","End":"21:26.740","Text":"divided by 2PirL Epsilon naught."},{"Start":"21:26.930 ","End":"21:32.070","Text":"Of course this is also in the radial direction."},{"Start":"21:32.070 ","End":"21:37.035","Text":"Finally, let\u0027s take the electric field when we\u0027re outside of 4R,"},{"Start":"21:37.035 ","End":"21:39.074","Text":"so we\u0027re outside of this whole area."},{"Start":"21:39.074 ","End":"21:43.065","Text":"Now, we already said in the previous section when we were answering question 1,"},{"Start":"21:43.065 ","End":"21:46.499","Text":"that the potential over here is equal to 0 and that"},{"Start":"21:46.499 ","End":"21:51.645","Text":"the potential over here infinitely far as also equal to 0."},{"Start":"21:51.645 ","End":"21:57.150","Text":"Therefore, we can say that the potential from this exterior shell up"},{"Start":"21:57.150 ","End":"22:02.595","Text":"until infinity is the same and is constant and it\u0027s constantly equal to 0."},{"Start":"22:02.595 ","End":"22:08.970","Text":"Therefore the negative derivative of the potential as a function of r,"},{"Start":"22:08.970 ","End":"22:12.659","Text":"which is equal to the E field,"},{"Start":"22:12.659 ","End":"22:15.000","Text":"has to be equal to 0."},{"Start":"22:15.000 ","End":"22:22.980","Text":"Therefore we can say that here the electric field is equal to 0."},{"Start":"22:22.980 ","End":"22:28.990","Text":"Just like if this was a very thick conductor."},{"Start":"22:30.320 ","End":"22:33.719","Text":"Now we have the electric field everywhere,"},{"Start":"22:33.719 ","End":"22:36.990","Text":"so we can go on to find the potential."},{"Start":"22:36.990 ","End":"22:38.580","Text":"The potential, as we know,"},{"Start":"22:38.580 ","End":"22:44.655","Text":"is equal to the negative integral of E.dr."},{"Start":"22:44.655 ","End":"22:50.445","Text":"We know that the dr vector is simply dr in the r-hat direction."},{"Start":"22:50.445 ","End":"22:56.160","Text":"When we take the dot product between our E fields with the r hat,"},{"Start":"22:56.160 ","End":"23:02.790","Text":"we\u0027re just going to get E dr without the vectors, the negative integral."},{"Start":"23:02.790 ","End":"23:04.950","Text":"This is going to be equal to,"},{"Start":"23:04.950 ","End":"23:06.405","Text":"so let\u0027s do this."},{"Start":"23:06.405 ","End":"23:11.519","Text":"In the region where r is larger, sorry,"},{"Start":"23:11.519 ","End":"23:14.610","Text":"here we\u0027re meant to have that r is smaller than"},{"Start":"23:14.610 ","End":"23:20.800","Text":"R. An integral on 0 is just going to give us 0."},{"Start":"23:20.960 ","End":"23:25.739","Text":"I forgot to say that because we\u0027re doing an indefinite integral so we\u0027re"},{"Start":"23:25.739 ","End":"23:29.895","Text":"going to have to add on our integrating constants."},{"Start":"23:29.895 ","End":"23:32.324","Text":"Here let\u0027s say that this constant is"},{"Start":"23:32.324 ","End":"23:35.895","Text":"C_1 and this is in the region where r is smaller than"},{"Start":"23:35.895 ","End":"23:43.049","Text":"R. Then in the next region we\u0027re dealing with this electric field over here,"},{"Start":"23:43.049 ","End":"23:47.260","Text":"so we\u0027re going to have an electric field of."},{"Start":"23:49.260 ","End":"23:55.495","Text":"It will be equal to Q divided by"},{"Start":"23:55.495 ","End":"24:02.860","Text":"2pi L Epsilon naught multiplied by ln r,"},{"Start":"24:02.860 ","End":"24:06.820","Text":"and then of course plus C_2, this integrating constant."},{"Start":"24:06.820 ","End":"24:10.855","Text":"This is between R and 2R."},{"Start":"24:10.855 ","End":"24:12.489","Text":"Then our next region again,"},{"Start":"24:12.489 ","End":"24:13.930","Text":"we\u0027re integrating along 0."},{"Start":"24:13.930 ","End":"24:17.275","Text":"It\u0027s going to be 0 plus another integration constant, C_3."},{"Start":"24:17.275 ","End":"24:18.640","Text":"This is of course,"},{"Start":"24:18.640 ","End":"24:24.535","Text":"in the region between 2R and 3R."},{"Start":"24:24.535 ","End":"24:28.269","Text":"Then we\u0027re going to this region over here."},{"Start":"24:28.269 ","End":"24:34.360","Text":"We\u0027re going to have a potential of 5Q divided by"},{"Start":"24:34.360 ","End":"24:43.375","Text":"2pi L Epsilon naught multiplied by ln r plus C_4 integrating constant."},{"Start":"24:43.375 ","End":"24:49.765","Text":"This is, of course, in the region between 3R and 4R."},{"Start":"24:49.765 ","End":"24:51.760","Text":"Then our final region again,"},{"Start":"24:51.760 ","End":"24:54.144","Text":"we have a 0E field."},{"Start":"24:54.144 ","End":"24:56.020","Text":"When we integrate along that,"},{"Start":"24:56.020 ","End":"24:59.065","Text":"we\u0027ll have 0 plus C_5."},{"Start":"24:59.065 ","End":"25:04.285","Text":"This is where we\u0027re outside of the whole system."},{"Start":"25:04.285 ","End":"25:05.709","Text":"It\u0027s important to note,"},{"Start":"25:05.709 ","End":"25:11.140","Text":"just remember when you\u0027re integrating along 1 divided by r,"},{"Start":"25:11.140 ","End":"25:13.300","Text":"the integral is just ln(r)."},{"Start":"25:13.300 ","End":"25:18.115","Text":"Of course we have a minus here,"},{"Start":"25:18.115 ","End":"25:21.080","Text":"but we\u0027re taking the negative integral."},{"Start":"25:21.360 ","End":"25:23.605","Text":"We also have a minus here,"},{"Start":"25:23.605 ","End":"25:25.375","Text":"but we\u0027re taking the negative integral,"},{"Start":"25:25.375 ","End":"25:31.699","Text":"so they cancel out and we get these positive values for the potential."},{"Start":"25:33.120 ","End":"25:37.915","Text":"Now what we need to do is we have to find our constants."},{"Start":"25:37.915 ","End":"25:42.880","Text":"We\u0027re using calibration and knowing that our potential function is constant."},{"Start":"25:42.880 ","End":"25:48.400","Text":"First of all, we know that the potential at 4R and greater."},{"Start":"25:48.400 ","End":"25:50.650","Text":"We already spoke about that over here."},{"Start":"25:50.650 ","End":"25:56.125","Text":"Remember, the potential is 0 at infinity and also 0 at 4R."},{"Start":"25:56.125 ","End":"26:01.029","Text":"We can just already say that C_5 is equal to 0."},{"Start":"26:01.029 ","End":"26:05.870","Text":"We can just rub this out and say that this is equal to 0."},{"Start":"26:06.750 ","End":"26:09.129","Text":"Now let\u0027s work on the rest."},{"Start":"26:09.129 ","End":"26:13.030","Text":"We\u0027re using the idea of continuous function."},{"Start":"26:13.030 ","End":"26:18.340","Text":"We can see that at every overlap point for this to be a continuous function,"},{"Start":"26:18.340 ","End":"26:22.810","Text":"the potential at 4R using this equation over here,"},{"Start":"26:22.810 ","End":"26:26.875","Text":"and the potential at 4R using this equation over here,"},{"Start":"26:26.875 ","End":"26:33.490","Text":"where these 2 regions overlap has to be equal to the same thing."},{"Start":"26:33.490 ","End":"26:39.355","Text":"Otherwise, our potential will be a discontinuous function and that cannot be."},{"Start":"26:39.355 ","End":"26:45.700","Text":"Therefore, we can see these 2 equations involve the region 4R."},{"Start":"26:45.700 ","End":"26:48.955","Text":"We\u0027re going to plug 4R into the 2 equations."},{"Start":"26:48.955 ","End":"26:52.419","Text":"4R, we can\u0027t really substitute it in here,"},{"Start":"26:52.419 ","End":"26:56.335","Text":"so we\u0027re just going to say 0 and this is equal to"},{"Start":"26:56.335 ","End":"27:03.250","Text":"5Q divided by 2pi L Epsilon naught ln(r),"},{"Start":"27:03.250 ","End":"27:05.110","Text":"where r here is 4R,"},{"Start":"27:05.110 ","End":"27:12.530","Text":"so ln(4r) plus C_4."},{"Start":"27:14.160 ","End":"27:20.034","Text":"Now if we just rearrange this to isolate out our C_4,"},{"Start":"27:20.034 ","End":"27:31.160","Text":"we\u0027ll get that C_4 is simply equal to negative 5Q divided by 2pi L Epsilon naught ln(4r)."},{"Start":"27:31.290 ","End":"27:34.135","Text":"This is rC-4."},{"Start":"27:34.135 ","End":"27:36.580","Text":"Then we can go on to C_3."},{"Start":"27:36.580 ","End":"27:42.520","Text":"Again, we are in the next region and we see that there\u0027s an overlap between 3R,"},{"Start":"27:42.520 ","End":"27:48.790","Text":"which means that using these 2 equations which apply to this mutual 0.3R,"},{"Start":"27:48.790 ","End":"27:51.010","Text":"we have to get the exact same value for"},{"Start":"27:51.010 ","End":"27:54.220","Text":"potential because potential is a continuous function."},{"Start":"27:54.220 ","End":"28:00.350","Text":"What we\u0027re going to do, is we\u0027re going to plug in 3R to both of these equations."},{"Start":"28:00.450 ","End":"28:06.339","Text":"Here we\u0027re going to get 5Q divided by"},{"Start":"28:06.339 ","End":"28:12.750","Text":"2pi L Epsilon naught multiplied by ln(R),"},{"Start":"28:12.750 ","End":"28:16.150","Text":"where r over here is 3R."},{"Start":"28:16.710 ","End":"28:19.780","Text":"ln(3R) plus our C_4."},{"Start":"28:19.780 ","End":"28:21.564","Text":"I\u0027m not going to rewrite it out."},{"Start":"28:21.564 ","End":"28:22.990","Text":"I\u0027m just going to write C_4."},{"Start":"28:22.990 ","End":"28:24.834","Text":"You can just plug this in."},{"Start":"28:24.834 ","End":"28:28.960","Text":"This has to be equal to 0 plus C_3,"},{"Start":"28:28.960 ","End":"28:30.370","Text":"where we substitute in 3R,"},{"Start":"28:30.370 ","End":"28:33.625","Text":"where here we don\u0027t have a place to substitute that in."},{"Start":"28:33.625 ","End":"28:36.190","Text":"We just substitute in C_3."},{"Start":"28:36.190 ","End":"28:39.640","Text":"Now we have what C_3 is equal to,"},{"Start":"28:39.640 ","End":"28:41.230","Text":"it\u0027s equal to this constant."},{"Start":"28:41.230 ","End":"28:43.749","Text":"We can just plug that in over here."},{"Start":"28:43.749 ","End":"28:48.924","Text":"I can just rub out that 0 and I\u0027ll just write in C3."},{"Start":"28:48.924 ","End":"28:51.055","Text":"You\u0027ll see here that it\u0027s equal to this."},{"Start":"28:51.055 ","End":"28:54.500","Text":"Now we can do C_2."},{"Start":"28:55.620 ","End":"29:02.740","Text":"Again, we have this mutual points between these 2 regions, which is 2R."},{"Start":"29:02.740 ","End":"29:08.930","Text":"The point 2R applies to this equation and also to this equation over here."},{"Start":"29:09.600 ","End":"29:14.620","Text":"What we\u0027re going to do is we\u0027re going to plug in 2R into these 2 equations."},{"Start":"29:14.620 ","End":"29:20.109","Text":"What we\u0027re going to get is that C_ 3 is equal to"},{"Start":"29:20.109 ","End":"29:26.890","Text":"Q divided by 2pi L Epsilon naught multiplied by ln(r),"},{"Start":"29:26.890 ","End":"29:28.494","Text":"where r here is 2R,"},{"Start":"29:28.494 ","End":"29:32.350","Text":"our mutual point, plus C_2."},{"Start":"29:32.350 ","End":"29:39.564","Text":"Therefore, we get that C_2 is simply equal to C_3 minus"},{"Start":"29:39.564 ","End":"29:48.400","Text":"Q divided by 2pi L Epsilon naught multiplied by ln(2R)."},{"Start":"29:48.400 ","End":"29:50.470","Text":"Now we\u0027ve found C_2,"},{"Start":"29:50.470 ","End":"29:51.760","Text":"then we want to find C_1."},{"Start":"29:51.760 ","End":"29:58.584","Text":"C_1 is here and we see that r is a point which is mutual to 2 regions."},{"Start":"29:58.584 ","End":"29:59.965","Text":"It\u0027s mutual to here,"},{"Start":"29:59.965 ","End":"30:03.685","Text":"and it\u0027s mutual to here where we have these 2 equations."},{"Start":"30:03.685 ","End":"30:10.150","Text":"What we\u0027ll get, we plug in r to each equation is that C_1 is equal to"},{"Start":"30:10.150 ","End":"30:18.745","Text":"Q divided by 2pi L Epsilon naught In(r),"},{"Start":"30:18.745 ","End":"30:24.940","Text":"which here our mutual point is R plus C_2,"},{"Start":"30:24.940 ","End":"30:27.820","Text":"where of course we found C_2 over here."},{"Start":"30:27.820 ","End":"30:34.699","Text":"Then we just plug in all of these constants and we find out what C_1 is equal to."},{"Start":"30:35.820 ","End":"30:39.475","Text":"Now what I\u0027m going to do is I\u0027m going to substitute in"},{"Start":"30:39.475 ","End":"30:42.625","Text":"all of my constants and write them out over here."},{"Start":"30:42.625 ","End":"30:46.645","Text":"I\u0027m not going to show the working out because it\u0027s simply algebra."},{"Start":"30:46.645 ","End":"30:50.500","Text":"Feel free to work this out on your"},{"Start":"30:50.500 ","End":"30:55.160","Text":"own and pause the video and see if you get the same answers."},{"Start":"30:55.950 ","End":"30:59.199","Text":"This is the final answer for the potential."},{"Start":"30:59.199 ","End":"31:04.854","Text":"The potential is equal to Q divided by 2pi L Epsilon naught,"},{"Start":"31:04.854 ","End":"31:07.419","Text":"which is a common factor of everything."},{"Start":"31:07.419 ","End":"31:11.185","Text":"Then in the region r is smaller than R,"},{"Start":"31:11.185 ","End":"31:15.580","Text":"we have this multiplied by ln(1/2) plus 5ln(3/4)."},{"Start":"31:15.580 ","End":"31:22.540","Text":"Then in the next region of r\u0027s between R and 2 times R,"},{"Start":"31:22.540 ","End":"31:29.830","Text":"we have this multiplied by ln(r) divided by 2R plus 5ln(3/4)."},{"Start":"31:29.830 ","End":"31:31.735","Text":"In the next region,"},{"Start":"31:31.735 ","End":"31:35.260","Text":"we have this multiplied by 5ln(3/4)."},{"Start":"31:35.260 ","End":"31:37.600","Text":"In the next region,"},{"Start":"31:37.600 ","End":"31:45.320","Text":"we have that the potential is equal to this multiplied by 5ln(r/4R)."},{"Start":"31:45.570 ","End":"31:48.519","Text":"Outside of the whole system,"},{"Start":"31:48.519 ","End":"31:52.609","Text":"we have that the potential is just equal to 0."},{"Start":"31:54.030 ","End":"31:56.830","Text":"This was our answer to Question 2,"},{"Start":"31:56.830 ","End":"32:00.565","Text":"and now let\u0027s solve Question 3."},{"Start":"32:00.565 ","End":"32:04.405","Text":"In questions we\u0027re being told that"},{"Start":"32:04.405 ","End":"32:10.435","Text":"a proton of mass Mp has a charge of the absolute value of e,"},{"Start":"32:10.435 ","End":"32:16.464","Text":"the electron charge, and it\u0027s released from rest at a distance of"},{"Start":"32:16.464 ","End":"32:24.505","Text":"2 times R. We\u0027re being asked what is the protons velocity after a distance of R?"},{"Start":"32:24.505 ","End":"32:30.430","Text":"That means that our proton starts somewhere at a radius of 2R."},{"Start":"32:30.430 ","End":"32:34.405","Text":"We can say somewhere along this line over here,"},{"Start":"32:34.405 ","End":"32:38.770","Text":"and it can go 1 of 2 directions."},{"Start":"32:38.770 ","End":"32:41.619","Text":"It can either travel here and then it will end up over"},{"Start":"32:41.619 ","End":"32:45.009","Text":"here at a distance r and then we need to find the velocity over here."},{"Start":"32:45.009 ","End":"32:49.615","Text":"Or it can travel in this direction over here."},{"Start":"32:49.615 ","End":"32:55.730","Text":"Then we\u0027ll be asked what its velocity is over here at this end."},{"Start":"32:57.060 ","End":"33:03.655","Text":"First of all, we\u0027re being told that our initial velocity,"},{"Start":"33:03.655 ","End":"33:06.205","Text":"Vi, is equal to 0."},{"Start":"33:06.205 ","End":"33:08.494","Text":"It\u0027s starting from rest."},{"Start":"33:08.494 ","End":"33:12.045","Text":"Now we just said that it can move 1 of 2 directions."},{"Start":"33:12.045 ","End":"33:16.455","Text":"Let\u0027s take a look at this first direction over here where it\u0027s moving outwards."},{"Start":"33:16.455 ","End":"33:24.490","Text":"Because we know here we have a 0 electric field inside this thick gray line."},{"Start":"33:24.490 ","End":"33:27.009","Text":"There\u0027s no reason why,"},{"Start":"33:27.009 ","End":"33:30.504","Text":"especially if it\u0027s starting off with a 0 velocity,"},{"Start":"33:30.504 ","End":"33:35.230","Text":"that it will move in this direction because there\u0027s no electric field."},{"Start":"33:35.230 ","End":"33:39.024","Text":"Here our electric field is equal to 0."},{"Start":"33:39.024 ","End":"33:42.685","Text":"If there\u0027s no electric field,"},{"Start":"33:42.685 ","End":"33:50.140","Text":"we can say that our electrical force in this direction is also equal to 0."},{"Start":"33:50.140 ","End":"33:53.020","Text":"Our proton isn\u0027t going to be moving in"},{"Start":"33:53.020 ","End":"33:56.710","Text":"this direction so we can just take out that option."},{"Start":"33:56.710 ","End":"33:58.419","Text":"We can see that it\u0027s going to be moving in"},{"Start":"33:58.419 ","End":"34:01.570","Text":"this direction which also makes sense because we\u0027re being"},{"Start":"34:01.570 ","End":"34:06.865","Text":"told that our proton has a positive charge."},{"Start":"34:06.865 ","End":"34:13.240","Text":"We know that the charge of this inner cylinder over here is a negative Q."},{"Start":"34:13.240 ","End":"34:17.260","Text":"Of course we know that positive is attracted to negative."},{"Start":"34:17.260 ","End":"34:20.649","Text":"It makes sense that a proton released from rest at this point over"},{"Start":"34:20.649 ","End":"34:25.940","Text":"here will be attracted to this point over here."},{"Start":"34:27.120 ","End":"34:33.235","Text":"We\u0027re going to solve this via using the idea of conservation of energy."},{"Start":"34:33.235 ","End":"34:38.245","Text":"We know that the potential energy is equal to Q,"},{"Start":"34:38.245 ","End":"34:41.890","Text":"the charge multiplied by the potential."},{"Start":"34:41.890 ","End":"34:46.840","Text":"We know that this is a conservative value."},{"Start":"34:46.840 ","End":"34:51.880","Text":"This is really what we workout the potential for."},{"Start":"34:51.880 ","End":"34:54.580","Text":"We work very hard in order to find the potential at"},{"Start":"34:54.580 ","End":"34:57.595","Text":"every single point and then using this,"},{"Start":"34:57.595 ","End":"35:01.970","Text":"we can find the energy at these different points."},{"Start":"35:02.610 ","End":"35:05.770","Text":"Let\u0027s use conservation of energy."},{"Start":"35:05.770 ","End":"35:10.134","Text":"Our initial energy Ei is consisting of,"},{"Start":"35:10.134 ","End":"35:11.499","Text":"just like in mechanics,"},{"Start":"35:11.499 ","End":"35:15.655","Text":"we have our potential energy and we have our kinetic energy."},{"Start":"35:15.655 ","End":"35:21.280","Text":"Our kinetic energy is equal to 1/2 multiplied by the mass of the proton,"},{"Start":"35:21.280 ","End":"35:24.220","Text":"multiplied by its initial velocity,"},{"Start":"35:24.220 ","End":"35:25.735","Text":"which we were told,"},{"Start":"35:25.735 ","End":"35:27.190","Text":"because it\u0027s released from rest,"},{"Start":"35:27.190 ","End":"35:30.079","Text":"Vi is equal to 0."},{"Start":"35:30.810 ","End":"35:35.740","Text":"Then we\u0027re going to add in our potential energy,"},{"Start":"35:35.740 ","End":"35:42.055","Text":"so plus our potential energy at our initial radius,"},{"Start":"35:42.055 ","End":"35:48.189","Text":"which is at 2R."},{"Start":"35:48.189 ","End":"35:58.570","Text":"All of this is equal to 0 plus our potential energy so that is our charge Q."},{"Start":"35:58.570 ","End":"36:02.934","Text":"We were told that it has a charge of the absolute value of E,"},{"Start":"36:02.934 ","End":"36:04.929","Text":"which is the charge of an electron,"},{"Start":"36:04.929 ","End":"36:09.535","Text":"and then multiplied by Phi at this radius 2R."},{"Start":"36:09.535 ","End":"36:13.659","Text":"We can see that 2R we can plug it in"},{"Start":"36:13.659 ","End":"36:17.980","Text":"over here or over here because our potential is continuous."},{"Start":"36:17.980 ","End":"36:21.204","Text":"The function is continuous so we can use either equation."},{"Start":"36:21.204 ","End":"36:25.299","Text":"This equation is of course shorter than writing all of this out,"},{"Start":"36:25.299 ","End":"36:26.590","Text":"so we can just do that."},{"Start":"36:26.590 ","End":"36:35.050","Text":"We have e multiplied by the common factor Q divided by 2piL Epsilon Naught"},{"Start":"36:35.050 ","End":"36:44.980","Text":"multiplied by 5 In(3/4)."},{"Start":"36:44.980 ","End":"36:51.835","Text":"Then our e final is equal to this point over here."},{"Start":"36:51.835 ","End":"36:56.109","Text":"We have the kinetic energy plus the potential energy at this point."},{"Start":"36:56.109 ","End":"37:02.680","Text":"We have the kinetic energy which is half times mass of the proton multiplied by Vf,"},{"Start":"37:02.680 ","End":"37:05.755","Text":"which is what we\u0027re trying to find."},{"Start":"37:05.755 ","End":"37:08.335","Text":"Of course, these are both squared."},{"Start":"37:08.335 ","End":"37:10.059","Text":"This is equal to 0, so it doesn\u0027t matter,"},{"Start":"37:10.059 ","End":"37:12.234","Text":"but it\u0027s squared over here."},{"Start":"37:12.234 ","End":"37:15.475","Text":"1/2mv squared is the kinetic energy."},{"Start":"37:15.475 ","End":"37:20.785","Text":"Vf is what we\u0027re trying to find plus the potential energy at that point."},{"Start":"37:20.785 ","End":"37:27.130","Text":"At that point our radius is R. What is the protons velocity after a distance of R?"},{"Start":"37:27.130 ","End":"37:28.870","Text":"We\u0027ve moved in a distance of R,"},{"Start":"37:28.870 ","End":"37:37.705","Text":"so we\u0027re located at a radius R. This is going to be equal to 1/2"},{"Start":"37:37.705 ","End":"37:45.219","Text":"(mpVf)^2 plus our charge absolute value"},{"Start":"37:45.219 ","End":"37:50.814","Text":"of e multiplied by the potential at r. The potential at r,"},{"Start":"37:50.814 ","End":"37:53.650","Text":"we can write in these 2 regions."},{"Start":"37:53.650 ","End":"37:57.475","Text":"Let\u0027s just use this equation over here."},{"Start":"37:57.475 ","End":"38:05.305","Text":"Multiplied by Q divided by 2PiL Epsilon_0,"},{"Start":"38:05.305 ","End":"38:10.195","Text":"and this is multiplied by ln(1.5)."},{"Start":"38:10.195 ","End":"38:19.010","Text":"We could have used either 1 of these equations plus 5 ln (3/4)."},{"Start":"38:20.070 ","End":"38:22.764","Text":"From conservation of energy,"},{"Start":"38:22.764 ","End":"38:25.345","Text":"we know that these 2 are equal to each other."},{"Start":"38:25.345 ","End":"38:26.980","Text":"Ei is equal to Ef."},{"Start":"38:26.980 ","End":"38:29.335","Text":"Therefore, we can just write this out."},{"Start":"38:29.335 ","End":"38:33.819","Text":"We have the absolute value of e multiplied by Q divided"},{"Start":"38:33.819 ","End":"38:39.249","Text":"by 2PiL Epsilon_0 multiplied by 5ln"},{"Start":"38:39.249 ","End":"38:42.669","Text":"(3/4) is equal to"},{"Start":"38:42.669 ","End":"38:50.919","Text":"1/2(mVf)^2 plus e multiplied"},{"Start":"38:50.919 ","End":"38:56.365","Text":"by Q divided by 2 PiL Epsilon_0."},{"Start":"38:56.365 ","End":"38:58.539","Text":"All of this is multiplied by"},{"Start":"38:58.539 ","End":"39:07.390","Text":"ln (1/2) plus 5ln(3/4)."},{"Start":"39:07.390 ","End":"39:11.409","Text":"We can see that we have eQ divided by"},{"Start":"39:11.409 ","End":"39:16.705","Text":"2 PiL Epsilon_0 multiplied by 5ln (3/4) on both sides."},{"Start":"39:16.705 ","End":"39:20.439","Text":"We can subtract that from both sides so let\u0027s just do that."},{"Start":"39:20.439 ","End":"39:22.615","Text":"We can just cross this out."},{"Start":"39:22.615 ","End":"39:26.950","Text":"Then let\u0027s scroll down to give us a bit more space."},{"Start":"39:26.950 ","End":"39:30.700","Text":"What we will have is that negative of"},{"Start":"39:30.700 ","End":"39:36.369","Text":"a 1/2 mass of the proton multiplied by (Vf)^2 is equal to"},{"Start":"39:36.369 ","End":"39:40.750","Text":"the absolute value of this multiplied by Q divided by"},{"Start":"39:40.750 ","End":"39:47.120","Text":"2PiL Epsilon_0 multiplied by ln(1.5)."},{"Start":"39:47.820 ","End":"39:55.315","Text":"Now we can do a trick and we can move this negative over to this side over here."},{"Start":"39:55.315 ","End":"39:58.780","Text":"Then if we have a negative over here,"},{"Start":"39:58.780 ","End":"40:05.000","Text":"then we can change this ln(1.5) to ln(2)."},{"Start":"40:05.400 ","End":"40:08.679","Text":"Then what we\u0027ll get is that this is equal to"},{"Start":"40:08.679 ","End":"40:12.940","Text":"the absolute value of e multiplied by Q divided"},{"Start":"40:12.940 ","End":"40:20.410","Text":"by 2PiL Epsilon_0 multiplied by ln(2)."},{"Start":"40:20.410 ","End":"40:25.390","Text":"If you don\u0027t know how I took this minus over here and made this ln(1/2) to ln/2"},{"Start":"40:25.390 ","End":"40:31.150","Text":"please review the laws of logs and lns and how to play around with them."},{"Start":"40:31.150 ","End":"40:36.020","Text":"Will help you in lots of different questions that deals with lns and logs."},{"Start":"40:36.720 ","End":"40:41.710","Text":"Now, all we have to do is we have to isolate out this Vf."},{"Start":"40:41.710 ","End":"40:44.350","Text":"We\u0027ll isolate out (Vf)^2 and then we\u0027ll take"},{"Start":"40:44.350 ","End":"40:47.844","Text":"the square root of everything. I\u0027ll just do it now."},{"Start":"40:47.844 ","End":"40:50.889","Text":"We\u0027ll get that Vf is simply equal to"},{"Start":"40:50.889 ","End":"40:55.029","Text":"the square root of the absolute value of e multiplied by"},{"Start":"40:55.029 ","End":"40:59.079","Text":"Q ln(2) divided by"},{"Start":"40:59.079 ","End":"41:06.290","Text":"PiL Epsilon_0 multiplied by the mass of the proton."},{"Start":"41:07.920 ","End":"41:15.590","Text":"This is the answer to question Number 3 and that is the end of our lesson."}],"ID":14209},{"Watched":false,"Name":"Exercise 3","Duration":"14m 14s","ChapterTopicVideoID":12143,"CourseChapterTopicPlaylistID":157337,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12143.jpeg","UploadDate":"2018-06-28T04:48:00.7730000","DurationForVideoObject":"PT14M14S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.905","Text":"Hello. In this lesson,"},{"Start":"00:01.905 ","End":"00:04.230","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.230 ","End":"00:10.080","Text":"A full sphere of radius R has a uniform charge density per unit volume Rho."},{"Start":"00:10.080 ","End":"00:15.480","Text":"Calculate the electric field and the potential at each point."},{"Start":"00:15.480 ","End":"00:19.185","Text":"Because we can see that we have spherical symmetry,"},{"Start":"00:19.185 ","End":"00:24.375","Text":"we\u0027re going to use Gauss\u0027s law in order to solve this question."},{"Start":"00:24.375 ","End":"00:27.780","Text":"First, we\u0027re going to find the electric field at each point."},{"Start":"00:27.780 ","End":"00:33.070","Text":"Then using the electric field we\u0027re going to find the potential at each point."},{"Start":"00:33.920 ","End":"00:40.195","Text":"First, let\u0027s find the electric field in the region inside the sphere."},{"Start":"00:40.195 ","End":"00:43.205","Text":"When r is smaller than capital R,"},{"Start":"00:43.205 ","End":"00:49.370","Text":"So we\u0027re going to draw this spherical Gaussian surface over here,"},{"Start":"00:49.370 ","End":"00:56.555","Text":"and it\u0027s a radius r. Because we know that the charge density is uniform,"},{"Start":"00:56.555 ","End":"00:59.150","Text":"we have the spherical symmetry."},{"Start":"00:59.150 ","End":"01:02.630","Text":"We know that the electric field coming out of"},{"Start":"01:02.630 ","End":"01:07.685","Text":"every single point here is going to be equal or uniform."},{"Start":"01:07.685 ","End":"01:10.130","Text":"We have a uniform electric field."},{"Start":"01:10.130 ","End":"01:16.670","Text":"That means that we can say that the closed integral on E dot ds,"},{"Start":"01:16.670 ","End":"01:20.060","Text":"where ds is a unit of area,"},{"Start":"01:20.060 ","End":"01:23.390","Text":"is simply going to be equal to the electric field at"},{"Start":"01:23.390 ","End":"01:28.480","Text":"each point multiplied by the total surface area."},{"Start":"01:30.290 ","End":"01:39.900","Text":"We have E and the surface area of a sphere is simply equal to 4 Pi r squared."},{"Start":"01:39.900 ","End":"01:43.250","Text":"This is equal to the other side of our E dot S equation,"},{"Start":"01:43.250 ","End":"01:47.605","Text":"which is Qin divided by Epsilon naught."},{"Start":"01:47.605 ","End":"01:50.305","Text":"Now let\u0027s calculate Qin."},{"Start":"01:50.305 ","End":"01:52.485","Text":"Here we have a full sphere."},{"Start":"01:52.485 ","End":"01:53.780","Text":"We\u0027re not given the charge,"},{"Start":"01:53.780 ","End":"01:56.605","Text":"we are given the charge density."},{"Start":"01:56.605 ","End":"02:01.310","Text":"What we have to do is we have to work out how much of"},{"Start":"02:01.310 ","End":"02:05.850","Text":"this charge density is enclosed within this sphere."},{"Start":"02:05.850 ","End":"02:09.740","Text":"Generally speaking, Qin when dealing with"},{"Start":"02:09.740 ","End":"02:16.400","Text":"a volumetric charge density is equal to the integral of Rho dv,"},{"Start":"02:16.400 ","End":"02:20.750","Text":"where dv is a small piece of volume."},{"Start":"02:20.750 ","End":"02:23.900","Text":"Now because we\u0027re being told that Rho,"},{"Start":"02:23.900 ","End":"02:28.595","Text":"so that\u0027s the charge density per unit volume is uniform."},{"Start":"02:28.595 ","End":"02:31.595","Text":"We don\u0027t have to integrate."},{"Start":"02:31.595 ","End":"02:35.540","Text":"We can just say that this is equal to simply Rho multiplied"},{"Start":"02:35.540 ","End":"02:40.155","Text":"by the total volume in this sphere."},{"Start":"02:40.155 ","End":"02:49.570","Text":"Here specifically that\u0027s going to be Rho multiplied by 4/3s Pi r cubed."},{"Start":"02:50.150 ","End":"02:57.230","Text":"If we had non-uniform charge density then the amount of charge enclosed in"},{"Start":"02:57.230 ","End":"03:05.720","Text":"a specific sphere would be dependent on the radius or on the angle or something else."},{"Start":"03:05.720 ","End":"03:09.560","Text":"In that case, we would have to use an integration in order to"},{"Start":"03:09.560 ","End":"03:14.345","Text":"calculate how much charge is enclosed in different types of spheres."},{"Start":"03:14.345 ","End":"03:17.100","Text":"However, because here we\u0027re being told that it\u0027s a uniform."},{"Start":"03:17.100 ","End":"03:20.165","Text":"We have the same charge density at every single point."},{"Start":"03:20.165 ","End":"03:27.050","Text":"We can just multiply the charge density by the total volume enclosed,"},{"Start":"03:27.050 ","End":"03:28.925","Text":"which is what we have here."},{"Start":"03:28.925 ","End":"03:33.330","Text":"Now we can plug this into here."},{"Start":"03:33.330 ","End":"03:43.690","Text":"Qin is Rho multiplied by 4/3s Pi r cubed and divided by Epsilon naught."},{"Start":"03:44.690 ","End":"03:51.095","Text":"Now we can say that E multiplied by 4 Pi r"},{"Start":"03:51.095 ","End":"03:59.460","Text":"squared is equal to Rho Pi r cubed multiplied by 4,"},{"Start":"03:59.460 ","End":"04:02.220","Text":"divided by 3 Epsilon naught."},{"Start":"04:02.220 ","End":"04:05.235","Text":"Now we can cross out the 4 Pi."},{"Start":"04:05.235 ","End":"04:07.665","Text":"We can cross out the r squared."},{"Start":"04:07.665 ","End":"04:14.540","Text":"Here we\u0027re left with an r. Therefore we get that the E field in this region over here,"},{"Start":"04:14.540 ","End":"04:17.465","Text":"it\u0027s still located inside the sphere,"},{"Start":"04:17.465 ","End":"04:25.635","Text":"is equal to Rho r divided by 3 Epsilon naught."},{"Start":"04:25.635 ","End":"04:30.660","Text":"Of course, this is going to be in the radial direction."},{"Start":"04:31.580 ","End":"04:34.745","Text":"Here\u0027s the electric field for this region."},{"Start":"04:34.745 ","End":"04:40.575","Text":"Now, let\u0027s take a look at the region where we\u0027re outside of the sphere."},{"Start":"04:40.575 ","End":"04:45.525","Text":"R is bigger than r. Let\u0027s just rub all of this out."},{"Start":"04:45.525 ","End":"04:51.565","Text":"Now I\u0027m doing my Gaussian surface in this region where we\u0027re located outside."},{"Start":"04:51.565 ","End":"04:54.590","Text":"Again, this is a radius."},{"Start":"04:55.040 ","End":"04:57.815","Text":"Again, just like before,"},{"Start":"04:57.815 ","End":"05:01.010","Text":"we have this uniform charge density,"},{"Start":"05:01.010 ","End":"05:04.880","Text":"which means that we\u0027re going to have a uniform electric field throughout."},{"Start":"05:04.880 ","End":"05:08.810","Text":"Therefore, we can just say that E dot S is equal to"},{"Start":"05:08.810 ","End":"05:14.945","Text":"E multiplied by the surface area of this sphere,"},{"Start":"05:14.945 ","End":"05:18.965","Text":"which is 4 Pi r squared just this time,"},{"Start":"05:18.965 ","End":"05:26.540","Text":"r is bigger than R and this is of course equal to Qin divided by Epsilon naught."},{"Start":"05:26.540 ","End":"05:30.090","Text":"Let\u0027s see what Qin is equal to."},{"Start":"05:30.290 ","End":"05:34.955","Text":"Just like before, we don\u0027t have to do an integration in this case."},{"Start":"05:34.955 ","End":"05:40.070","Text":"Because we know that our charge density is uniform,"},{"Start":"05:40.070 ","End":"05:45.575","Text":"we can just say that if E is equal to Rho multiplied by the total volume."},{"Start":"05:45.575 ","End":"05:50.010","Text":"Now notice the charge density is just within this sphere."},{"Start":"05:50.010 ","End":"05:53.765","Text":"Even if I took my radius to be over here,"},{"Start":"05:53.765 ","End":"05:57.665","Text":"my charge density is still encompassed within this radius"},{"Start":"05:57.665 ","End":"06:03.620","Text":"of R. If it\u0027s encompassed by this radius of R,"},{"Start":"06:03.620 ","End":"06:06.095","Text":"so when I put in my volume,"},{"Start":"06:06.095 ","End":"06:09.500","Text":"I don\u0027t care how big this lowercase r is."},{"Start":"06:09.500 ","End":"06:11.165","Text":"I just care about this."},{"Start":"06:11.165 ","End":"06:16.155","Text":"I have Rho multiplied by 4/3s Pi,"},{"Start":"06:16.155 ","End":"06:23.560","Text":"and then this time it\u0027s R cubed because there\u0027s no charge outside of the sphere."},{"Start":"06:23.560 ","End":"06:25.275","Text":"That\u0027s why I\u0027m taking this."},{"Start":"06:25.275 ","End":"06:27.075","Text":"Now I can plug this in here."},{"Start":"06:27.075 ","End":"06:28.665","Text":"I\u0027ll have that Qin,"},{"Start":"06:28.665 ","End":"06:31.980","Text":"which is equal to Rho multiplied by"},{"Start":"06:31.980 ","End":"06:40.900","Text":"4 Pi R cubed divided by this 3 over here and Epsilon naught."},{"Start":"06:42.110 ","End":"06:46.575","Text":"Therefore, I can take my E field,"},{"Start":"06:46.575 ","End":"06:52.590","Text":"that is multiplied by 4 Pi r squared."},{"Start":"06:52.590 ","End":"07:00.535","Text":"This is equal to 4 Rho Pi R cubed divided by 3 Epsilon naught."},{"Start":"07:00.535 ","End":"07:03.815","Text":"[inaudible] and then I can divide both sides by 4,"},{"Start":"07:03.815 ","End":"07:05.945","Text":"divide both sides by Pi."},{"Start":"07:05.945 ","End":"07:11.240","Text":"But now notice that my rs don\u0027t cancel out because these are 2 different rs."},{"Start":"07:11.240 ","End":"07:13.565","Text":"This is my radius which is changing."},{"Start":"07:13.565 ","End":"07:15.455","Text":"I can lengthen and shorten it."},{"Start":"07:15.455 ","End":"07:18.770","Text":"But this R^3 is a constant,"},{"Start":"07:18.770 ","End":"07:22.280","Text":"it\u0027s the radius of the sphere and it\u0027s the charge that I\u0027ll have"},{"Start":"07:22.280 ","End":"07:26.675","Text":"no matter how far away I am outside of the sphere."},{"Start":"07:26.675 ","End":"07:32.675","Text":"Therefore, I\u0027ll get that my E field is equal to"},{"Start":"07:32.675 ","End":"07:41.705","Text":"Rho R^3 divided by 3r^2 Epsilon naught."},{"Start":"07:41.705 ","End":"07:46.380","Text":"Of course, this is also in the radial direction."},{"Start":"07:47.440 ","End":"07:52.295","Text":"Let\u0027s write out the electric field neatly."},{"Start":"07:52.295 ","End":"07:54.470","Text":"We get that our electric field,"},{"Start":"07:54.470 ","End":"08:03.590","Text":"which is a vector is equal to Rho r divided by 3 Epsilon naught in"},{"Start":"08:03.590 ","End":"08:07.010","Text":"the radial direction in the region where r is smaller than"},{"Start":"08:07.010 ","End":"08:12.920","Text":"capital R and is equal to Rho R^3"},{"Start":"08:12.920 ","End":"08:16.865","Text":"divided by 3 Epsilon naught r"},{"Start":"08:16.865 ","End":"08:21.860","Text":"squared in the radial direction when we\u0027re in the region where r is"},{"Start":"08:21.860 ","End":"08:31.010","Text":"bigger than capital R. This is the answer to the question about the electric field."},{"Start":"08:31.010 ","End":"08:35.270","Text":"Now let\u0027s work out the potential. Let\u0027s do it here."},{"Start":"08:35.270 ","End":"08:41.550","Text":"We know that our potential is equal to the negative integral E.dr."},{"Start":"08:44.320 ","End":"08:50.840","Text":"Now what we can do is we can say that the potential for the first region,"},{"Start":"08:50.840 ","End":"08:55.190","Text":"so that\u0027s where r is smaller than R is going to be equal to."},{"Start":"08:55.190 ","End":"08:58.100","Text":"Let\u0027s take the indefinite integrals."},{"Start":"08:58.100 ","End":"09:00.830","Text":"That means that we are not including bounds and"},{"Start":"09:00.830 ","End":"09:03.965","Text":"that means that we have to include an integrating constant."},{"Start":"09:03.965 ","End":"09:10.370","Text":"We\u0027re taking the integral on Rho r divided by 3 Epsilon naught and because it\u0027s in"},{"Start":"09:10.370 ","End":"09:19.430","Text":"the r hat direction and dr vector is just dr in the r-hat direction."},{"Start":"09:19.430 ","End":"09:24.065","Text":"When we take the dot product between r hat and r hat, we get 1."},{"Start":"09:24.065 ","End":"09:26.510","Text":"We just get this."},{"Start":"09:26.510 ","End":"09:29.510","Text":"Then what we\u0027ll get is that this is equal to"},{"Start":"09:29.510 ","End":"09:37.624","Text":"negative Rho r^2 divided by 6 Epsilon naught plus an integration constant."},{"Start":"09:37.624 ","End":"09:39.930","Text":"Let\u0027s call it c_1."},{"Start":"09:39.930 ","End":"09:46.090","Text":"Now let\u0027s take the integral in the region where r is bigger than R. Again,"},{"Start":"09:46.090 ","End":"09:49.260","Text":"we\u0027re taking an indefinite integral."},{"Start":"09:49.260 ","End":"09:53.930","Text":"We have Rho r cubed."},{"Start":"09:53.930 ","End":"09:55.430","Text":"Notice this is a constant."},{"Start":"09:55.430 ","End":"10:03.380","Text":"We\u0027re not integrating along R divided by 3 Epsilon naught r^2."},{"Start":"10:03.380 ","End":"10:04.595","Text":"Again, dr,"},{"Start":"10:04.595 ","End":"10:09.740","Text":"because dr hat dot-product with this r had from the dr just gives us this."},{"Start":"10:09.740 ","End":"10:12.725","Text":"Then we get negative from here."},{"Start":"10:12.725 ","End":"10:18.126","Text":"Then because we\u0027re integrating by 1 divided by r squared,"},{"Start":"10:18.126 ","End":"10:26.075","Text":"we\u0027re going to get that this is equal to negative Rho r cubed divided by"},{"Start":"10:26.075 ","End":"10:32.330","Text":"3 Epsilon naught r. Then we\u0027re"},{"Start":"10:32.330 ","End":"10:38.990","Text":"adding on another constant, c_2."},{"Start":"10:38.990 ","End":"10:46.040","Text":"Right now we have that our potential is equal to this function over here,"},{"Start":"10:46.040 ","End":"10:50.720","Text":"which is equal to negative Rho r^2 divided by"},{"Start":"10:50.720 ","End":"10:58.415","Text":"6 Epsilon naught plus c_1 in the region where r is smaller than R,"},{"Start":"10:58.415 ","End":"11:08.270","Text":"and we have Rho R cubed divided by 3 Epsilon naught r plus c_2 in"},{"Start":"11:08.270 ","End":"11:13.970","Text":"the region where r is bigger than R. Now we\u0027re going"},{"Start":"11:13.970 ","End":"11:20.075","Text":"to use the idea of calibration and the facts that are potential as a continuous function."},{"Start":"11:20.075 ","End":"11:27.600","Text":"In calibration, we know that the potential at infinity is equal to 0."},{"Start":"11:27.760 ","End":"11:32.075","Text":"That means when we\u0027re infinitely far away from the sphere,"},{"Start":"11:32.075 ","End":"11:35.510","Text":"so that\u0027s going to be in this region over here."},{"Start":"11:35.510 ","End":"11:42.530","Text":"Then if we substitute an infinity for r we\u0027ll get that Rho R^3 divided by"},{"Start":"11:42.530 ","End":"11:49.820","Text":"3 Epsilon naught multiplied by infinity plus c_2 is equal to 0."},{"Start":"11:49.820 ","End":"11:53.375","Text":"Here the denominator is really big,"},{"Start":"11:53.375 ","End":"11:59.210","Text":"which means that this whole fraction is approaching 0 or is equal to 0."},{"Start":"11:59.210 ","End":"12:02.255","Text":"0 plus c_2 has to equal to 0."},{"Start":"12:02.255 ","End":"12:06.965","Text":"Therefore, we get that c_2 is equal to 0."},{"Start":"12:06.965 ","End":"12:09.530","Text":"We can rub out our c_2 already."},{"Start":"12:09.530 ","End":"12:12.380","Text":"Now we want to calculate out c_1."},{"Start":"12:12.380 ","End":"12:17.075","Text":"Now we\u0027re going to use the fact that the potential is a continuous function."},{"Start":"12:17.075 ","End":"12:21.935","Text":"That means that at the point which is mutual to these 2 regions,"},{"Start":"12:21.935 ","End":"12:24.710","Text":"the function has to be equal."},{"Start":"12:24.710 ","End":"12:27.980","Text":"The point that\u0027s mutual to these 2 regions is at"},{"Start":"12:27.980 ","End":"12:33.860","Text":"R. That\u0027s right at the point that joins these 2 regions."},{"Start":"12:33.860 ","End":"12:36.635","Text":"In case you write on the surface of the sphere,"},{"Start":"12:36.635 ","End":"12:40.640","Text":"that means that negative Rho and then we\u0027re going to substitute in"},{"Start":"12:40.640 ","End":"12:45.290","Text":"this R negative Rho R squared divided by"},{"Start":"12:45.290 ","End":"12:49.790","Text":"6 Epsilon naught plus c_1 has to be equal"},{"Start":"12:49.790 ","End":"12:55.505","Text":"to when we substitute in R into this equation for this region."},{"Start":"12:55.505 ","End":"13:04.190","Text":"That will be Rho R^3 divided by 3 Epsilon naught and again"},{"Start":"13:04.190 ","End":"13:12.590","Text":"here R. Then we can cross this out and say that this is squared and cross this out."},{"Start":"13:12.590 ","End":"13:17.400","Text":"Now, all we have to do is isolate out c_1."},{"Start":"13:19.180 ","End":"13:22.385","Text":"We\u0027ll just do some quick algebra."},{"Start":"13:22.385 ","End":"13:28.340","Text":"We\u0027ll get that c_1 is equal to Rho R^2 divided by"},{"Start":"13:28.340 ","End":"13:36.755","Text":"3 Epsilon naught plus Rho R squared divided by 6 Epsilon naught."},{"Start":"13:36.755 ","End":"13:42.200","Text":"Then to get this common denominator will change this 3 into a 6,"},{"Start":"13:42.200 ","End":"13:46.205","Text":"which means that here we have to multiply by 2,"},{"Start":"13:46.205 ","End":"13:53.930","Text":"and then we have 3 Rho R^2 divided by 6 Epsilon naught,"},{"Start":"13:53.930 ","End":"14:01.745","Text":"which is simply equal to Rho R squared divided by 3 Epsilon naught."},{"Start":"14:01.745 ","End":"14:04.955","Text":"This is what c_1 is equal to."},{"Start":"14:04.955 ","End":"14:08.630","Text":"Then we just plug that in here and that\u0027s it."},{"Start":"14:08.630 ","End":"14:15.120","Text":"Here is our potential function and that is the end of our lesson."}],"ID":14210},{"Watched":false,"Name":"Third Method - Potential Difference","Duration":"14m 43s","ChapterTopicVideoID":12144,"CourseChapterTopicPlaylistID":157337,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12144.jpeg","UploadDate":"2018-06-28T04:50:58.8530000","DurationForVideoObject":"PT14M43S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.665","Text":"Hello. In this lesson we\u0027re going to be going over the third method"},{"Start":"00:04.665 ","End":"00:10.200","Text":"for finding the potential and that is by using the idea of potential difference."},{"Start":"00:10.200 ","End":"00:12.960","Text":"We saw the equation that that means that the potential at"},{"Start":"00:12.960 ","End":"00:15.990","Text":"A minus the potential at B is equal to the"},{"Start":"00:15.990 ","End":"00:19.065","Text":"integral from A to B of"},{"Start":"00:19.065 ","End":"00:22.490","Text":"E dot dr. We\u0027re going to go over this"},{"Start":"00:22.490 ","End":"00:26.370","Text":"by answering a question and then we will already see how to do it."},{"Start":"00:26.370 ","End":"00:33.710","Text":"We\u0027re being told that we have an infinite plane that has a charge per unit area of Sigma."},{"Start":"00:33.710 ","End":"00:37.430","Text":"Then we\u0027re being told that a distance D above the plane,"},{"Start":"00:37.430 ","End":"00:39.610","Text":"there\u0027s a conducting sphere,"},{"Start":"00:39.610 ","End":"00:45.110","Text":"so this is a conductor of radius r and charge Q and"},{"Start":"00:45.110 ","End":"00:50.630","Text":"we\u0027re being asked what is the potential difference between the plane and the sphere."},{"Start":"00:50.630 ","End":"00:56.570","Text":"If we say that on the plane we have this point A and on the sphere we have this point B,"},{"Start":"00:56.570 ","End":"01:00.965","Text":"so we know that we have a uniform charge"},{"Start":"01:00.965 ","End":"01:06.110","Text":"along the plane and along the sphere so we could choose from any point to any point"},{"Start":"01:06.110 ","End":"01:15.110","Text":"and therefore we can find the change in potential from A to B and that is simply equal to"},{"Start":"01:15.110 ","End":"01:20.285","Text":"the potential difference which is the potential at B minus the potential"},{"Start":"01:20.285 ","End":"01:26.090","Text":"at A and this is of course equal to V,"},{"Start":"01:26.090 ","End":"01:30.030","Text":"the voltage between A and B."},{"Start":"01:31.280 ","End":"01:37.835","Text":"What we\u0027re going to be doing is we are going to be integrating along this line,"},{"Start":"01:37.835 ","End":"01:39.905","Text":"joining A and B."},{"Start":"01:39.905 ","End":"01:45.180","Text":"Let\u0027s say that this line lies on the z axis."},{"Start":"01:46.160 ","End":"01:49.660","Text":"Let\u0027s choose some points along this line,"},{"Start":"01:49.660 ","End":"01:52.810","Text":"let\u0027s say over here and let\u0027s say that this is"},{"Start":"01:52.810 ","End":"01:58.590","Text":"a height of z away from the infinite plane."},{"Start":"01:58.590 ","End":"02:03.640","Text":"Now we want to know what the electric field is along here. We\u0027re at this point z."},{"Start":"02:03.640 ","End":"02:11.485","Text":"The electric field of an infinite plane is equal to Sigma divided by 2 Epsilon naught and"},{"Start":"02:11.485 ","End":"02:14.730","Text":"we can say that it\u0027s going in the positive z direction if Sigma is"},{"Start":"02:14.730 ","End":"02:20.210","Text":"positive and if Sigma is negative then it will be in the negatives, that direction."},{"Start":"02:21.580 ","End":"02:28.348","Text":"This is the E field due to the infinite plane and"},{"Start":"02:28.348 ","End":"02:34.670","Text":"now we\u0027re going to look at the E field at the same point z over here due to our sphere."},{"Start":"02:34.670 ","End":"02:38.370","Text":"That\u0027s going to be this point over here."},{"Start":"02:38.370 ","End":"02:43.520","Text":"We have the electric field coming all the way over here acting on this point."},{"Start":"02:43.520 ","End":"02:46.610","Text":"Then we can add it just onto here."},{"Start":"02:46.610 ","End":"02:52.490","Text":"The electric field of a sphere is similar to that of a point charge and that\u0027s just kQ,"},{"Start":"02:52.490 ","End":"02:58.890","Text":"where here Q is capital Q divided by I squared."},{"Start":"02:58.890 ","End":"03:03.800","Text":"Where I squared is this distance from"},{"Start":"03:03.800 ","End":"03:11.180","Text":"the center of the sphere all the way up until r point."},{"Start":"03:11.180 ","End":"03:15.935","Text":"This is r and then what we can see is"},{"Start":"03:15.935 ","End":"03:20.885","Text":"although the electric field coming out of the sphere is acting in a radial direction,"},{"Start":"03:20.885 ","End":"03:24.680","Text":"we can see that if we\u0027re speaking specifically about this point on"},{"Start":"03:24.680 ","End":"03:30.810","Text":"the plane then it\u0027s acting over here in the negative z direction."},{"Start":"03:31.850 ","End":"03:35.773","Text":"What is our I? Let\u0027s just write it over here."},{"Start":"03:35.773 ","End":"03:40.400","Text":"So r is equal to the total distance between the center of the sphere up until the plane"},{"Start":"03:40.400 ","End":"03:46.470","Text":"which is d minus this distance over here z."},{"Start":"03:46.470 ","End":"03:53.790","Text":"Then d minus z will leave us with this distance over here r. Great."},{"Start":"03:53.790 ","End":"03:59.010","Text":"Now we have our E field and now what we want to do is we want to find the potential."},{"Start":"03:59.010 ","End":"04:03.005","Text":"We\u0027re going to do this equation over here."},{"Start":"04:03.005 ","End":"04:04.895","Text":"Let\u0027s scroll down."},{"Start":"04:04.895 ","End":"04:10.365","Text":"We\u0027re trying to find the potential difference between B and A."},{"Start":"04:10.365 ","End":"04:16.985","Text":"The potential at B minus the potential at A which is the integral from A"},{"Start":"04:16.985 ","End":"04:25.330","Text":"until B of E dot dr. What is dr?"},{"Start":"04:25.330 ","End":"04:27.495","Text":"Let\u0027s just write it over here."},{"Start":"04:27.495 ","End":"04:31.290","Text":"So dr in general,"},{"Start":"04:31.290 ","End":"04:36.345","Text":"it\u0027s equal to dx in the x direction plus dy"},{"Start":"04:36.345 ","End":"04:42.780","Text":"in the y direction plus dz in the z direction."},{"Start":"04:43.000 ","End":"04:46.520","Text":"Now, our example is specifically we\u0027re only"},{"Start":"04:46.520 ","End":"04:49.670","Text":"speaking about the electric field along the z axis."},{"Start":"04:49.670 ","End":"04:53.585","Text":"We can see we only have it in the positive and negative z directions."},{"Start":"04:53.585 ","End":"04:59.120","Text":"That means that in our example over here we can cross out the x and y directions of"},{"Start":"04:59.120 ","End":"05:05.870","Text":"dr and we\u0027re left with dr vector is just dz in the z direction."},{"Start":"05:06.950 ","End":"05:10.290","Text":"Now we can plug this in."},{"Start":"05:10.290 ","End":"05:15.925","Text":"We\u0027re going to have this integral and then we have our E field."},{"Start":"05:15.925 ","End":"05:21.220","Text":"That is Sigma divided by 2 Epsilon naught minus,"},{"Start":"05:21.220 ","End":"05:22.570","Text":"we\u0027ll take the minus over here,"},{"Start":"05:22.570 ","End":"05:27.430","Text":"minus kQ divided by r squared where r is"},{"Start":"05:27.430 ","End":"05:33.810","Text":"d minus z and all of this is dz."},{"Start":"05:33.810 ","End":"05:36.130","Text":"When I took the dot product between"},{"Start":"05:36.130 ","End":"05:40.165","Text":"this z hat and this z hat over here, this crossed out."},{"Start":"05:40.165 ","End":"05:45.520","Text":"It just turned into 1 and between this z hat and this z hat it also turned into 1,"},{"Start":"05:45.520 ","End":"05:49.504","Text":"the dot-product, but I\u0027m still left with this minus over here."},{"Start":"05:49.504 ","End":"05:52.980","Text":"Now what\u0027s left is to plug in our bounds."},{"Start":"05:52.980 ","End":"06:00.115","Text":"A is located at z is equal to 0 and B is located over here."},{"Start":"06:00.115 ","End":"06:03.370","Text":"We see that this distance d is up until the center of"},{"Start":"06:03.370 ","End":"06:06.910","Text":"the sphere but B is at the edge of the sphere."},{"Start":"06:06.910 ","End":"06:11.680","Text":"The sphere has a radius r. We\u0027re just going to say that that is at"},{"Start":"06:11.680 ","End":"06:16.870","Text":"d minus r. If we take a d minus r,"},{"Start":"06:16.870 ","End":"06:20.140","Text":"the radius of the sphere we\u0027re located at point B."},{"Start":"06:20.140 ","End":"06:24.885","Text":"Now all we have to do is we have to do this integration."},{"Start":"06:24.885 ","End":"06:27.100","Text":"First of all, let\u0027s do this first one,"},{"Start":"06:27.100 ","End":"06:28.705","Text":"let\u0027s split it up into 2."},{"Start":"06:28.705 ","End":"06:30.610","Text":"Let\u0027s integrate here d z."},{"Start":"06:30.610 ","End":"06:34.720","Text":"Here we don\u0027t have a z so just going to integrate."},{"Start":"06:34.720 ","End":"06:36.100","Text":"We have a constant,"},{"Start":"06:36.100 ","End":"06:40.850","Text":"so Sigma divided by 2 Epsilon naught multiplied by z and then"},{"Start":"06:40.850 ","End":"06:45.650","Text":"we substitute in our bounds d minus r so that we\u0027ll do in a"},{"Start":"06:45.650 ","End":"06:52.475","Text":"second and then we have to do this integration from 0 to d minus r"},{"Start":"06:52.475 ","End":"07:02.220","Text":"of kQ divided by d minus z squared d z."},{"Start":"07:03.480 ","End":"07:09.910","Text":"Let\u0027s just scroll down a little bit and carry this on over here."},{"Start":"07:09.910 ","End":"07:15.040","Text":"Here we\u0027ll have Sigma divided by 2 Episilon naught,"},{"Start":"07:15.040 ","End":"07:19.465","Text":"and when we substitute in d minus R minus 0."},{"Start":"07:19.465 ","End":"07:21.985","Text":"We have the first bound as d minus R,"},{"Start":"07:21.985 ","End":"07:24.445","Text":"and then we minus the second bound."},{"Start":"07:24.445 ","End":"07:26.725","Text":"Then we have negative."},{"Start":"07:26.725 ","End":"07:31.750","Text":"The integral of kQ divided by d minus"},{"Start":"07:31.750 ","End":"07:37.720","Text":"z^2 is going to be kQ divided by d minus z."},{"Start":"07:37.720 ","End":"07:44.335","Text":"Then we have to take out a minus because this is like 1 divided by x."},{"Start":"07:44.335 ","End":"07:47.755","Text":"Then we multiply this by minus 1."},{"Start":"07:47.755 ","End":"07:53.635","Text":"But then we also have to multiply it by another minus 1 or negative 1,"},{"Start":"07:53.635 ","End":"07:57.535","Text":"because we have a negative over here before the z."},{"Start":"07:57.535 ","End":"08:00.490","Text":"The z has a negative coefficient,"},{"Start":"08:00.490 ","End":"08:03.220","Text":"we multiply it by another minus 1."},{"Start":"08:03.220 ","End":"08:06.895","Text":"Then we have negative 1 multiplied by negative 1 which is positive."},{"Start":"08:06.895 ","End":"08:09.370","Text":"But then we had this negative over here,"},{"Start":"08:09.370 ","End":"08:12.055","Text":"its still will turn out negative."},{"Start":"08:12.055 ","End":"08:15.650","Text":"We can rub all of this out."},{"Start":"08:16.110 ","End":"08:25.285","Text":"Here, we also put in the bounds of 0 to d minus R. Then if we do that,"},{"Start":"08:25.285 ","End":"08:31.960","Text":"we\u0027ll have kQ divided by d minus z,"},{"Start":"08:31.960 ","End":"08:37.375","Text":"which is d minus R. Then"},{"Start":"08:37.375 ","End":"08:46.520","Text":"minus kQ divided by d minus z, which is 0."},{"Start":"08:49.830 ","End":"08:54.835","Text":"Now we\u0027re going to plug that in over here."},{"Start":"08:54.835 ","End":"09:02.050","Text":"We\u0027ll have Sigma divided by 2 Episilon naught multiplied by d minus R,"},{"Start":"09:02.050 ","End":"09:06.925","Text":"and then minus this over here."},{"Start":"09:06.925 ","End":"09:10.210","Text":"I can cross this out,"},{"Start":"09:10.210 ","End":"09:11.650","Text":"minus comes from here."},{"Start":"09:11.650 ","End":"09:19.988","Text":"We have minus kQ divided by d minus d minus minus R,"},{"Start":"09:19.988 ","End":"09:24.750","Text":"divided by R. Then we have the minus and this minus,"},{"Start":"09:24.750 ","End":"09:34.364","Text":"that becomes a plus kQ divided by d. This is the answer,"},{"Start":"09:34.364 ","End":"09:38.595","Text":"and this is the difference between our potentials,"},{"Start":"09:38.595 ","End":"09:40.170","Text":"or our potential difference,"},{"Start":"09:40.170 ","End":"09:44.680","Text":"or the voltage between A and B."},{"Start":"09:45.870 ","End":"09:52.735","Text":"Now what I\u0027m going to do is I\u0027m going to show you another way of solving this question."},{"Start":"09:52.735 ","End":"09:56.080","Text":"That involves finding the potential at A,"},{"Start":"09:56.080 ","End":"09:57.940","Text":"finding the potential at B,"},{"Start":"09:57.940 ","End":"10:00.830","Text":"and then subtracting them."},{"Start":"10:00.930 ","End":"10:03.025","Text":"How I\u0027m going to do that?"},{"Start":"10:03.025 ","End":"10:08.350","Text":"I\u0027m going to find the potential at A. I\u0027m going to use superposition."},{"Start":"10:08.350 ","End":"10:14.695","Text":"I\u0027m going to superimpose the potential at A due to the charges on the infinite plane."},{"Start":"10:14.695 ","End":"10:21.235","Text":"I\u0027m going to add to that the potential at A due to the electric charges on the sphere."},{"Start":"10:21.235 ","End":"10:27.085","Text":"The potential from a sphere is like the potential of a point charge."},{"Start":"10:27.085 ","End":"10:31.240","Text":"That is equal to k multiplied by the charge,"},{"Start":"10:31.240 ","End":"10:32.515","Text":"which is Q,"},{"Start":"10:32.515 ","End":"10:38.995","Text":"divided by r, where r is the distance from the center of the sphere."},{"Start":"10:38.995 ","End":"10:43.360","Text":"Of course, this is in the region where we\u0027re outside of the sphere,"},{"Start":"10:43.360 ","End":"10:50.290","Text":"so r is bigger than R. If we were located within the sphere,"},{"Start":"10:50.290 ","End":"10:54.980","Text":"then of course the potential would be 0 because it\u0027s a conducting sphere."},{"Start":"10:55.320 ","End":"11:02.050","Text":"Therefore, if we want to work out the potential at our point B,"},{"Start":"11:02.050 ","End":"11:07.600","Text":"point B is located at a distance R from the center,"},{"Start":"11:07.600 ","End":"11:10.780","Text":"because it\u0027s right on the surface of the sphere and the sphere has a radius"},{"Start":"11:10.780 ","End":"11:15.640","Text":"r. It\u0027s a distance r away from the center of the sphere."},{"Start":"11:15.640 ","End":"11:24.160","Text":"We can write this out as kQ divided by R. Then the potential at"},{"Start":"11:24.160 ","End":"11:28.990","Text":"A is simply going to be equal to"},{"Start":"11:28.990 ","End":"11:36.730","Text":"kQ divided by the distance from the center of the sphere up until point A,"},{"Start":"11:36.730 ","End":"11:38.725","Text":"which is a distance of d away,"},{"Start":"11:38.725 ","End":"11:45.260","Text":"because D begins from the center of the sphere all the way to the infinite plane."},{"Start":"11:46.860 ","End":"11:53.840","Text":"These are the potentials from the sphere."},{"Start":"11:54.600 ","End":"12:02.930","Text":"Now, let\u0027s deal with the potentials from the plane at these 2 points."},{"Start":"12:03.150 ","End":"12:07.645","Text":"This is the equation for the potential of a sphere in general."},{"Start":"12:07.645 ","End":"12:10.390","Text":"We can just rub this out because I wrote it here."},{"Start":"12:10.390 ","End":"12:18.775","Text":"The potential for a plane in general is equal to negative Sigma,"},{"Start":"12:18.775 ","End":"12:21.350","Text":"this is an infinite plane."},{"Start":"12:21.540 ","End":"12:25.870","Text":"Negative Sigma, the charge density per unit area divided by"},{"Start":"12:25.870 ","End":"12:30.835","Text":"2 Episilon naught multiplied by z."},{"Start":"12:30.835 ","End":"12:34.795","Text":"This is the potential of an infinite plane."},{"Start":"12:34.795 ","End":"12:40.880","Text":"Therefore, the potential at point B due to the plane"},{"Start":"12:41.100 ","End":"12:49.645","Text":"is simply going to be equal to negative Sigma divided by 2 Episilon naught."},{"Start":"12:49.645 ","End":"12:53.395","Text":"Point B is this distance over here."},{"Start":"12:53.395 ","End":"13:01.840","Text":"We go up d till the center of the sphere and then minus this R,"},{"Start":"13:01.840 ","End":"13:06.010","Text":"the radius of the sphere in order to get to the surface of the sphere."},{"Start":"13:06.010 ","End":"13:14.510","Text":"We just put d minus R over here and the potential at point A."},{"Start":"13:14.850 ","End":"13:17.740","Text":"Then we\u0027ll have negative Sigma divided by"},{"Start":"13:17.740 ","End":"13:20.710","Text":"2 Episilon naught multiplied by its z coordinate."},{"Start":"13:20.710 ","End":"13:22.530","Text":"We can see that point A,"},{"Start":"13:22.530 ","End":"13:24.805","Text":"Z coordinate is at 0,"},{"Start":"13:24.805 ","End":"13:31.430","Text":"the potential at A due to the infinite plane is going to be equal to 0."},{"Start":"13:31.980 ","End":"13:36.024","Text":"Now in order to find the potential difference,"},{"Start":"13:36.024 ","End":"13:39.760","Text":"I\u0027m going to superimpose all of these values."},{"Start":"13:39.760 ","End":"13:47.290","Text":"Our potential difference is delta Phi from B to A,"},{"Start":"13:47.290 ","End":"13:51.820","Text":"and this is equal to the potential at B."},{"Start":"13:51.820 ","End":"13:54.700","Text":"We have the potential due to the infinite plane,"},{"Start":"13:54.700 ","End":"13:56.485","Text":"which is negative Sigma."},{"Start":"13:56.485 ","End":"14:01.480","Text":"This is Sigma divided by 2 Episilon naught multiplied by d"},{"Start":"14:01.480 ","End":"14:07.960","Text":"minus R. We also have this potential from the sphere,"},{"Start":"14:07.960 ","End":"14:14.950","Text":"plus kQ divided by R. Then we minus the potential at A."},{"Start":"14:14.950 ","End":"14:21.580","Text":"The potential at A from the infinite plane is equal to 0 plus the potential A due to"},{"Start":"14:21.580 ","End":"14:29.065","Text":"the sphere which is equal to kQ divided by d. This is the final answer."},{"Start":"14:29.065 ","End":"14:35.885","Text":"If you look at the answer that we got for the previous method earlier on in the video,"},{"Start":"14:35.885 ","End":"14:40.835","Text":"you\u0027ll see that we get the exact same answer for the potential difference."},{"Start":"14:40.835 ","End":"14:43.860","Text":"That\u0027s the end of this lesson."}],"ID":14211},{"Watched":false,"Name":"Exercise 4","Duration":"7m 49s","ChapterTopicVideoID":12145,"CourseChapterTopicPlaylistID":157337,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12145.jpeg","UploadDate":"2018-06-28T04:52:33.2470000","DurationForVideoObject":"PT7M49S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.920","Text":"Hello. In this lesson,"},{"Start":"00:01.920 ","End":"00:04.380","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.380 ","End":"00:09.300","Text":"We\u0027re being told to calculate the potential difference between these 2 planes,"},{"Start":"00:09.300 ","End":"00:12.810","Text":"where 1 plane has uniform charge density of"},{"Start":"00:12.810 ","End":"00:17.730","Text":"Sigma and the other has a uniform charge density of negative Sigma."},{"Start":"00:17.730 ","End":"00:21.240","Text":"The planes are a distance of d from one another and"},{"Start":"00:21.240 ","End":"00:26.250","Text":"their area A is significantly larger than the distance between them."},{"Start":"00:26.250 ","End":"00:28.680","Text":"A is much, much larger than d. Therefore,"},{"Start":"00:28.680 ","End":"00:32.590","Text":"we can consider them as infinite planes."},{"Start":"00:33.110 ","End":"00:37.470","Text":"What we\u0027re doing is we\u0027re finding the potential difference."},{"Start":"00:37.470 ","End":"00:39.600","Text":"The first thing that we\u0027re going to do is we\u0027re going to find"},{"Start":"00:39.600 ","End":"00:44.295","Text":"the electric field between the 2 planes."},{"Start":"00:44.295 ","End":"00:46.616","Text":"This is what we\u0027re going to calculate."},{"Start":"00:46.616 ","End":"00:50.750","Text":"Then we\u0027re going to integrate along this electric field between"},{"Start":"00:50.750 ","End":"00:56.540","Text":"these 2 points over here and then we will find the potential difference."},{"Start":"00:56.540 ","End":"00:59.420","Text":"What we\u0027ll find is the potential at this point."},{"Start":"00:59.420 ","End":"01:01.820","Text":"Let\u0027s say it\u0027s A over here."},{"Start":"01:01.820 ","End":"01:03.650","Text":"The potential at this point,"},{"Start":"01:03.650 ","End":"01:05.510","Text":"let\u0027s say it\u0027s a B over here."},{"Start":"01:05.510 ","End":"01:08.990","Text":"We\u0027ll integrate from A to B along"},{"Start":"01:08.990 ","End":"01:13.430","Text":"this line and then we will get the potential difference."},{"Start":"01:13.430 ","End":"01:21.900","Text":"Let\u0027s define that this is the positive y-direction and this is the positive x-direction."},{"Start":"01:21.940 ","End":"01:25.280","Text":"As we said, because the area of"},{"Start":"01:25.280 ","End":"01:28.670","Text":"each plane is significantly larger than the distance between them,"},{"Start":"01:28.670 ","End":"01:32.855","Text":"we can say that the 2 planes are both infinite."},{"Start":"01:32.855 ","End":"01:35.720","Text":"We know that the electric field of"},{"Start":"01:35.720 ","End":"01:40.010","Text":"an infinite plane is equal to if we\u0027re taking Sigma to be positive,"},{"Start":"01:40.010 ","End":"01:43.830","Text":"it\u0027s going to be Sigma divided by 2 Epsilon naught,"},{"Start":"01:43.830 ","End":"01:46.730","Text":"and then here it\u0027s going to be in the upwards direction,"},{"Start":"01:46.730 ","End":"01:50.370","Text":"which is the positive y-direction as we defined it."},{"Start":"01:50.690 ","End":"01:53.340","Text":"This is for the positive Sigma."},{"Start":"01:53.340 ","End":"01:55.745","Text":"We can draw that over here."},{"Start":"01:55.745 ","End":"01:59.434","Text":"For the positively charged planes,"},{"Start":"01:59.434 ","End":"02:04.525","Text":"we\u0027re going to have the electric field coming out of it, like so."},{"Start":"02:04.525 ","End":"02:07.525","Text":"This is the E field."},{"Start":"02:07.525 ","End":"02:12.670","Text":"Now, of course, this region below the plane doesn\u0027t really interest us."},{"Start":"02:12.670 ","End":"02:14.260","Text":"It\u0027s in the negative y-direction,"},{"Start":"02:14.260 ","End":"02:17.020","Text":"but it also doesn\u0027t interest us because we\u0027re trying to find"},{"Start":"02:17.020 ","End":"02:20.025","Text":"the electric field between these 2 points,"},{"Start":"02:20.025 ","End":"02:22.755","Text":"that means between the 2 planes."},{"Start":"02:22.755 ","End":"02:29.110","Text":"Now we also have the other plane which has a charge density of negative Sigma."},{"Start":"02:29.110 ","End":"02:33.520","Text":"That means that above the plane,"},{"Start":"02:33.520 ","End":"02:38.770","Text":"we\u0027re going to have an electric field in this direction into the plane."},{"Start":"02:38.770 ","End":"02:43.990","Text":"Below the plane, we\u0027re also going to have an electric field in the plane."},{"Start":"02:43.990 ","End":"02:49.925","Text":"However, it\u0027s going to be here when we\u0027re looking between the 2 planes."},{"Start":"02:49.925 ","End":"02:52.760","Text":"This electric field is in the positive y-direction"},{"Start":"02:52.760 ","End":"02:56.750","Text":"and above the plane which is in the region that doesn\u0027t interest us,"},{"Start":"02:56.750 ","End":"02:59.300","Text":"it\u0027s in the negative y-direction, but this doesn\u0027t matter."},{"Start":"02:59.300 ","End":"03:02.555","Text":"But what does matter is that we can see that we have"},{"Start":"03:02.555 ","End":"03:07.940","Text":"this green arrow representing the electric field from this plane,"},{"Start":"03:07.940 ","End":"03:11.450","Text":"the bottom plane in the positive y-direction."},{"Start":"03:11.450 ","End":"03:17.885","Text":"We have this red arrow representing the electric field from the negatively charged plane,"},{"Start":"03:17.885 ","End":"03:21.860","Text":"where the top line also in positive y-direction."},{"Start":"03:21.860 ","End":"03:25.895","Text":"These 2 arrows don\u0027t interest us."},{"Start":"03:25.895 ","End":"03:27.440","Text":"Because that\u0027s not in the region."},{"Start":"03:27.440 ","End":"03:32.255","Text":"But what we can see is that we have 2 times the electric field, and obviously,"},{"Start":"03:32.255 ","End":"03:35.540","Text":"it\u0027s the same electric fields because they\u0027re both"},{"Start":"03:35.540 ","End":"03:39.433","Text":"have a charge of Sigma or charge density of Sigma,"},{"Start":"03:39.433 ","End":"03:41.590","Text":"1 is positive and 1 is negative."},{"Start":"03:41.590 ","End":"03:43.940","Text":"But what\u0027s important is that"},{"Start":"03:43.940 ","End":"03:50.330","Text":"the total electric field between the 2 planes is simply equal"},{"Start":"03:50.330 ","End":"04:00.665","Text":"to 2 times the absolute value of the electric field in the y-direction."},{"Start":"04:00.665 ","End":"04:05.660","Text":"That\u0027s just going to be Sigma divided by 2 Epsilon naught times 2."},{"Start":"04:05.660 ","End":"04:10.295","Text":"Sigma divided by Epsilon naught in the y-direction."},{"Start":"04:10.295 ","End":"04:14.905","Text":"This is the total electric field between these 2 planes."},{"Start":"04:14.905 ","End":"04:17.630","Text":"Now what we\u0027re going to do is we\u0027re going to find"},{"Start":"04:17.630 ","End":"04:20.240","Text":"the potential difference between these 2 points,"},{"Start":"04:20.240 ","End":"04:21.950","Text":"or in other words, the voltage."},{"Start":"04:21.950 ","End":"04:29.385","Text":"We\u0027re doing an integral between from point A to point B of E.dr vector."},{"Start":"04:29.385 ","End":"04:32.120","Text":"I\u0027m just going to write over here what dr vector is."},{"Start":"04:32.120 ","End":"04:38.615","Text":"It\u0027s dx in the x-direction plus dy in the y-direction,"},{"Start":"04:38.615 ","End":"04:42.470","Text":"plus dz in the z-direction."},{"Start":"04:42.470 ","End":"04:47.570","Text":"What we can see here is that the only direction that is relevant is the y-direction."},{"Start":"04:47.570 ","End":"04:50.330","Text":"We\u0027re integrating along this line which goes up"},{"Start":"04:50.330 ","End":"04:54.065","Text":"the y-axis and our E fields are in the y-direction."},{"Start":"04:54.065 ","End":"04:57.230","Text":"We can cross out this and this."},{"Start":"04:57.230 ","End":"05:01.685","Text":"Then when we do the dot product between E,"},{"Start":"05:01.685 ","End":"05:04.415","Text":"which has this y hat vector, and our dr,"},{"Start":"05:04.415 ","End":"05:07.850","Text":"which also has this y hat vector,"},{"Start":"05:07.850 ","End":"05:14.315","Text":"the dot product between y vector and y vector or y hat and y hat is simply 1."},{"Start":"05:14.315 ","End":"05:18.305","Text":"We\u0027re going to be integrating along the electric field,"},{"Start":"05:18.305 ","End":"05:23.113","Text":"which is Sigma divided by Epsilon naught,"},{"Start":"05:23.113 ","End":"05:24.530","Text":"and then just dr,"},{"Start":"05:24.530 ","End":"05:28.160","Text":"which over here specifically is dy."},{"Start":"05:28.160 ","End":"05:30.495","Text":"Now we have to set up our bounds."},{"Start":"05:30.495 ","End":"05:36.620","Text":"We\u0027re going from A to B and our point A is at a height of 0."},{"Start":"05:36.620 ","End":"05:42.675","Text":"Our point B is a distance d away from point A."},{"Start":"05:42.675 ","End":"05:45.660","Text":"We can see it\u0027s just this over here,"},{"Start":"05:45.660 ","End":"05:49.620","Text":"d. Now, this is a very easy integral."},{"Start":"05:49.620 ","End":"05:53.855","Text":"We have Sigma divided by Epsilon naught multiplied by"},{"Start":"05:53.855 ","End":"05:59.435","Text":"y between the bounds of 0 to d. This is just simply Sigma"},{"Start":"05:59.435 ","End":"06:04.280","Text":"divided by Epsilon naught multiplied by d. This is"},{"Start":"06:04.280 ","End":"06:12.015","Text":"the potential difference or the voltage between these 2 plates or between these 2 planes."},{"Start":"06:12.015 ","End":"06:16.535","Text":"What do we can notice over here is that we get that our voltage"},{"Start":"06:16.535 ","End":"06:21.065","Text":"or potential difference is equal to Sigma divided by Epsilon naught,"},{"Start":"06:21.065 ","End":"06:26.605","Text":"which is also what we saw is the total electric field between the 2 plates."},{"Start":"06:26.605 ","End":"06:31.155","Text":"This was the electric field without the direction."},{"Start":"06:31.155 ","End":"06:35.300","Text":"We\u0027re just taking the magnitude of the electric field multiplied by"},{"Start":"06:35.300 ","End":"06:40.990","Text":"d. d is the distance between these 2 plates or these infinite planes."},{"Start":"06:40.990 ","End":"06:45.440","Text":"What we can see is that the voltage or the potential difference is always"},{"Start":"06:45.440 ","End":"06:49.619","Text":"going to be equal to the electric field."},{"Start":"06:49.619 ","End":"06:54.650","Text":"Just the magnitude of the electric field multiplied by d,"},{"Start":"06:54.650 ","End":"06:59.580","Text":"where d represents the distance between the 2 infinite plates."},{"Start":"07:00.100 ","End":"07:08.110","Text":"In other words, we can say that the voltage is equal to the integral on E, dy."},{"Start":"07:08.110 ","End":"07:10.925","Text":"But we can see that the electric field is"},{"Start":"07:10.925 ","End":"07:13.550","Text":"always going to be constant between these 2 plates."},{"Start":"07:13.550 ","End":"07:18.380","Text":"We can take the E-field out from the integration sign."},{"Start":"07:18.380 ","End":"07:23.815","Text":"Then dy we\u0027re obviously just integrating along the distance between the 2 plates."},{"Start":"07:23.815 ","End":"07:27.440","Text":"Then we\u0027re just going to get that this is Ed."},{"Start":"07:27.440 ","End":"07:30.575","Text":"This is a very useful equation to remember,"},{"Start":"07:30.575 ","End":"07:36.050","Text":"but we can only use it when the electric field is constant throughout,"},{"Start":"07:36.050 ","End":"07:38.230","Text":"when it isn\u0027t changing."},{"Start":"07:38.230 ","End":"07:41.210","Text":"Of course, it represents the magnitude of the electric field and"},{"Start":"07:41.210 ","End":"07:44.875","Text":"that d is the distance between the 2 plates."},{"Start":"07:44.875 ","End":"07:50.010","Text":"This is the answer to the question and that is the end of the lesson."}],"ID":14212}],"Thumbnail":null,"ID":157337}]

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