Method Of Images
0/14 completed

{"Free":0,"Sample":1,"Paid":2}

[{"Name":"Method Of Images","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Method Of Images And Picard_s Existence Theorem","Duration":"18m 15s","ChapterTopicVideoID":12110,"CourseChapterTopicPlaylistID":161612,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12110.jpeg","UploadDate":"2018-06-28T03:20:08.6300000","DurationForVideoObject":"PT18M15S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.980","Text":"Hello. In this lesson,"},{"Start":"00:01.980 ","End":"00:08.775","Text":"we\u0027re going to learn about the method of images and of Picard\u0027s existence theorem."},{"Start":"00:08.775 ","End":"00:12.495","Text":"The method of images is a way of finding"},{"Start":"00:12.495 ","End":"00:19.750","Text":"a potential when the other methods for finding the potential are impossible to use."},{"Start":"00:20.510 ","End":"00:26.985","Text":"The way I\u0027m going to explain the method of images is by explaining it via an example."},{"Start":"00:26.985 ","End":"00:34.890","Text":"This is a classic example of how the method of images can be used to find the potential."},{"Start":"00:34.890 ","End":"00:40.845","Text":"In front of us, we have a conducting infinite plane."},{"Start":"00:40.845 ","End":"00:43.230","Text":"It\u0027s infinite, obviously,"},{"Start":"00:43.230 ","End":"00:45.570","Text":"in the xy-plane,"},{"Start":"00:45.570 ","End":"00:48.880","Text":"and it\u0027s also grounded."},{"Start":"00:49.880 ","End":"00:58.075","Text":"The first thing that we have to do is we set up our axes somewhere in the infinite plane."},{"Start":"00:58.075 ","End":"01:02.910","Text":"Then we place a point charge, q."},{"Start":"01:02.910 ","End":"01:04.990","Text":"Let\u0027s say we know the value of q,"},{"Start":"01:04.990 ","End":"01:06.965","Text":"and we place it,"},{"Start":"01:06.965 ","End":"01:11.660","Text":"a value of z, in the z direction."},{"Start":"01:11.660 ","End":"01:16.955","Text":"Now that we have our point charge placed at height z,"},{"Start":"01:16.955 ","End":"01:19.100","Text":"what we want to know is,"},{"Start":"01:19.100 ","End":"01:26.645","Text":"what is our potential in the region where is z is larger than 0?"},{"Start":"01:26.645 ","End":"01:29.836","Text":"This is the question. This is what we\u0027re trying to find,"},{"Start":"01:29.836 ","End":"01:36.350","Text":"so that means in the entire region of space which is above our infinite plane."},{"Start":"01:36.350 ","End":"01:38.600","Text":"Now, below our infinite plane,"},{"Start":"01:38.600 ","End":"01:41.450","Text":"we\u0027re going to have 0 potential because we always"},{"Start":"01:41.450 ","End":"01:44.555","Text":"know that the potential at infinity is equal to 0,"},{"Start":"01:44.555 ","End":"01:48.635","Text":"and because our plane has been grounded,"},{"Start":"01:48.635 ","End":"01:53.030","Text":"so we know that the potential at the plane is also going to be equal to 0,"},{"Start":"01:53.030 ","End":"01:56.000","Text":"and then we can see that there\u0027s no potential difference"},{"Start":"01:56.000 ","End":"01:59.615","Text":"between infinity and on the plane,"},{"Start":"01:59.615 ","End":"02:04.800","Text":"which means that there\u0027s going to be 0 potential below the plane."},{"Start":"02:05.210 ","End":"02:10.340","Text":"We\u0027re looking to find the potential above our infinite plane, and also,"},{"Start":"02:10.340 ","End":"02:11.780","Text":"once we find our potential,"},{"Start":"02:11.780 ","End":"02:13.760","Text":"then we can take the derivative of it,"},{"Start":"02:13.760 ","End":"02:17.250","Text":"and we can also find the electric field."},{"Start":"02:17.260 ","End":"02:22.760","Text":"Now, a lot of people think that we can find the potential by working out"},{"Start":"02:22.760 ","End":"02:28.520","Text":"the potential of a point charge and the potential of a conducting infinite plane."},{"Start":"02:28.520 ","End":"02:32.090","Text":"Now, we can\u0027t do that because due to"},{"Start":"02:32.090 ","End":"02:36.140","Text":"this point charge over here and because this is a conducting plane,"},{"Start":"02:36.140 ","End":"02:40.370","Text":"the charge carriers on the plane with the electrons"},{"Start":"02:40.370 ","End":"02:46.105","Text":"are going to be attracted towards this point over here."},{"Start":"02:46.105 ","End":"02:50.870","Text":"That means that the charge density on"},{"Start":"02:50.870 ","End":"02:56.045","Text":"the infinite conducting plane is not going to be uniform."},{"Start":"02:56.045 ","End":"03:00.910","Text":"There\u0027ll be a greater concentration of charge density over here."},{"Start":"03:00.910 ","End":"03:07.325","Text":"That means that we\u0027re not going to have symmetry in our xy-plane anymore."},{"Start":"03:07.325 ","End":"03:09.710","Text":"Before, our point charge was round,"},{"Start":"03:09.710 ","End":"03:15.065","Text":"we could shift our infinite plane in any direction on the xy-plane,"},{"Start":"03:15.065 ","End":"03:16.310","Text":"and we would have symmetry."},{"Start":"03:16.310 ","End":"03:18.200","Text":"It would look like the exact same problem."},{"Start":"03:18.200 ","End":"03:21.140","Text":"But now that we\u0027ve brought this point charge over here,"},{"Start":"03:21.140 ","End":"03:27.630","Text":"and now that we have a concentration of charge carriers over here on the conductor,"},{"Start":"03:27.850 ","End":"03:30.920","Text":"we don\u0027t have a case of symmetry anymore."},{"Start":"03:30.920 ","End":"03:34.160","Text":"That\u0027s why we need to use this method of images"},{"Start":"03:34.160 ","End":"03:38.970","Text":"in order to calculate the potential because there\u0027s no other way to do it."},{"Start":"03:40.430 ","End":"03:42.680","Text":"Due to this point charge,"},{"Start":"03:42.680 ","End":"03:49.235","Text":"the charge carriers on the conductor move and are attracted to this area over here,"},{"Start":"03:49.235 ","End":"03:50.570","Text":"which means that our Sigma,"},{"Start":"03:50.570 ","End":"03:56.390","Text":"our charge density per unit area is going to be non-uniform."},{"Start":"03:56.390 ","End":"03:59.810","Text":"That means that the E-field due to an infinite plane,"},{"Start":"03:59.810 ","End":"04:05.030","Text":"which we know as being equal to Sigma divided by 2Epsilon_0 isn\u0027t"},{"Start":"04:05.030 ","End":"04:11.175","Text":"going to work here because our Sigma is non-uniform."},{"Start":"04:11.175 ","End":"04:15.542","Text":"What we\u0027re going to see, afterwards we\u0027ll find out exactly what our Sigma is,"},{"Start":"04:15.542 ","End":"04:20.850","Text":"but we\u0027re going to have rings of charge density."},{"Start":"04:22.880 ","End":"04:29.330","Text":"To sum up, because we don\u0027t know our charge distribution because it\u0027s non-uniform,"},{"Start":"04:29.330 ","End":"04:35.720","Text":"so we cannot use Gauss or Coulomb\u0027s law in order to find the electric field."},{"Start":"04:36.170 ","End":"04:40.340","Text":"In order to solve problems exactly like these,"},{"Start":"04:40.340 ","End":"04:45.020","Text":"where Sigma is non-uniform and we don\u0027t know it or just"},{"Start":"04:45.020 ","End":"04:50.509","Text":"anytime you see some conductor which is grounded and then some other charge,"},{"Start":"04:50.509 ","End":"04:55.910","Text":"the ways to find the potential is by using the method of images."},{"Start":"04:55.910 ","End":"04:58.500","Text":"Now let\u0027s talk about it."},{"Start":"04:59.150 ","End":"05:03.845","Text":"The way we solve this type of question is by playing a game,"},{"Start":"05:03.845 ","End":"05:06.800","Text":"and this game has 2 rules."},{"Start":"05:06.800 ","End":"05:13.505","Text":"The rules in general are stating that what we have to do is we have to take this problem"},{"Start":"05:13.505 ","End":"05:20.510","Text":"and form a very similar problem that has the same conditions."},{"Start":"05:20.510 ","End":"05:23.017","Text":"However, it\u0027s built in a different way,"},{"Start":"05:23.017 ","End":"05:25.670","Text":"and via this different way,"},{"Start":"05:25.670 ","End":"05:29.610","Text":"we can solve this question over here."},{"Start":"05:29.780 ","End":"05:34.535","Text":"The first rule of the game states that"},{"Start":"05:34.535 ","End":"05:39.650","Text":"in the area where I\u0027m trying to find my potential, so specifically here,"},{"Start":"05:39.650 ","End":"05:43.490","Text":"in the region where z is larger than 0,"},{"Start":"05:43.490 ","End":"05:50.400","Text":"I have to keep everything to do with my charge distribution the same,"},{"Start":"05:50.400 ","End":"05:51.900","Text":"the exact same thing."},{"Start":"05:51.900 ","End":"05:54.470","Text":"So if my problem started with this point charge, q,"},{"Start":"05:54.470 ","End":"05:58.980","Text":"located z above the grounded conducting plane,"},{"Start":"05:58.980 ","End":"06:02.990","Text":"so that has to be the only charge that I have,"},{"Start":"06:02.990 ","End":"06:04.805","Text":"and it cannot move."},{"Start":"06:04.805 ","End":"06:07.000","Text":"That stays exactly the same."},{"Start":"06:07.000 ","End":"06:10.220","Text":"That\u0027s the only thing that needs to stay exactly the same."},{"Start":"06:10.220 ","End":"06:13.530","Text":"Every other region in space,"},{"Start":"06:13.530 ","End":"06:19.189","Text":"so including on the surface of my conductor and below my conductor,"},{"Start":"06:19.189 ","End":"06:22.910","Text":"I can change the charge distribution however I"},{"Start":"06:22.910 ","End":"06:29.165","Text":"want in any way that can make it easier for me to solve the question."},{"Start":"06:29.165 ","End":"06:32.690","Text":"So I can add point charges here, here, here."},{"Start":"06:32.690 ","End":"06:34.760","Text":"I can add a wire,"},{"Start":"06:34.760 ","End":"06:36.800","Text":"another wire, it doesn\u0027t matter."},{"Start":"06:36.800 ","End":"06:41.010","Text":"I can do whatever I want and also on the surface of the conductor,"},{"Start":"06:41.010 ","End":"06:42.400","Text":"and that\u0027s allowed,"},{"Start":"06:42.400 ","End":"06:47.630","Text":"as long as I\u0027m not in the region where I\u0027m trying to find my potential."},{"Start":"06:47.630 ","End":"06:50.195","Text":"That\u0027s Rule Number 1."},{"Start":"06:50.195 ","End":"06:52.940","Text":"Rule Number 2 is that,"},{"Start":"06:52.940 ","End":"06:56.375","Text":"as we can see here on our plane, because it\u0027s grounded,"},{"Start":"06:56.375 ","End":"07:04.585","Text":"I know that the potential on the plane or the potential at z=0 is equal to 0."},{"Start":"07:04.585 ","End":"07:10.205","Text":"The second rule is that this condition has to remain."},{"Start":"07:10.205 ","End":"07:14.345","Text":"So I can do whatever I want under the plane or on the plane,"},{"Start":"07:14.345 ","End":"07:22.690","Text":"so long as the potential in the area where the plane once was stays equal to 0."},{"Start":"07:24.200 ","End":"07:28.230","Text":"Here is Rule Number 2 as well."},{"Start":"07:28.230 ","End":"07:30.555","Text":"So these are the rules of the game,"},{"Start":"07:30.555 ","End":"07:32.640","Text":"and that\u0027s how we have to play."},{"Start":"07:32.640 ","End":"07:37.560","Text":"Let\u0027s see what we would do in a question like this."},{"Start":"07:37.940 ","End":"07:41.135","Text":"Of course, just like here,"},{"Start":"07:41.135 ","End":"07:44.240","Text":"the potential at infinity is equal to 0."},{"Start":"07:44.240 ","End":"07:46.250","Text":"In the new problem that we find,"},{"Start":"07:46.250 ","End":"07:49.790","Text":"our potential at infinity must also remain equal"},{"Start":"07:49.790 ","End":"07:58.910","Text":"to 0, so analogous."},{"Start":"07:58.910 ","End":"08:03.745","Text":"A similar problem is that if we replace"},{"Start":"08:03.745 ","End":"08:10.010","Text":"our conducting infinite plane with nothing,"},{"Start":"08:10.500 ","End":"08:14.049","Text":"so imagine now our plane is non-existent,"},{"Start":"08:14.049 ","End":"08:19.780","Text":"and then we place a point charge which has a charge of negative q,"},{"Start":"08:19.780 ","End":"08:22.540","Text":"so the opposite charge to what we have over here,"},{"Start":"08:22.540 ","End":"08:30.290","Text":"and its distance from the origin is equal to negative z."},{"Start":"08:31.050 ","End":"08:35.410","Text":"So now we have this type of problem,"},{"Start":"08:35.410 ","End":"08:39.050","Text":"and let\u0027s see if it abides by these rules."},{"Start":"08:39.690 ","End":"08:45.190","Text":"Rule Number 1 is that the charge distribution in z is bigger than naught,"},{"Start":"08:45.190 ","End":"08:47.050","Text":"specifically, than 0,"},{"Start":"08:47.050 ","End":"08:54.025","Text":"specifically, in this question so that the charge distribution cannot change."},{"Start":"08:54.025 ","End":"08:57.085","Text":"As we can see in z is bigger than 0,"},{"Start":"08:57.085 ","End":"08:59.215","Text":"we still have our point charge, q,"},{"Start":"08:59.215 ","End":"09:03.745","Text":"located z above where we used to have our conducting plane,"},{"Start":"09:03.745 ","End":"09:07.820","Text":"so perfect, Rule 1, we can check."},{"Start":"09:08.430 ","End":"09:10.840","Text":"Now let\u0027s check Rule Number 2."},{"Start":"09:10.840 ","End":"09:16.270","Text":"So I need my potential where my plane used to be to be equal to 0,"},{"Start":"09:16.270 ","End":"09:22.975","Text":"and also my potential at infinity must be equal to 0 as well."},{"Start":"09:22.975 ","End":"09:25.675","Text":"So we have 2 point charges over here,"},{"Start":"09:25.675 ","End":"09:27.580","Text":"so we know that, at infinity,"},{"Start":"09:27.580 ","End":"09:30.415","Text":"our potential is going to be equal to 0."},{"Start":"09:30.415 ","End":"09:33.520","Text":"Now what we have to check is that our potential in"},{"Start":"09:33.520 ","End":"09:38.180","Text":"the region where the plane used to be is also equal to 0."},{"Start":"09:39.000 ","End":"09:46.195","Text":"Now let\u0027s check the potential where the conducting surface used to be."},{"Start":"09:46.195 ","End":"09:49.780","Text":"Let\u0027s choose this point over here."},{"Start":"09:49.780 ","End":"09:53.150","Text":"Let\u0027s choose it in red."},{"Start":"09:53.610 ","End":"10:02.470","Text":"Now we can say that the potential due to our point charge up here on"},{"Start":"10:02.470 ","End":"10:05.020","Text":"this is going to be equal to"},{"Start":"10:05.020 ","End":"10:10.780","Text":"kq divided by r. That\u0027s the potential due to this point charge."},{"Start":"10:10.780 ","End":"10:15.520","Text":"The potential due to this point charge is going to be equal to"},{"Start":"10:15.520 ","End":"10:21.010","Text":"k multiplied by negative q divided by r. Now we can"},{"Start":"10:21.010 ","End":"10:24.010","Text":"see that because our r for both of"},{"Start":"10:24.010 ","End":"10:28.690","Text":"these point charges is going to be equal because they\u0027re"},{"Start":"10:28.690 ","End":"10:37.315","Text":"both z steps along the z direction in either direction,"},{"Start":"10:37.315 ","End":"10:44.185","Text":"so we can see that the potential at this point over here is going to just cancel out."},{"Start":"10:44.185 ","End":"10:47.785","Text":"Then we can do that anywhere we want along"},{"Start":"10:47.785 ","End":"10:51.190","Text":"our area where the conducting plane used to be,"},{"Start":"10:51.190 ","End":"10:54.505","Text":"and we\u0027ll see that our potential along"},{"Start":"10:54.505 ","End":"11:01.310","Text":"our infinite conducting plane in this parallel problem is equal to 0."},{"Start":"11:02.310 ","End":"11:07.735","Text":"So, great, we\u0027re abiding by both of the rules to the game."},{"Start":"11:07.735 ","End":"11:15.760","Text":"Now I can try and work out the potential in the region where z is larger than 0."},{"Start":"11:15.760 ","End":"11:18.505","Text":"Let\u0027s choose this point over here,"},{"Start":"11:18.505 ","End":"11:21.740","Text":"and here we want to measure our potential."},{"Start":"11:21.900 ","End":"11:26.890","Text":"The potential at this point is going to be equal to"},{"Start":"11:26.890 ","End":"11:32.560","Text":"the potential due to this point charge plus the potential due to this point charge,"},{"Start":"11:32.560 ","End":"11:37.615","Text":"so that\u0027s going to be equal to kq divided by r_1 plus"},{"Start":"11:37.615 ","End":"11:44.368","Text":"k multiplied by negative q divided by r_2."},{"Start":"11:44.368 ","End":"11:49.465","Text":"Of course, r_1 is this,"},{"Start":"11:49.465 ","End":"11:53.720","Text":"and r_2 is this."},{"Start":"11:56.340 ","End":"12:03.325","Text":"Now I can find my potential at this point over here in the region that I wanted,"},{"Start":"12:03.325 ","End":"12:07.480","Text":"and the magic is that because I\u0027ve abided by"},{"Start":"12:07.480 ","End":"12:12.985","Text":"these 2 rules when forming my analogous system,"},{"Start":"12:12.985 ","End":"12:19.840","Text":"so the potential that I found over here is the exact same potential that I"},{"Start":"12:19.840 ","End":"12:22.180","Text":"would get from the original problem where I had"},{"Start":"12:22.180 ","End":"12:27.530","Text":"a point charge along a grounded conducting infinite plane."},{"Start":"12:27.530 ","End":"12:29.955","Text":"Now, how does this work?"},{"Start":"12:29.955 ","End":"12:34.215","Text":"This comes from Picard\u0027s existence theorem."},{"Start":"12:34.215 ","End":"12:37.185","Text":"What is Picard\u0027s existence theorem?"},{"Start":"12:37.185 ","End":"12:43.230","Text":"If I found a function for potential in a certain domain that upholds, 1,"},{"Start":"12:43.230 ","End":"12:47.865","Text":"Poisson\u0027s equation in the domain where I\u0027ll speak about that in a second, and 2,"},{"Start":"12:47.865 ","End":"12:49.530","Text":"the boundary conditions,"},{"Start":"12:49.530 ","End":"12:52.590","Text":"so that\u0027s the value of the function at the boundary"},{"Start":"12:52.590 ","End":"12:55.965","Text":"of the domain is equal to the value of the potential there."},{"Start":"12:55.965 ","End":"12:58.660","Text":"So for instance, in our previous example,"},{"Start":"12:58.660 ","End":"13:04.330","Text":"we saw that our potential at z is equal to 0 is meant to be equal to 0,"},{"Start":"13:04.330 ","End":"13:09.715","Text":"so if the value of our function at z is equal to 0 is also equal to 0,"},{"Start":"13:09.715 ","End":"13:12.644","Text":"then we uphold the boundary conditions,"},{"Start":"13:12.644 ","End":"13:19.672","Text":"then the function that I found is the only possible potential function."},{"Start":"13:19.672 ","End":"13:20.860","Text":"So that\u0027s what that means."},{"Start":"13:20.860 ","End":"13:23.095","Text":"Now, let\u0027s see what Poisson\u0027s equation is."},{"Start":"13:23.095 ","End":"13:30.640","Text":"Poisson\u0027s equation is speaking about our potential difference,"},{"Start":"13:30.640 ","End":"13:36.535","Text":"and that is equal to Nabla squared multiplied by the potential,"},{"Start":"13:36.535 ","End":"13:43.105","Text":"and we say that that is meant to be equal to negative Rho divided by Epsilon_0."},{"Start":"13:43.105 ","End":"13:46.434","Text":"Now, what is this Nabla squared multiplied by the potential?"},{"Start":"13:46.434 ","End":"13:54.175","Text":"That is our Nabla function multiplied by our Nabla function Phi."},{"Start":"13:54.175 ","End":"14:03.170","Text":"Nabla squared Phi is the same as doing the divergence of the gradient of Phi."},{"Start":"14:04.110 ","End":"14:06.685","Text":"What is this equal to?"},{"Start":"14:06.685 ","End":"14:10.360","Text":"What we have over here in our brackets,"},{"Start":"14:10.360 ","End":"14:12.400","Text":"so this is the gradient of Phi."},{"Start":"14:12.400 ","End":"14:17.680","Text":"So we know that the gradient of Phi is equal to negative E,"},{"Start":"14:17.680 ","End":"14:20.161","Text":"the negative electric field."},{"Start":"14:20.161 ","End":"14:22.660","Text":"So that means that this is going to be equal to"},{"Start":"14:22.660 ","End":"14:28.195","Text":"the divergence of the negative electric field,"},{"Start":"14:28.195 ","End":"14:39.355","Text":"so that\u0027s going to be equal to negative dot E. Then as we know from Gauss\u0027s law,"},{"Start":"14:39.355 ","End":"14:47.420","Text":"this is equal to Rho divided by Epsilon_0, and we get that."},{"Start":"14:47.700 ","End":"14:51.235","Text":"If we go back to our example over here,"},{"Start":"14:51.235 ","End":"14:55.870","Text":"the solution to this problem of the point charge above the conducting"},{"Start":"14:55.870 ","End":"15:02.725","Text":"infinite plane is going to be this function for our potential."},{"Start":"15:02.725 ","End":"15:06.970","Text":"Of course, if I have this function for the potential,"},{"Start":"15:06.970 ","End":"15:09.160","Text":"if I go the opposite direction,"},{"Start":"15:09.160 ","End":"15:12.025","Text":"I go backwards, and I take the derivative of this,"},{"Start":"15:12.025 ","End":"15:16.321","Text":"I\u0027ll get this charge distribution."},{"Start":"15:16.321 ","End":"15:19.775","Text":"This works in every single question like this."},{"Start":"15:19.775 ","End":"15:21.295","Text":"We can go backwards and forwards."},{"Start":"15:21.295 ","End":"15:27.475","Text":"If I take some charge distribution in the region that I\u0027m trying to find,"},{"Start":"15:27.475 ","End":"15:29.680","Text":"the potential for it is"},{"Start":"15:29.680 ","End":"15:37.225","Text":"the same charge distribution as my original question or as my original problem."},{"Start":"15:37.225 ","End":"15:47.690","Text":"Then its Poisson equation is going to adhere or uphold the exact same conditions."},{"Start":"15:48.540 ","End":"15:52.795","Text":"That\u0027s why if I don\u0027t change"},{"Start":"15:52.795 ","End":"15:58.075","Text":"the charge distribution in my domain where I\u0027m trying to find the potential for it,"},{"Start":"15:58.075 ","End":"16:03.640","Text":"then I\u0027m adhering to the first rule over"},{"Start":"16:03.640 ","End":"16:10.975","Text":"there or the first condition for my Picard\u0027s theory of uniqueness."},{"Start":"16:10.975 ","End":"16:15.475","Text":"That is Rule Number 1 over here,"},{"Start":"16:15.475 ","End":"16:19.250","Text":"and that is why this rule exists."},{"Start":"16:20.970 ","End":"16:28.085","Text":"Number 1 just basically means don\u0027t change the charge distribution in the domain,"},{"Start":"16:28.085 ","End":"16:33.360","Text":"and Rule Number 2 is speaking about adhering to the same boundary conditions."},{"Start":"16:33.360 ","End":"16:35.035","Text":"So let\u0027s go back here."},{"Start":"16:35.035 ","End":"16:36.530","Text":"Here, again,"},{"Start":"16:36.530 ","End":"16:37.850","Text":"as I briefly explained,"},{"Start":"16:37.850 ","End":"16:42.935","Text":"the boundary condition is that over here and below my infinite plane,"},{"Start":"16:42.935 ","End":"16:44.840","Text":"my potential is equal to 0,"},{"Start":"16:44.840 ","End":"16:48.450","Text":"and my potential at infinity is equal to 0 as well."},{"Start":"16:48.450 ","End":"16:50.840","Text":"Those are my boundary conditions,"},{"Start":"16:50.840 ","End":"16:55.820","Text":"and so I have to adhere to them as well in here."},{"Start":"16:55.820 ","End":"17:01.190","Text":"I have to check that my potential function that I found also adheres to"},{"Start":"17:01.190 ","End":"17:07.855","Text":"those exact boundary conditions and that is of course what this says also over here."},{"Start":"17:07.855 ","End":"17:11.480","Text":"We can see that this will really be at"},{"Start":"17:11.480 ","End":"17:15.950","Text":"infinity equal to 0 because our r_1 and r_2 would be extremely large,"},{"Start":"17:15.950 ","End":"17:18.260","Text":"so that will cancel out."},{"Start":"17:18.260 ","End":"17:20.400","Text":"Also when we\u0027re at 0,"},{"Start":"17:20.400 ","End":"17:24.470","Text":"and so also our charges will cancel out,"},{"Start":"17:24.470 ","End":"17:26.089","Text":"and we\u0027ll have a 0 over here,"},{"Start":"17:26.089 ","End":"17:29.660","Text":"just like we saw with our red dotted lines."},{"Start":"17:29.660 ","End":"17:33.600","Text":"So, great, we\u0027re adhering to boundary conditions."},{"Start":"17:34.060 ","End":"17:38.570","Text":"The quick final note about that Number 2 is that we saw that,"},{"Start":"17:38.570 ","End":"17:39.770","Text":"below the infinite plane,"},{"Start":"17:39.770 ","End":"17:43.623","Text":"our potential always has to be equal to 0 as well."},{"Start":"17:43.623 ","End":"17:44.720","Text":"But this answer,"},{"Start":"17:44.720 ","End":"17:46.450","Text":"we can see that if we go below the plane,"},{"Start":"17:46.450 ","End":"17:48.470","Text":"it\u0027s not equal to 0."},{"Start":"17:48.470 ","End":"17:52.760","Text":"But what we need to know is that we\u0027re looking for the potential equations"},{"Start":"17:52.760 ","End":"17:57.870","Text":"such that it is in our requested region or domain."},{"Start":"17:57.870 ","End":"18:00.350","Text":"So if we say that this is the potential,"},{"Start":"18:00.350 ","End":"18:03.975","Text":"and z is bigger or equal to 0,"},{"Start":"18:03.975 ","End":"18:08.699","Text":"then it matches all of our boundary conditions,"},{"Start":"18:08.699 ","End":"18:12.710","Text":"and it\u0027s correct for the domain that we\u0027re looking for."},{"Start":"18:12.710 ","End":"18:15.900","Text":"That\u0027s the end of this lesson."}],"ID":14213},{"Watched":false,"Name":"Exercise 1","Duration":"9m 26s","ChapterTopicVideoID":12111,"CourseChapterTopicPlaylistID":161612,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12111.jpeg","UploadDate":"2018-06-28T03:24:05.2500000","DurationForVideoObject":"PT9M26S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.845","Text":"Hello. In this question,"},{"Start":"00:01.845 ","End":"00:05.265","Text":"we have two infinite planes."},{"Start":"00:05.265 ","End":"00:09.000","Text":"The first plane is on the x,"},{"Start":"00:09.000 ","End":"00:11.820","Text":"y plane and it\u0027s infinite,"},{"Start":"00:11.820 ","End":"00:14.100","Text":"and the second plane is on the x,"},{"Start":"00:14.100 ","End":"00:18.285","Text":"z plane and it\u0027s infinite, up and across."},{"Start":"00:18.285 ","End":"00:23.760","Text":"Now, both of these planes are joined and they\u0027re both grounded."},{"Start":"00:23.760 ","End":"00:25.754","Text":"Now, if they\u0027re joined,"},{"Start":"00:25.754 ","End":"00:30.645","Text":"it doesn\u0027t matter if only one of them is grounded or this one means the same thing."},{"Start":"00:30.645 ","End":"00:34.610","Text":"But they can also both be grounded. It\u0027s the same."},{"Start":"00:34.610 ","End":"00:41.650","Text":"These two infinite planes are joined at 90 degrees to one another."},{"Start":"00:41.650 ","End":"00:46.370","Text":"Then we\u0027re being told that some point charge,"},{"Start":"00:46.370 ","End":"00:48.905","Text":"that we know its charge, its charge is q,"},{"Start":"00:48.905 ","End":"00:57.320","Text":"is located a distance d away from our infinite grounded plane on the x,"},{"Start":"00:57.320 ","End":"01:01.100","Text":"y axes and it\u0027s also located at"},{"Start":"01:01.100 ","End":"01:07.205","Text":"distance d from the grounded infinite plane on these z, x axis."},{"Start":"01:07.205 ","End":"01:14.795","Text":"Then our question is to find the potential in the region where z is bigger than 0,"},{"Start":"01:14.795 ","End":"01:21.665","Text":"so here, and also where y is bigger than 0, so here."},{"Start":"01:21.665 ","End":"01:26.015","Text":"What we\u0027re trying to do is we\u0027re trying to find"},{"Start":"01:26.015 ","End":"01:34.980","Text":"the potential in all of this region here."},{"Start":"01:37.790 ","End":"01:42.875","Text":"The first thing that we can see from this question is that we have"},{"Start":"01:42.875 ","End":"01:49.655","Text":"some grounded planes or some body which is grounded and it\u0027s a conductor,"},{"Start":"01:49.655 ","End":"01:52.640","Text":"if it\u0027s grounded, it\u0027s also a conductor,"},{"Start":"01:52.640 ","End":"01:57.965","Text":"and we have some charge distribution in the region that we want to know."},{"Start":"01:57.965 ","End":"02:03.360","Text":"We\u0027re being asked to find the potential in the specific region."},{"Start":"02:03.360 ","End":"02:07.520","Text":"These three things automatically whenever you see them,"},{"Start":"02:07.520 ","End":"02:12.305","Text":"this tells you that you have to use the method of images in order to find the potential."},{"Start":"02:12.305 ","End":"02:14.680","Text":"That\u0027s straight away."},{"Start":"02:14.680 ","End":"02:16.845","Text":"Let\u0027s begin doing that."},{"Start":"02:16.845 ","End":"02:20.135","Text":"First of all, we know that our boundary conditions,"},{"Start":"02:20.135 ","End":"02:22.775","Text":"because these planes are grounded."},{"Start":"02:22.775 ","End":"02:27.525","Text":"We know that the potential on the z,"},{"Start":"02:27.525 ","End":"02:31.425","Text":"x plane is going to be equal to 0 and similarly,"},{"Start":"02:31.425 ","End":"02:33.345","Text":"the potential, the y,"},{"Start":"02:33.345 ","End":"02:35.730","Text":"x plane is also going to be equal to 0."},{"Start":"02:35.730 ","End":"02:40.400","Text":"Of course, because this is how we usually define it and this is what usually happens,"},{"Start":"02:40.400 ","End":"02:46.065","Text":"the potential at infinity is also equal to 0."},{"Start":"02:46.065 ","End":"02:51.725","Text":"Now what we have to do is we have to find an identical problem to this one"},{"Start":"02:51.725 ","End":"02:58.960","Text":"using our method of images and then we can find our potential."},{"Start":"03:00.620 ","End":"03:07.010","Text":"The analogous problem to this is going to be to,"},{"Start":"03:07.010 ","End":"03:14.000","Text":"first of all, add in some charge distribution in this region over here."},{"Start":"03:14.000 ","End":"03:16.745","Text":"I can do whatever I want over here."},{"Start":"03:16.745 ","End":"03:20.360","Text":"Now let\u0027s see which charge distribution I should add."},{"Start":"03:20.360 ","End":"03:22.535","Text":"I have my point charge here, q,"},{"Start":"03:22.535 ","End":"03:25.700","Text":"at a distance d from each of these planes."},{"Start":"03:25.700 ","End":"03:31.850","Text":"The first clue that you should have is that if I have a point charge q,"},{"Start":"03:31.850 ","End":"03:34.880","Text":"I\u0027m going to have at least another point charge in"},{"Start":"03:34.880 ","End":"03:38.210","Text":"this region that\u0027s going to be negative q."},{"Start":"03:38.210 ","End":"03:43.270","Text":"It always has to have the opposite charge, so negative q."},{"Start":"03:43.270 ","End":"03:48.065","Text":"Now let\u0027s see, my q is located at a distance d from this plane,"},{"Start":"03:48.065 ","End":"03:53.430","Text":"so in order to cancel out the potential on this plane,"},{"Start":"03:53.630 ","End":"03:56.980","Text":"because remember, now we\u0027re ignoring is if"},{"Start":"03:56.980 ","End":"03:59.650","Text":"this plane doesn\u0027t exist and we\u0027re trying to find"},{"Start":"03:59.650 ","End":"04:05.764","Text":"a problem that has charged distribution in this region that we don\u0027t really care about,"},{"Start":"04:05.764 ","End":"04:09.070","Text":"such that our potential along"},{"Start":"04:09.070 ","End":"04:14.140","Text":"the areas where the plane was is still going to be equal to 0."},{"Start":"04:14.140 ","End":"04:18.160","Text":"In order to cancel out the potential at this point over here,"},{"Start":"04:18.160 ","End":"04:23.610","Text":"we\u0027d probably need to put a charge negative q over here."},{"Start":"04:23.610 ","End":"04:27.085","Text":"Of course, the distance between negative q,"},{"Start":"04:27.085 ","End":"04:30.170","Text":"and this is also going to be d,"},{"Start":"04:30.920 ","End":"04:35.740","Text":"and then we can see that the potential at this point due to this charge and"},{"Start":"04:35.740 ","End":"04:39.895","Text":"the potential at the point due to this charge will be equal to 0."},{"Start":"04:39.895 ","End":"04:43.610","Text":"Cool. They\u0027ll cancel out, both of them together."},{"Start":"04:44.000 ","End":"04:47.165","Text":"That\u0027s great. We\u0027ve dealt with this infinite plane."},{"Start":"04:47.165 ","End":"04:49.355","Text":"Now what about this infinite plane?"},{"Start":"04:49.355 ","End":"04:51.080","Text":"Again, with distance d,"},{"Start":"04:51.080 ","End":"04:56.630","Text":"so we have to add in another charge negative q over here and it\u0027s also"},{"Start":"04:56.630 ","End":"05:02.300","Text":"going to be a distance of d away from where the grounded plane used to be."},{"Start":"05:02.300 ","End":"05:06.230","Text":"Because then the potential at this point due to this charge will"},{"Start":"05:06.230 ","End":"05:10.880","Text":"cancel out with the potential due to this charge and then again,"},{"Start":"05:10.880 ","End":"05:15.460","Text":"we have 0 along this infinite grounded plane."},{"Start":"05:15.460 ","End":"05:17.075","Text":"We\u0027re almost done."},{"Start":"05:17.075 ","End":"05:20.960","Text":"What we can see here is that along this plane,"},{"Start":"05:20.960 ","End":"05:26.110","Text":"the charge from here and the charge from here cause the potential here to cancel out."},{"Start":"05:26.110 ","End":"05:28.640","Text":"On this plane, the charge from here and the child from"},{"Start":"05:28.640 ","End":"05:31.370","Text":"here cause the potential on this plane to cancel out."},{"Start":"05:31.370 ","End":"05:36.950","Text":"However, we still have an interaction between this charge negative q on"},{"Start":"05:36.950 ","End":"05:43.460","Text":"this plane and it has no end to cancel it out and this charge negative q on this plane,"},{"Start":"05:43.460 ","End":"05:46.265","Text":"because it has no one here to cancel it out."},{"Start":"05:46.265 ","End":"05:50.310","Text":"We have some crisscross between our charges,"},{"Start":"05:50.540 ","End":"05:53.430","Text":"so what do I need to do?"},{"Start":"05:53.430 ","End":"05:59.870","Text":"Here I can see that I have a negative q affecting here and a negative q affecting here."},{"Start":"05:59.870 ","End":"06:02.390","Text":"In order to cancel them out,"},{"Start":"06:02.390 ","End":"06:06.860","Text":"I\u0027m going to have to add in an oppositely charged to my negative"},{"Start":"06:06.860 ","End":"06:12.155","Text":"q\u0027s because they\u0027re the ones that are causing some charge crisscrossing over here."},{"Start":"06:12.155 ","End":"06:17.030","Text":"I\u0027m going to add in a charge q over here,"},{"Start":"06:17.030 ","End":"06:19.020","Text":"this as a positive q,"},{"Start":"06:19.020 ","End":"06:23.465","Text":"and it\u0027s obviously going to be a distance from each one of 2d."},{"Start":"06:23.465 ","End":"06:30.930","Text":"We could see that it\u0027s a square of distance of where each side is of length 2d."},{"Start":"06:31.570 ","End":"06:42.940","Text":"This is d, d and this is also a distance of 2d."},{"Start":"06:43.030 ","End":"06:49.059","Text":"Now we can see that this charge is going to balance out"},{"Start":"06:49.059 ","End":"06:56.450","Text":"what we had from this charge on this plane and this charge on this plane."},{"Start":"06:58.280 ","End":"07:03.490","Text":"Now let\u0027s check if we\u0027ve played by the rules of the game."},{"Start":"07:03.490 ","End":"07:09.670","Text":"We can see that the potential along this plane is canceled out due to"},{"Start":"07:09.670 ","End":"07:15.580","Text":"the negative q here and the positive q here and the potential on this plane,"},{"Start":"07:15.580 ","End":"07:16.810","Text":"the y, x plane,"},{"Start":"07:16.810 ","End":"07:19.315","Text":"is canceled out due to this charge here,"},{"Start":"07:19.315 ","End":"07:22.670","Text":"and then this charge here as well."},{"Start":"07:22.670 ","End":"07:28.090","Text":"We can see that the potential along the grounded plane is still equal to 0."},{"Start":"07:28.090 ","End":"07:33.625","Text":"For point charges, the potential at infinity is still going to be equal to 0."},{"Start":"07:33.625 ","End":"07:39.130","Text":"Great. We\u0027ve checked off that row and then the next rule that we have to"},{"Start":"07:39.130 ","End":"07:40.810","Text":"see is that we haven\u0027t changed"},{"Start":"07:40.810 ","End":"07:45.185","Text":"the charge distribution in the region that we\u0027re interested in."},{"Start":"07:45.185 ","End":"07:48.445","Text":"Our region is z is bigger than 0,"},{"Start":"07:48.445 ","End":"07:51.475","Text":"so that\u0027s here, and y is bigger than 0, so that\u0027s here."},{"Start":"07:51.475 ","End":"07:55.023","Text":"We can see that are charged or distribution began as 1."},{"Start":"07:55.023 ","End":"08:00.580","Text":"charge of charge q and located at this position over here,"},{"Start":"08:00.580 ","End":"08:02.440","Text":"and we can see we haven\u0027t changed that."},{"Start":"08:02.440 ","End":"08:06.700","Text":"We only added charge distributions in the region where we\u0027re allowed to."},{"Start":"08:06.700 ","End":"08:15.225","Text":"Great. The question was to find this potential in this region."},{"Start":"08:15.225 ","End":"08:19.565","Text":"Let\u0027s choose some random point in the regions that is bigger than 0,"},{"Start":"08:19.565 ","End":"08:21.980","Text":"y is bigger than 0, and let\u0027s see what"},{"Start":"08:21.980 ","End":"08:24.785","Text":"our potential in this region is going to be equal to."},{"Start":"08:24.785 ","End":"08:28.190","Text":"It\u0027s just going to be equal to the superposition of all of"},{"Start":"08:28.190 ","End":"08:32.765","Text":"the potentials arising from these four point charges."},{"Start":"08:32.765 ","End":"08:38.375","Text":"From here, I potential is going to be kq divided by r_1,"},{"Start":"08:38.375 ","End":"08:42.020","Text":"where r_1 is the distance from this point to this point,"},{"Start":"08:42.020 ","End":"08:44.015","Text":"plus from here,"},{"Start":"08:44.015 ","End":"08:49.100","Text":"so k multiplied by negative q divided by r_2,"},{"Start":"08:49.100 ","End":"08:51.380","Text":"where r_2 is from here to here,"},{"Start":"08:51.380 ","End":"08:53.695","Text":"plus now from here,"},{"Start":"08:53.695 ","End":"08:59.955","Text":"kq divided by r_3 plus k"},{"Start":"08:59.955 ","End":"09:04.460","Text":"negative q divided by r_4 and this is"},{"Start":"09:04.460 ","End":"09:09.095","Text":"the potential here due to these for point charges and that is also,"},{"Start":"09:09.095 ","End":"09:17.015","Text":"if we rub out these three point charges and bring back our grounded infinite planes,"},{"Start":"09:17.015 ","End":"09:20.975","Text":"the potential function is going to be the same."},{"Start":"09:20.975 ","End":"09:23.240","Text":"This is what it\u0027s going to be."},{"Start":"09:23.240 ","End":"09:26.460","Text":"That\u0027s the end of this question."}],"ID":14214},{"Watched":false,"Name":"Exercise 2","Duration":"12m 26s","ChapterTopicVideoID":12112,"CourseChapterTopicPlaylistID":161612,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12112.jpeg","UploadDate":"2018-06-28T03:27:52.4170000","DurationForVideoObject":"PT12M26S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.010","Text":"Hello. This question is very similar to the last question that we saw."},{"Start":"00:05.010 ","End":"00:09.524","Text":"Again, we have 2 conducting infinite planes which have been grounded."},{"Start":"00:09.524 ","End":"00:11.925","Text":"The only difference is that here,"},{"Start":"00:11.925 ","End":"00:15.780","Text":"the angle between the 2 planes is Alpha."},{"Start":"00:15.780 ","End":"00:18.930","Text":"Here we have a point charge q,"},{"Start":"00:18.930 ","End":"00:20.430","Text":"and it\u0027s a distance,"},{"Start":"00:20.430 ","End":"00:22.410","Text":"just like in the previous question, d,"},{"Start":"00:22.410 ","End":"00:30.885","Text":"away from this infinite plane and it\u0027s also a distance d away from this infinite plane."},{"Start":"00:30.885 ","End":"00:33.585","Text":"Now, we\u0027re also given another piece of information"},{"Start":"00:33.585 ","End":"00:37.020","Text":"which soon we\u0027ll understand why it\u0027s important."},{"Start":"00:37.020 ","End":"00:40.410","Text":"But we\u0027re being told that this angle, Alpha,"},{"Start":"00:40.410 ","End":"00:48.610","Text":"divided by 360 is equal to an even number."},{"Start":"00:48.800 ","End":"00:51.700","Text":"It\u0027s some even number."},{"Start":"00:51.700 ","End":"00:55.854","Text":"This will explain why this is important later on."},{"Start":"00:55.854 ","End":"01:05.340","Text":"This question is asking us to find the potential in this region over here,"},{"Start":"01:05.340 ","End":"01:09.955","Text":"the whole region between the 2 infinite planes."},{"Start":"01:09.955 ","End":"01:15.080","Text":"Now again, anytime you see some body, plane, 2 planes,"},{"Start":"01:15.080 ","End":"01:18.485","Text":"whatever it might be, which is grounded,"},{"Start":"01:18.485 ","End":"01:19.940","Text":"which means it\u0027s also conducting,"},{"Start":"01:19.940 ","End":"01:26.155","Text":"but it\u0027s grounded, and we have some charge distribution within the region,"},{"Start":"01:26.155 ","End":"01:32.720","Text":"between the bodies and we\u0027re being told to find the potential in that region,"},{"Start":"01:32.720 ","End":"01:36.355","Text":"we\u0027re using a method of images."},{"Start":"01:36.355 ","End":"01:40.655","Text":"Now what we have to do is we have to find a problem"},{"Start":"01:40.655 ","End":"01:44.825","Text":"which is similar to what we have here in order to find the potential."},{"Start":"01:44.825 ","End":"01:47.435","Text":"Because we have these 2 grounded infinite planes,"},{"Start":"01:47.435 ","End":"01:53.005","Text":"we can\u0027t just use Gauss\u0027s law in order to find the potential of this point charge."},{"Start":"01:53.005 ","End":"01:55.200","Text":"How are we going to do this?"},{"Start":"01:55.200 ","End":"02:00.105","Text":"The first thing that we\u0027re going to do is we\u0027re going to complete"},{"Start":"02:00.105 ","End":"02:07.140","Text":"a circle with lots of little different wedges of angle Alpha."},{"Start":"02:07.140 ","End":"02:09.705","Text":"Let\u0027s draw that."},{"Start":"02:09.705 ","End":"02:14.550","Text":"We\u0027ll have this over here. We\u0027re just drawing some pizza shape."},{"Start":"02:14.550 ","End":"02:21.000","Text":"Something like this."},{"Start":"02:21.000 ","End":"02:22.620","Text":"This angle is Alpha,"},{"Start":"02:22.620 ","End":"02:25.470","Text":"this is Alpha, this is Alpha,"},{"Start":"02:25.470 ","End":"02:27.042","Text":"this is Alpha,"},{"Start":"02:27.042 ","End":"02:30.130","Text":"this, and this."},{"Start":"02:31.000 ","End":"02:35.150","Text":"I added in another wedge over here."},{"Start":"02:35.150 ","End":"02:39.500","Text":"Now, it doesn\u0027t really matter how many wedges I put in here."},{"Start":"02:39.500 ","End":"02:44.330","Text":"All I did really was follow these lines, but it doesn\u0027t matter."},{"Start":"02:44.330 ","End":"02:46.355","Text":"I could have put in less or more."},{"Start":"02:46.355 ","End":"02:49.640","Text":"What is important here is that if we can see"},{"Start":"02:49.640 ","End":"02:53.750","Text":"how many wedges of angle Alpha we have, here we have 8."},{"Start":"02:53.750 ","End":"02:56.915","Text":"Now what\u0027s important is that 8 is an even number,"},{"Start":"02:56.915 ","End":"02:59.435","Text":"which is where this comes in over here."},{"Start":"02:59.435 ","End":"03:06.365","Text":"We know that this angle divided by 360 is equal to an even number of wedges."},{"Start":"03:06.365 ","End":"03:08.705","Text":"We could have had 6, we could have had 4."},{"Start":"03:08.705 ","End":"03:14.210","Text":"It didn\u0027t really matter as long as we\u0027re completing a full circle because it\u0027s 360,"},{"Start":"03:14.210 ","End":"03:18.660","Text":"so a full circle, and then we have an even number of wedges."},{"Start":"03:19.220 ","End":"03:25.950","Text":"Now, in order to balance out this positive charge over here,"},{"Start":"03:26.020 ","End":"03:31.555","Text":"remember that now we\u0027re taking away our grounded planes and we\u0027re"},{"Start":"03:31.555 ","End":"03:36.715","Text":"adding in charge distribution in this region where we\u0027re allowed to."},{"Start":"03:36.715 ","End":"03:38.650","Text":"As we saw before,"},{"Start":"03:38.650 ","End":"03:41.365","Text":"if here we have a positive q,"},{"Start":"03:41.365 ","End":"03:44.530","Text":"so we\u0027re going to add a point charge over here,"},{"Start":"03:44.530 ","End":"03:46.420","Text":"a charge negative q,"},{"Start":"03:46.420 ","End":"03:50.170","Text":"where its distance away from this plane is"},{"Start":"03:50.170 ","End":"03:54.700","Text":"also going to be equal to d. Then the same over here."},{"Start":"03:54.700 ","End":"03:57.670","Text":"We\u0027re going to add another charge,"},{"Start":"03:57.670 ","End":"04:00.175","Text":"negative q, a distance d away."},{"Start":"04:00.175 ","End":"04:04.130","Text":"Then the negative q and the positive q\u0027s here cancel out."},{"Start":"04:04.130 ","End":"04:08.590","Text":"Then just like before, where we had our 2 planes at 90 degrees,"},{"Start":"04:08.590 ","End":"04:12.170","Text":"we had to add in positive q in order to"},{"Start":"04:12.170 ","End":"04:17.180","Text":"cancel out this negative q interacting with this plane over here,"},{"Start":"04:17.180 ","End":"04:21.380","Text":"now we\u0027re going to have to add in a positive q over here,"},{"Start":"04:21.380 ","End":"04:23.495","Text":"and then again to cancel that out,"},{"Start":"04:23.495 ","End":"04:25.520","Text":"a negative q over here,"},{"Start":"04:25.520 ","End":"04:27.100","Text":"and then to cancel that,"},{"Start":"04:27.100 ","End":"04:33.660","Text":"positive q, negative q, positive q."},{"Start":"04:33.660 ","End":"04:39.330","Text":"What we can see is that we\u0027re going with alternating charges around a circle."},{"Start":"04:39.550 ","End":"04:51.120","Text":"What\u0027s important is that each charge is a distance d away from these dotted lines."},{"Start":"04:53.030 ","End":"04:57.080","Text":"This is what I am assuming is going to be"},{"Start":"04:57.080 ","End":"05:01.960","Text":"a parallel problem to what we had when we did have the infinite grounded planes."},{"Start":"05:01.960 ","End":"05:05.063","Text":"Now let\u0027s see if this is a parallel problem."},{"Start":"05:05.063 ","End":"05:07.980","Text":"One of the things that we had to adhere to,"},{"Start":"05:07.980 ","End":"05:09.635","Text":"or one of the rules of the game,"},{"Start":"05:09.635 ","End":"05:14.720","Text":"is that the potential along each plane will remain equal to 0."},{"Start":"05:14.720 ","End":"05:16.910","Text":"Because when the planes were around,"},{"Start":"05:16.910 ","End":"05:21.260","Text":"we saw that the potential at each plane was equal to 0,"},{"Start":"05:21.260 ","End":"05:25.385","Text":"and also that the potential at infinity is also equal to 0."},{"Start":"05:25.385 ","End":"05:30.500","Text":"Let\u0027s see. Let\u0027s choose this random point over here on this plane over here."},{"Start":"05:30.500 ","End":"05:33.020","Text":"We can see that the potential at this point is going"},{"Start":"05:33.020 ","End":"05:36.205","Text":"to cancel out due to these 2 charges."},{"Start":"05:36.205 ","End":"05:38.430","Text":"But then what about these 2 charges?"},{"Start":"05:38.430 ","End":"05:40.910","Text":"We can see we have a negative q with the potential here,"},{"Start":"05:40.910 ","End":"05:43.705","Text":"and a plus q with the potential over here."},{"Start":"05:43.705 ","End":"05:46.865","Text":"They\u0027re going to be equal because of the symmetry,"},{"Start":"05:46.865 ","End":"05:49.760","Text":"and opposite because of their opposite charges."},{"Start":"05:49.760 ","End":"05:52.730","Text":"The potential is going to cancel out here because of them."},{"Start":"05:52.730 ","End":"05:56.409","Text":"Then what about the potential here due to these? It\u0027s the same."},{"Start":"05:56.409 ","End":"06:00.470","Text":"They\u0027re an equal distance away from the point and they\u0027re of opposite charges."},{"Start":"06:00.470 ","End":"06:02.255","Text":"The potential will cancel out."},{"Start":"06:02.255 ","End":"06:07.875","Text":"Also, the potential from these 2 will also cancel out over here."},{"Start":"06:07.875 ","End":"06:11.045","Text":"We can see that every single charge has"},{"Start":"06:11.045 ","End":"06:14.000","Text":"an equal and opposite charge that will also"},{"Start":"06:14.000 ","End":"06:17.870","Text":"exert a potential that will be equal and opposite at this point."},{"Start":"06:17.870 ","End":"06:19.895","Text":"Meaning that the potential along"},{"Start":"06:19.895 ","End":"06:25.550","Text":"every single point along this plane is going to be equal to 0."},{"Start":"06:25.550 ","End":"06:31.466","Text":"Now, let\u0027s again look at a random point on our other grounded plane."},{"Start":"06:31.466 ","End":"06:34.250","Text":"Of course, if this is true at this point,"},{"Start":"06:34.250 ","End":"06:37.640","Text":"then it\u0027s of course true along every point on the plane."},{"Start":"06:37.640 ","End":"06:39.740","Text":"Again, due to symmetry,"},{"Start":"06:39.740 ","End":"06:44.725","Text":"the potential due to this charge and this charge cancels out over here."},{"Start":"06:44.725 ","End":"06:46.940","Text":"Due to this and this,"},{"Start":"06:46.940 ","End":"06:48.015","Text":"they will cancel out,"},{"Start":"06:48.015 ","End":"06:50.280","Text":"and these 2 will pair up to cancel out."},{"Start":"06:50.280 ","End":"06:54.815","Text":"These 2 will pair up to have the potential here cancel out."},{"Start":"06:54.815 ","End":"07:00.985","Text":"That means that the potential on the 2 charged,"},{"Start":"07:00.985 ","End":"07:06.980","Text":"or the 2 conducting and grounded infinite planes cancels out."},{"Start":"07:06.980 ","End":"07:10.865","Text":"Then we have also 8 point charges."},{"Start":"07:10.865 ","End":"07:16.635","Text":"The potential of point charges does in fact cancel out at infinity."},{"Start":"07:16.635 ","End":"07:21.850","Text":"That\u0027s great. We\u0027re adhering to our boundary conditions."},{"Start":"07:22.490 ","End":"07:26.720","Text":"This is also exactly why we had to create"},{"Start":"07:26.720 ","End":"07:30.815","Text":"this rule that Alpha divided by 360 is equal to an even number."},{"Start":"07:30.815 ","End":"07:33.095","Text":"Because if we would have an odd number,"},{"Start":"07:33.095 ","End":"07:37.520","Text":"then at some stage we would have a negative q and a negative q,"},{"Start":"07:37.520 ","End":"07:40.640","Text":"and then we wouldn\u0027t be able to do this canceling out effect."},{"Start":"07:40.640 ","End":"07:45.148","Text":"Because we have our charges alternating between plus and minus,"},{"Start":"07:45.148 ","End":"07:55.330","Text":"we can create this overall net 0 potential along the grounded planes."},{"Start":"07:56.420 ","End":"08:01.130","Text":"Now, we can look and see also the next row"},{"Start":"08:01.130 ","End":"08:04.910","Text":"is that we don\u0027t change the charge distribution within our region."},{"Start":"08:04.910 ","End":"08:07.520","Text":"We started off with this point charge q"},{"Start":"08:07.520 ","End":"08:10.560","Text":"located a distance d away from each of the planes."},{"Start":"08:10.560 ","End":"08:17.390","Text":"We can see that we still only have 1 point charge of q at the exact same position."},{"Start":"08:17.390 ","End":"08:21.590","Text":"Then if we wanted to find the potential at this point over here,"},{"Start":"08:21.590 ","End":"08:25.865","Text":"we\u0027re always finding the potential just in the region that we\u0027ve been asked to."},{"Start":"08:25.865 ","End":"08:30.655","Text":"We just add up the potential due to each point charge."},{"Start":"08:30.655 ","End":"08:34.455","Text":"Then obviously, you just change the radius."},{"Start":"08:34.455 ","End":"08:35.975","Text":"Here, you will have r_1,"},{"Start":"08:35.975 ","End":"08:39.095","Text":"r_2, r_3, and so on and so forth."},{"Start":"08:39.095 ","End":"08:43.380","Text":"Now, let\u0027s make this question a little bit harder."},{"Start":"08:43.840 ","End":"08:48.110","Text":"What would happen if our original point charge q,"},{"Start":"08:48.110 ","End":"08:51.455","Text":"which was located at distance d away from each plane,"},{"Start":"08:51.455 ","End":"08:55.325","Text":"wasn\u0027t a distance d away from each plane,"},{"Start":"08:55.325 ","End":"08:58.085","Text":"or it wasn\u0027t equidistant from each plane,"},{"Start":"08:58.085 ","End":"09:02.345","Text":"but was rather closer to one of the planes than the other?"},{"Start":"09:02.345 ","End":"09:05.120","Text":"How would we solve this when"},{"Start":"09:05.120 ","End":"09:08.344","Text":"our point charge is let\u0027s say closer to this plane over here?"},{"Start":"09:08.344 ","End":"09:10.100","Text":"One of the tricks,"},{"Start":"09:10.100 ","End":"09:15.875","Text":"which is, there\u0027s a clue in the name of this method,"},{"Start":"09:15.875 ","End":"09:19.610","Text":"it\u0027s the method of images or another name"},{"Start":"09:19.610 ","End":"09:24.050","Text":"for this is the mirror method or method of mirrors."},{"Start":"09:24.050 ","End":"09:27.910","Text":"Both of these methods are speaking about exactly what we\u0027re doing now."},{"Start":"09:27.910 ","End":"09:29.740","Text":"How would we do this?"},{"Start":"09:29.740 ","End":"09:35.345","Text":"What we do is we imagine that each one of these grounded planes is"},{"Start":"09:35.345 ","End":"09:42.810","Text":"a mirror and then we imagine what this charge would look like in the mirror image."},{"Start":"09:43.600 ","End":"09:46.445","Text":"In a case like this,"},{"Start":"09:46.445 ","End":"09:52.865","Text":"our mirror image for all of these charges would look something like this."},{"Start":"09:52.865 ","End":"09:58.775","Text":"We can see that if this segment was in mirror,"},{"Start":"09:58.775 ","End":"10:02.165","Text":"the image of it would be a charge over here,"},{"Start":"10:02.165 ","End":"10:05.120","Text":"also close to the plane,"},{"Start":"10:05.120 ","End":"10:08.270","Text":"but with the opposite charge because this is positive q,"},{"Start":"10:08.270 ","End":"10:10.115","Text":"so this will be negative q."},{"Start":"10:10.115 ","End":"10:16.865","Text":"Then we can see that if we take this dotted line as a mirror,"},{"Start":"10:16.865 ","End":"10:20.030","Text":"our negative charge here is really far away from the mirror,"},{"Start":"10:20.030 ","End":"10:22.438","Text":"our positive charge,"},{"Start":"10:22.438 ","End":"10:24.990","Text":"the opposite charge to our negative q,"},{"Start":"10:24.990 ","End":"10:27.275","Text":"is also going to be very far away."},{"Start":"10:27.275 ","End":"10:29.330","Text":"Then again we have a mirror,"},{"Start":"10:29.330 ","End":"10:32.475","Text":"and again we\u0027re close and so on and so forth."},{"Start":"10:32.475 ","End":"10:33.950","Text":"Also, if we go this way,"},{"Start":"10:33.950 ","End":"10:39.065","Text":"we can see that our original charge q is far away from this mirror."},{"Start":"10:39.065 ","End":"10:41.600","Text":"Its opposite charge, negative q,"},{"Start":"10:41.600 ","End":"10:45.148","Text":"is also going to be far away from the mirror just on the other side,"},{"Start":"10:45.148 ","End":"10:48.500","Text":"and then we have another mirror where it\u0027s close to that mirror."},{"Start":"10:48.500 ","End":"10:53.180","Text":"Then this charge is also going to be close to the mirror which is on the other side,"},{"Start":"10:53.180 ","End":"10:55.315","Text":"and we just repeat that all the way around."},{"Start":"10:55.315 ","End":"11:03.905","Text":"Then let\u0027s take this random point over here on our grounded infinite plane."},{"Start":"11:03.905 ","End":"11:13.155","Text":"We can see that these 2 charges are equidistant from this point as well."},{"Start":"11:13.155 ","End":"11:16.475","Text":"The potential at this point is going to cancel out due to them"},{"Start":"11:16.475 ","End":"11:19.715","Text":"being equal distance away and of opposite charge."},{"Start":"11:19.715 ","End":"11:21.815","Text":"Then the same with these 2,"},{"Start":"11:21.815 ","End":"11:24.485","Text":"and these 2 will pair up,"},{"Start":"11:24.485 ","End":"11:26.675","Text":"and these 2 will pair up,"},{"Start":"11:26.675 ","End":"11:28.196","Text":"and the potential here will cancel out."},{"Start":"11:28.196 ","End":"11:32.210","Text":"Of course, the same thing goes for this plane."},{"Start":"11:32.210 ","End":"11:35.320","Text":"Just the pairs will be slightly different."},{"Start":"11:35.320 ","End":"11:38.540","Text":"As we can see, the trick to answering this type of"},{"Start":"11:38.540 ","End":"11:42.440","Text":"question is to remember that if we have some angle Alpha,"},{"Start":"11:42.440 ","End":"11:46.370","Text":"that Alpha divided by 360 has to equal an even number,"},{"Start":"11:46.370 ","End":"11:48.835","Text":"so that means an even number of wedges,"},{"Start":"11:48.835 ","End":"11:51.810","Text":"including the original wedge."},{"Start":"11:51.810 ","End":"11:56.645","Text":"The second trick is to realize that whether"},{"Start":"11:56.645 ","End":"12:02.885","Text":"our original point charge or charge distribution is in the center of the wedge or not,"},{"Start":"12:02.885 ","End":"12:08.465","Text":"to use each side of the wedges and imagine them as mirrors,"},{"Start":"12:08.465 ","End":"12:11.725","Text":"and draw the mirror image of the point charge."},{"Start":"12:11.725 ","End":"12:14.430","Text":"Then you can solve this very easily."},{"Start":"12:14.430 ","End":"12:18.260","Text":"Then the potential in our area over here is just going to"},{"Start":"12:18.260 ","End":"12:23.000","Text":"be the superposition of all the potentials due to all of these point charges."},{"Start":"12:23.000 ","End":"12:25.980","Text":"That\u0027s the end of this lesson."}],"ID":14215},{"Watched":false,"Name":"Not Point Charges","Duration":"19m 41s","ChapterTopicVideoID":12113,"CourseChapterTopicPlaylistID":161612,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12113.jpeg","UploadDate":"2018-06-28T03:32:31.7400000","DurationForVideoObject":"PT19M41S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.665","Text":"Hello. In this lesson,"},{"Start":"00:01.665 ","End":"00:07.590","Text":"we\u0027re going to learn how to work with charges that aren\u0027t point charges."},{"Start":"00:07.590 ","End":"00:12.480","Text":"We\u0027re going to try and give it some intuition to see how this works,"},{"Start":"00:12.480 ","End":"00:16.720","Text":"and it\u0027s another way of thinking of these types of problems."},{"Start":"00:17.510 ","End":"00:22.920","Text":"Here we have our grounded infinite plane and we already know how to"},{"Start":"00:22.920 ","End":"00:28.605","Text":"answer a question where we have a single-point charge above this plane."},{"Start":"00:28.605 ","End":"00:35.490","Text":"Now we\u0027re going to look at how to solve a question where we have a finite wire."},{"Start":"00:35.490 ","End":"00:43.085","Text":"Here we have a wire and the wire has charged density Lambda,"},{"Start":"00:43.085 ","End":"00:49.195","Text":"and it is a distance of d away from the grounded infinite plane."},{"Start":"00:49.195 ","End":"00:51.380","Text":"Now what I want to do,"},{"Start":"00:51.380 ","End":"00:52.655","Text":"as with all these questions,"},{"Start":"00:52.655 ","End":"00:59.400","Text":"is I wanted to find the potential all the way in this space over here."},{"Start":"00:59.400 ","End":"01:03.785","Text":"We\u0027ve already seen that if I had some point charge,"},{"Start":"01:03.785 ","End":"01:06.410","Text":"then I would have to use a mirror image over"},{"Start":"01:06.410 ","End":"01:09.635","Text":"here on the opposite side of the grounded plane,"},{"Start":"01:09.635 ","End":"01:12.215","Text":"but with a negative charge."},{"Start":"01:12.215 ","End":"01:14.840","Text":"Now what I\u0027m going to do is I\u0027m going to"},{"Start":"01:14.840 ","End":"01:21.440","Text":"find a mirror image"},{"Start":"01:21.440 ","End":"01:26.310","Text":"of every single point along this finite wire."},{"Start":"01:26.870 ","End":"01:31.640","Text":"In other words, I\u0027m going to split up my wire into lots of little pieces."},{"Start":"01:31.640 ","End":"01:35.990","Text":"For each piece, I\u0027m going to pretend like it\u0027s a point charge and use the method of"},{"Start":"01:35.990 ","End":"01:41.970","Text":"images on that point charge and sum up all the way along the wire."},{"Start":"01:41.970 ","End":"01:45.413","Text":"If I take this piece over here,"},{"Start":"01:45.413 ","End":"01:51.285","Text":"its mirror image is going to be this piece over here."},{"Start":"01:51.285 ","End":"01:53.420","Text":"This piece is of course,"},{"Start":"01:53.420 ","End":"01:59.375","Text":"also a distance d away from the infinite plane."},{"Start":"01:59.375 ","End":"02:02.855","Text":"If I keep doing this for all of the pieces along the wire,"},{"Start":"02:02.855 ","End":"02:09.210","Text":"I\u0027m just going to get the wire or like so."},{"Start":"02:10.700 ","End":"02:15.350","Text":"Then if my charge density over here is Lambda,"},{"Start":"02:15.350 ","End":"02:22.020","Text":"so I have to have a charge density down here of negative Lambda."},{"Start":"02:22.490 ","End":"02:28.325","Text":"Now, if I look at the potential at every single point on this infinite plane,"},{"Start":"02:28.325 ","End":"02:31.100","Text":"I\u0027ll see that the potential along"},{"Start":"02:31.100 ","End":"02:34.880","Text":"here is going to be equal to 0 because it will be canceled"},{"Start":"02:34.880 ","End":"02:38.240","Text":"out by the positive value due to Lambda over"},{"Start":"02:38.240 ","End":"02:42.440","Text":"here and the negative value due to negative Lambda over here."},{"Start":"02:42.440 ","End":"02:46.805","Text":"I have a 0 potential along the grounded plane."},{"Start":"02:46.805 ","End":"02:53.435","Text":"Therefore I know that in order to work out the potential in this gray region over here,"},{"Start":"02:53.435 ","End":"03:00.380","Text":"all I have to do is work out the potential of two tiles where one has a charge of Lambda"},{"Start":"03:00.380 ","End":"03:03.620","Text":"and the other has a charge of negative Lambda and where"},{"Start":"03:03.620 ","End":"03:08.250","Text":"the two tiles are a distance of 2d away from one another."},{"Start":"03:09.170 ","End":"03:15.695","Text":"That\u0027s if we have a wire that is parallel to our infinite plane."},{"Start":"03:15.695 ","End":"03:21.120","Text":"What happens if the wire is at an angle to our infinite plane?"},{"Start":"03:22.310 ","End":"03:25.130","Text":"Here, we have some kind of angle."},{"Start":"03:25.130 ","End":"03:27.710","Text":"I\u0027m going to do the same thing as I did before."},{"Start":"03:27.710 ","End":"03:30.260","Text":"I\u0027m going to split this wire into lots of"},{"Start":"03:30.260 ","End":"03:35.375","Text":"little pieces and then I\u0027m going to pretend that each piece is a point charge."},{"Start":"03:35.375 ","End":"03:39.180","Text":"Then I\u0027m going to use the method of images."},{"Start":"03:39.830 ","End":"03:47.075","Text":"What we\u0027ll see is that let\u0027s say that this distance over here is d_2."},{"Start":"03:47.075 ","End":"03:51.440","Text":"What we\u0027ll see is it\u0027s just like the method of images,"},{"Start":"03:51.440 ","End":"03:53.060","Text":"just like with the point charge,"},{"Start":"03:53.060 ","End":"03:56.435","Text":"just like with the regular wire that\u0027s parallel."},{"Start":"03:56.435 ","End":"04:01.985","Text":"This will be the mirror image of it will be over here,"},{"Start":"04:01.985 ","End":"04:06.890","Text":"which is of course a distance of d_2 away from"},{"Start":"04:06.890 ","End":"04:14.505","Text":"the infinite plane and we\u0027ll keep going along until we get to this point over here."},{"Start":"04:14.505 ","End":"04:23.010","Text":"This is of course a distance of d away and then we can draw the entire wire like so."},{"Start":"04:23.010 ","End":"04:28.715","Text":"We can see that this wire is a mirror image of our original wire."},{"Start":"04:28.715 ","End":"04:33.605","Text":"Just like we would have expected in what we saw with our point charges and of course,"},{"Start":"04:33.605 ","End":"04:36.500","Text":"this will be charged with negative Lambda again."},{"Start":"04:36.500 ","End":"04:39.050","Text":"Then we can check each point along"},{"Start":"04:39.050 ","End":"04:43.550","Text":"this infinite plane and the potential at each point is going to be 0 because it"},{"Start":"04:43.550 ","End":"04:46.970","Text":"cancels out from the positive value of potential from"},{"Start":"04:46.970 ","End":"04:51.755","Text":"Lambda with a negative value over here from negative Lambda."},{"Start":"04:51.755 ","End":"04:57.515","Text":"Therefore, to work out the potential in this region with the gray lines,"},{"Start":"04:57.515 ","End":"05:01.430","Text":"we can just work out the potential of two wires"},{"Start":"05:01.430 ","End":"05:05.615","Text":"that are at a certain angle to one another,"},{"Start":"05:05.615 ","End":"05:14.530","Text":"and a distance starting from 2d until a maximum distance of d2 away from one another."},{"Start":"05:14.720 ","End":"05:18.640","Text":"Now, let\u0027s look at another example."},{"Start":"05:18.640 ","End":"05:23.075","Text":"Let\u0027s imagine that we have this arch shape."},{"Start":"05:23.075 ","End":"05:30.125","Text":"What we can see is that this point over here will be reflected over here."},{"Start":"05:30.125 ","End":"05:34.670","Text":"This point over here will be reflected in the mirror over"},{"Start":"05:34.670 ","End":"05:40.845","Text":"here and this point over here will be reflected over here."},{"Start":"05:40.845 ","End":"05:43.115","Text":"Of course, I didn\u0027t draw this to scale."},{"Start":"05:43.115 ","End":"05:46.745","Text":"But what will we get if we draw this properly is"},{"Start":"05:46.745 ","End":"05:54.370","Text":"the mirror image of this arch up top over here."},{"Start":"05:54.740 ","End":"05:59.750","Text":"Let\u0027s imagine that this is the mirror image and so it will look like this."},{"Start":"05:59.750 ","End":"06:02.345","Text":"Then in order to find the potential here,"},{"Start":"06:02.345 ","End":"06:08.595","Text":"we\u0027ll just find the potential of these two arches when they\u0027re in this formation."},{"Start":"06:08.595 ","End":"06:15.060","Text":"We can see that from symmetry the potential along here will be equal to 0."},{"Start":"06:15.380 ","End":"06:18.080","Text":"Now we have the general idea."},{"Start":"06:18.080 ","End":"06:22.770","Text":"Let\u0027s look at a slightly more difficult example."},{"Start":"06:23.350 ","End":"06:29.060","Text":"Here, we have a spherical shell of radius R and it\u0027s grounded."},{"Start":"06:29.060 ","End":"06:32.570","Text":"We\u0027ve already seen what happens or how to solve"},{"Start":"06:32.570 ","End":"06:37.400","Text":"a question where we have a point charge located outside of the spherical shell."},{"Start":"06:37.400 ","End":"06:44.700","Text":"But what happens if instead of a point charge we have a finite wire?"},{"Start":"06:45.380 ","End":"06:55.855","Text":"Let\u0027s imagine that this edge is a distance a away from the center of the sphere"},{"Start":"06:55.855 ","End":"07:02.110","Text":"and that the length of this wire"},{"Start":"07:02.110 ","End":"07:09.710","Text":"is L and that the wire is charged with charge density Lambda."},{"Start":"07:09.740 ","End":"07:16.345","Text":"Now, I want to find the image that I can put somewhere here"},{"Start":"07:16.345 ","End":"07:24.080","Text":"such that I\u0027ll get a 0 potential along the spherical shell."},{"Start":"07:24.170 ","End":"07:27.285","Text":"Remember when dealing with a sphere,"},{"Start":"07:27.285 ","End":"07:32.970","Text":"we know that the charge of our image inside the sphere."},{"Start":"07:32.970 ","End":"07:40.925","Text":"We call that q tag has to be equal to negative the radius of the sphere divided by a."},{"Start":"07:40.925 ","End":"07:44.105","Text":"The distance from the center of the sphere to"},{"Start":"07:44.105 ","End":"07:49.160","Text":"the closest point of our object multiplied by q,"},{"Start":"07:49.160 ","End":"07:54.755","Text":"and b, which is"},{"Start":"07:54.755 ","End":"08:01.980","Text":"the distance of the new q tag charge within the sphere."},{"Start":"08:01.980 ","End":"08:04.970","Text":"The distance of this q tag charge away from the center of"},{"Start":"08:04.970 ","End":"08:10.250","Text":"the sphere so that is equal to R divided by a"},{"Start":"08:10.250 ","End":"08:20.420","Text":"multiplied by R. Let\u0027s say if we\u0027re looking at the tip of this wire over here."},{"Start":"08:20.420 ","End":"08:22.820","Text":"We can say that here,"},{"Start":"08:22.820 ","End":"08:30.475","Text":"we\u0027ll have the charge q and this distance over here,"},{"Start":"08:30.475 ","End":"08:34.460","Text":"let\u0027s say in green or in blue over here,"},{"Start":"08:34.460 ","End":"08:41.905","Text":"will be b from the center until our new charge over here, our imaginary charge."},{"Start":"08:41.905 ","End":"08:46.930","Text":"Now, notice the relationship between b and a."},{"Start":"08:46.930 ","End":"08:49.780","Text":"As a becomes bigger,"},{"Start":"08:49.780 ","End":"08:54.430","Text":"so as the distance of this wire or the piece along the wire is"},{"Start":"08:54.430 ","End":"09:00.725","Text":"further away from the center of the sphere so b will become smaller."},{"Start":"09:00.725 ","End":"09:10.035","Text":"But the closer we look to the sphere the smallest b will be."},{"Start":"09:10.035 ","End":"09:17.270","Text":"That means that we\u0027ll get closer to the center of the sphere for our image charge."},{"Start":"09:18.360 ","End":"09:22.270","Text":"Let\u0027s say that this is our first particle,"},{"Start":"09:22.270 ","End":"09:24.300","Text":"or our first point charge."},{"Start":"09:24.300 ","End":"09:29.165","Text":"Let\u0027s say that we call this a distance of a_1 away."},{"Start":"09:29.165 ","End":"09:33.170","Text":"Then, if we look at our last piece over here,"},{"Start":"09:33.170 ","End":"09:35.945","Text":"let\u0027s call this a distance of a_2 away,"},{"Start":"09:35.945 ","End":"09:39.710","Text":"or really it\u0027s a distance of a plus L away."},{"Start":"09:39.710 ","End":"09:44.660","Text":"Here, this we can call a distance b_1."},{"Start":"09:44.660 ","End":"09:48.245","Text":"That\u0027s where a_1 will be reflected to."},{"Start":"09:48.245 ","End":"09:54.185","Text":"But a_2 will be reflected somewhere here."},{"Start":"09:54.185 ","End":"09:59.910","Text":"This will be our reflection of a_2."},{"Start":"10:01.140 ","End":"10:06.580","Text":"This small distance over here is distance b_2."},{"Start":"10:06.580 ","End":"10:12.520","Text":"We can see that we have this mirror image going on over here."},{"Start":"10:12.520 ","End":"10:19.000","Text":"The closest point over here to the sphere is also the closest to the sphere,"},{"Start":"10:19.000 ","End":"10:23.290","Text":"but from the inside and the furthest point away from the sphere,"},{"Start":"10:23.290 ","End":"10:26.800","Text":"from the outside is also the furthest point away from"},{"Start":"10:26.800 ","End":"10:31.340","Text":"the surface of the sphere from the inside."},{"Start":"10:31.710 ","End":"10:39.400","Text":"What this shows us is that if we have a wire of length L located outside the sphere,"},{"Start":"10:39.400 ","End":"10:42.145","Text":"then we can also see how long"},{"Start":"10:42.145 ","End":"10:50.515","Text":"this reflected wire inside the sphere is going to be."},{"Start":"10:50.515 ","End":"10:52.810","Text":"We can also find this length."},{"Start":"10:52.810 ","End":"10:55.570","Text":"This length is called L tag."},{"Start":"10:55.570 ","End":"11:05.881","Text":"The length inside the sphere and L tag is simply going to be equal to be b_1 minus b_2."},{"Start":"11:05.881 ","End":"11:08.080","Text":"Then we\u0027ll be left with this black length over here."},{"Start":"11:08.080 ","End":"11:18.055","Text":"B_1 is simply going to be R divided by a_1 multiplied by R minus b_2,"},{"Start":"11:18.055 ","End":"11:22.705","Text":"which is R, divided by a_2."},{"Start":"11:22.705 ","End":"11:27.880","Text":"So a_2, we already said is a plus L or a_1, sorry,"},{"Start":"11:27.880 ","End":"11:38.380","Text":"plus L multiplied by R. We can see that A1 is the same length as a."},{"Start":"11:38.380 ","End":"11:43.165","Text":"We can see that a_1 is equal to a from this diagram."},{"Start":"11:43.165 ","End":"11:49.255","Text":"Just to make this a slightly more general equation."},{"Start":"11:49.255 ","End":"11:52.730","Text":"We can go like so."},{"Start":"11:53.100 ","End":"11:56.110","Text":"If you want, you can simplify it further,"},{"Start":"11:56.110 ","End":"11:57.370","Text":"but it doesn\u0027t really matter."},{"Start":"11:57.370 ","End":"11:59.770","Text":"What we can see is that the L tag,"},{"Start":"11:59.770 ","End":"12:05.560","Text":"which is the reflected length of this wire,"},{"Start":"12:05.560 ","End":"12:10.180","Text":"of length L, is going to be a different length inside the sphere."},{"Start":"12:10.180 ","End":"12:16.460","Text":"We can see that the length isn\u0027t going to be conserved."},{"Start":"12:17.250 ","End":"12:21.025","Text":"The next thing that we want to do is,"},{"Start":"12:21.025 ","End":"12:29.140","Text":"we know that this wire over here outside has a charge density of Lambda."},{"Start":"12:29.140 ","End":"12:32.290","Text":"What does this wire inside the sphere,"},{"Start":"12:32.290 ","End":"12:34.630","Text":"what is its charged density?"},{"Start":"12:34.630 ","End":"12:38.635","Text":"Let\u0027s call its charge density Lambda tag."},{"Start":"12:38.635 ","End":"12:41.650","Text":"Right now we want to work out what it is."},{"Start":"12:41.650 ","End":"12:45.310","Text":"What we\u0027re going to do is we\u0027re going to look"},{"Start":"12:45.310 ","End":"12:49.600","Text":"at some random piece in the middle of the reflected wire."},{"Start":"12:49.600 ","End":"12:52.160","Text":"Let\u0027s say this over here."},{"Start":"12:52.770 ","End":"12:55.585","Text":"We have this random piece,"},{"Start":"12:55.585 ","End":"13:04.000","Text":"and let\u0027s say that it is a distance from here to this piece."},{"Start":"13:04.000 ","End":"13:07.645","Text":"This is a distance of, let\u0027s call it x."},{"Start":"13:07.645 ","End":"13:12.910","Text":"Then we know that this piece came from the reflection of,"},{"Start":"13:12.910 ","End":"13:15.325","Text":"let\u0027s say, this piece over here."},{"Start":"13:15.325 ","End":"13:19.855","Text":"Let\u0027s say that the distance to this piece over here,"},{"Start":"13:19.855 ","End":"13:23.750","Text":"let\u0027s call this X_2."},{"Start":"13:25.290 ","End":"13:31.840","Text":"My relationship between x and X_2 is the same relationship between b and a."},{"Start":"13:31.840 ","End":"13:34.285","Text":"Let\u0027s write it up here."},{"Start":"13:34.285 ","End":"13:37.390","Text":"I have that X, which is my b,"},{"Start":"13:37.390 ","End":"13:41.965","Text":"is equal to I divided by my a."},{"Start":"13:41.965 ","End":"13:47.335","Text":"My a over here is this X_2 multiplied by"},{"Start":"13:47.335 ","End":"13:57.085","Text":"R. This charge over here on my wire is equal to dq."},{"Start":"13:57.085 ","End":"13:59.320","Text":"What is my dq?"},{"Start":"13:59.320 ","End":"14:07.525","Text":"My dq is equal to the charge density Lambda multiplied by dl."},{"Start":"14:07.525 ","End":"14:11.080","Text":"Or in here because we\u0027re using x coordinate,"},{"Start":"14:11.080 ","End":"14:14.720","Text":"so I can say dx."},{"Start":"14:16.260 ","End":"14:18.430","Text":"This point charge over here,"},{"Start":"14:18.430 ","End":"14:20.634","Text":"which is the reflection of dq,"},{"Start":"14:20.634 ","End":"14:22.960","Text":"also has a charge."},{"Start":"14:22.960 ","End":"14:26.539","Text":"Let\u0027s call it dq tag."},{"Start":"14:26.539 ","End":"14:31.570","Text":"Dq tag is equal to the charge density of this point,"},{"Start":"14:31.570 ","End":"14:33.460","Text":"it\u0027s the reflected points and we don\u0027t know,"},{"Start":"14:33.460 ","End":"14:34.885","Text":"so it\u0027s Lambda tag."},{"Start":"14:34.885 ","End":"14:39.490","Text":"We\u0027re trying to find that and also dx."},{"Start":"14:39.490 ","End":"14:42.325","Text":"Dq tag is simply a charge."},{"Start":"14:42.325 ","End":"14:46.570","Text":"Now, we have this equation for q tag over here."},{"Start":"14:46.570 ","End":"14:49.240","Text":"We can see that dq tag,"},{"Start":"14:49.240 ","End":"14:54.895","Text":"which is equal to Lambda tag dx is also equal to."},{"Start":"14:54.895 ","End":"14:59.950","Text":"We have negative R divided by a."},{"Start":"14:59.950 ","End":"15:01.839","Text":"A, in this case,"},{"Start":"15:01.839 ","End":"15:10.810","Text":"is in fact x_2 divided by x_2 multiplied by q multiplied by,"},{"Start":"15:10.810 ","End":"15:12.490","Text":"in this case, dq,"},{"Start":"15:12.490 ","End":"15:15.760","Text":"because the charge of this point isn\u0027t q is dq."},{"Start":"15:15.760 ","End":"15:17.395","Text":"What is dq?"},{"Start":"15:17.395 ","End":"15:22.495","Text":"Dq is here Lambda dx."},{"Start":"15:22.495 ","End":"15:29.665","Text":"Then what I can do is I can divide both sides by dx."},{"Start":"15:29.665 ","End":"15:36.835","Text":"Then, therefore, what I\u0027m left with is that Lambda tag is equal to"},{"Start":"15:36.835 ","End":"15:44.360","Text":"negative R Lambda divided by x_2."},{"Start":"15:45.600 ","End":"15:53.845","Text":"That means that we have negative R Lambda and x_2."},{"Start":"15:53.845 ","End":"15:59.035","Text":"So x_2 is our a. I can isolate out my a over here."},{"Start":"15:59.035 ","End":"16:07.340","Text":"A is going to be b divided by R^2."},{"Start":"16:08.580 ","End":"16:10.975","Text":"Sorry my mistake,"},{"Start":"16:10.975 ","End":"16:20.455","Text":"a is equal to k equal to R^2 divided by b. X_2 over here is taking the place of our a."},{"Start":"16:20.455 ","End":"16:24.970","Text":"If we have negative R lambda divided by x_2, which is a."},{"Start":"16:24.970 ","End":"16:29.950","Text":"It\u0027s divided by R^2 divided by b."},{"Start":"16:29.950 ","End":"16:31.855","Text":"But what is b over here?"},{"Start":"16:31.855 ","End":"16:35.780","Text":"B and R specific example over here was x."},{"Start":"16:38.040 ","End":"16:43.585","Text":"Then what we\u0027ll get is that Lambda tag is in fact equal to"},{"Start":"16:43.585 ","End":"16:49.610","Text":"negative x divided by R multiplied by Lambda."},{"Start":"16:50.550 ","End":"16:53.740","Text":"To sum up, let\u0027s just put this here."},{"Start":"16:53.740 ","End":"16:56.860","Text":"We\u0027ll get that Lambda tag is equal to negative x,"},{"Start":"16:56.860 ","End":"17:03.535","Text":"which is in fact b divided by R multiplied by Lambda."},{"Start":"17:03.535 ","End":"17:09.400","Text":"We can see that the charge density of this reflected wire"},{"Start":"17:09.400 ","End":"17:16.250","Text":"within this spherical shell is also going to be different to the original charge density."},{"Start":"17:16.710 ","End":"17:20.830","Text":"Now, let\u0027s look at another example."},{"Start":"17:20.830 ","End":"17:27.595","Text":"In this example, we have this perpendicular line going like so."},{"Start":"17:27.595 ","End":"17:30.520","Text":"If we look at this point over here,"},{"Start":"17:30.520 ","End":"17:33.490","Text":"which is exactly a distance a away."},{"Start":"17:33.490 ","End":"17:38.140","Text":"We can say that it\u0027s b position,"},{"Start":"17:38.140 ","End":"17:43.030","Text":"it\u0027s reflected image is approximately here."},{"Start":"17:43.030 ","End":"17:46.660","Text":"However, if we look at this point over here,"},{"Start":"17:46.660 ","End":"17:53.590","Text":"we can see that its distance from the center of the sphere is incredibly long."},{"Start":"17:53.590 ","End":"17:55.360","Text":"It\u0027s longer than a,"},{"Start":"17:55.360 ","End":"18:03.910","Text":"because its length is going from the center of the sphere and like so in a diagonal."},{"Start":"18:03.910 ","End":"18:07.940","Text":"This is a_2 and it is longer than a."},{"Start":"18:09.840 ","End":"18:14.590","Text":"We can see that our a value is increasing over here."},{"Start":"18:14.590 ","End":"18:18.070","Text":"It\u0027s like a 90-degree triangle where this is the hypotenuse,"},{"Start":"18:18.070 ","End":"18:19.660","Text":"so it\u0027s the longest side."},{"Start":"18:19.660 ","End":"18:21.790","Text":"Here, our a is increasing,"},{"Start":"18:21.790 ","End":"18:25.030","Text":"so the denominator is increasing."},{"Start":"18:25.030 ","End":"18:28.210","Text":"This fraction as a whole is decreasing,"},{"Start":"18:28.210 ","End":"18:30.655","Text":"which means that b is decreasing."},{"Start":"18:30.655 ","End":"18:35.005","Text":"Then, we\u0027ll see that our point over here,"},{"Start":"18:35.005 ","End":"18:40.510","Text":"the reflection, which will be here around b_2,"},{"Start":"18:40.510 ","End":"18:43.000","Text":"is going to be much shorter."},{"Start":"18:43.000 ","End":"18:49.795","Text":"Our reflected charge b_2 is going to be much closer to the center of the sphere."},{"Start":"18:49.795 ","End":"18:52.000","Text":"What we can see is that instead of getting"},{"Start":"18:52.000 ","End":"18:57.370","Text":"some semicircle or part of a circle inside our sphere,"},{"Start":"18:57.370 ","End":"18:59.320","Text":"our radius is changing."},{"Start":"18:59.320 ","End":"19:01.435","Text":"We have b_2 which is short,"},{"Start":"19:01.435 ","End":"19:03.100","Text":"and then b which is longer,"},{"Start":"19:03.100 ","End":"19:05.875","Text":"and then it will go back to being short."},{"Start":"19:05.875 ","End":"19:07.930","Text":"Our radius is changing."},{"Start":"19:07.930 ","End":"19:16.700","Text":"We\u0027re going to get maybe some kind of curve that looks like so where this is the center."},{"Start":"19:17.550 ","End":"19:20.380","Text":"This is the center of the sphere so we\u0027ll get"},{"Start":"19:20.380 ","End":"19:27.305","Text":"some kind of R shape that looks something like so."},{"Start":"19:27.305 ","End":"19:30.040","Text":"That is the end of the lesson."},{"Start":"19:30.040 ","End":"19:34.120","Text":"I hope that you have some greater intuition on what to"},{"Start":"19:34.120 ","End":"19:39.145","Text":"do if you don\u0027t have a point charge when you\u0027re working out this type of question."},{"Start":"19:39.145 ","End":"19:42.290","Text":"That is the end of this lesson."}],"ID":14216},{"Watched":false,"Name":"Finding Charge Density on Surface of Conductors","Duration":"24m 22s","ChapterTopicVideoID":12114,"CourseChapterTopicPlaylistID":161612,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12114.jpeg","UploadDate":"2018-06-28T03:38:39.7530000","DurationForVideoObject":"PT24M22S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.875","Text":"Hello. In this lesson,"},{"Start":"00:01.875 ","End":"00:07.890","Text":"we\u0027re going to learn how to find the charge density on the surface of a conductor."},{"Start":"00:07.890 ","End":"00:12.240","Text":"The first thing that we\u0027re going to do is we\u0027re going to look at this example,"},{"Start":"00:12.240 ","End":"00:21.205","Text":"where we have an infinite plane which is grounded and we have a point charge on top."},{"Start":"00:21.205 ","End":"00:23.295","Text":"Then after we look at this example,"},{"Start":"00:23.295 ","End":"00:25.380","Text":"we\u0027ll look at the spherical example,"},{"Start":"00:25.380 ","End":"00:27.940","Text":"where we have a grounded sphere."},{"Start":"00:28.910 ","End":"00:35.219","Text":"Of course, we\u0027re looking at the charge density on the surface of the conductor."},{"Start":"00:35.219 ","End":"00:40.124","Text":"As we\u0027ve seen before that in-between the surface of the conductor,"},{"Start":"00:40.124 ","End":"00:42.240","Text":"so we won\u0027t have"},{"Start":"00:42.240 ","End":"00:47.690","Text":"a charge density and the charge density is concentrated along the surface."},{"Start":"00:47.690 ","End":"00:50.750","Text":"If we\u0027re dealing with a surface charge,"},{"Start":"00:50.750 ","End":"00:55.113","Text":"of course we know that we\u0027re going to be using Sigma."},{"Start":"00:55.113 ","End":"01:01.325","Text":"So the equation for the charge density per unit area is equal to"},{"Start":"01:01.325 ","End":"01:08.940","Text":"Epsilon naught multiplied by Delta E. This is the jump in the electric field."},{"Start":"01:08.940 ","End":"01:12.035","Text":"Of course we\u0027re speaking about the jump in the electric field,"},{"Start":"01:12.035 ","End":"01:15.940","Text":"which is perpendicular to our surface."},{"Start":"01:15.940 ","End":"01:21.830","Text":"What we can see over here is that our surface is on the x,"},{"Start":"01:21.830 ","End":"01:27.425","Text":"y plane, so our Sigma is going to be in the x, y plane."},{"Start":"01:27.425 ","End":"01:31.130","Text":"However, the electric fields that we\u0027re looking at is going to"},{"Start":"01:31.130 ","End":"01:36.120","Text":"be perpendicular to the plane in the z direction,"},{"Start":"01:36.120 ","End":"01:39.520","Text":"so we can call this E_z."},{"Start":"01:39.890 ","End":"01:44.645","Text":"In order to find the jump in the electric field,"},{"Start":"01:44.645 ","End":"01:48.745","Text":"I choose a point slightly above my surface."},{"Start":"01:48.745 ","End":"01:53.550","Text":"Then I look for the electrical field over here,"},{"Start":"01:53.550 ","End":"01:56.100","Text":"so this will be E_z plus."},{"Start":"01:56.100 ","End":"01:59.300","Text":"Then I look at a points slightly below my surface over"},{"Start":"01:59.300 ","End":"02:03.500","Text":"here and I look for the electric field over here,"},{"Start":"02:03.500 ","End":"02:06.960","Text":"so this will be E_z minus."},{"Start":"02:06.980 ","End":"02:11.225","Text":"In that case, my equation for Sigma turns into"},{"Start":"02:11.225 ","End":"02:15.620","Text":"epsilon naught multiplied by this jump in the electric field,"},{"Start":"02:15.620 ","End":"02:24.720","Text":"which means E_z plus minus E_z minus."},{"Start":"02:24.850 ","End":"02:33.020","Text":"Usually, my E_z minus is going to be equal to 0. Why is that?"},{"Start":"02:33.020 ","End":"02:35.940","Text":"We know that this is a grounded surface,"},{"Start":"02:35.940 ","End":"02:41.950","Text":"so that means that the potential over here at the surface is equal to 0."},{"Start":"02:41.950 ","End":"02:49.500","Text":"We also know that the potential at infinity is equal to 0."},{"Start":"02:49.500 ","End":"02:52.900","Text":"Therefore, if the potential at the surface is"},{"Start":"02:52.900 ","End":"02:55.915","Text":"equal to 0 and the potential at infinity is equal to"},{"Start":"02:55.915 ","End":"03:02.920","Text":"0 and we\u0027ve nothing in the space over here that could produce a potential,"},{"Start":"03:03.740 ","End":"03:10.520","Text":"we can therefore say that there was no electric field in this area over here."},{"Start":"03:10.820 ","End":"03:14.535","Text":"If my E_z minus is equal to 0,"},{"Start":"03:14.535 ","End":"03:22.830","Text":"I can say therefore that my Sigma is equal to Epsilon naught multiplied by E_z plus."},{"Start":"03:22.830 ","End":"03:27.340","Text":"What I want to do is, I want to work out the electric field over here in"},{"Start":"03:27.340 ","End":"03:32.020","Text":"this region at a point that is very close to my surface,"},{"Start":"03:32.020 ","End":"03:36.700","Text":"so close that the point is approaching being on the surface."},{"Start":"03:36.700 ","End":"03:38.510","Text":"Where is that?"},{"Start":"03:38.510 ","End":"03:45.025","Text":"Epsilon naught multiplied by the electric field in the z direction,"},{"Start":"03:45.025 ","End":"03:50.120","Text":"E_z and z is equal to 0."},{"Start":"03:50.720 ","End":"03:54.665","Text":"How do we solve a question like this?"},{"Start":"03:54.665 ","End":"03:58.775","Text":"What do we do is, we use the method of images."},{"Start":"03:58.775 ","End":"04:02.495","Text":"That\u0027s where I imagined that I don\u0027t have"},{"Start":"04:02.495 ","End":"04:06.785","Text":"this surface over here and I use my method of images."},{"Start":"04:06.785 ","End":"04:14.085","Text":"Let\u0027s represent the surface by this dotted line over here, it\u0027s invisible."},{"Start":"04:14.085 ","End":"04:16.550","Text":"Then we have my point charge over here,"},{"Start":"04:16.550 ","End":"04:19.340","Text":"q and then with my method of images."},{"Start":"04:19.340 ","End":"04:21.605","Text":"In the mirror image of this dotted line,"},{"Start":"04:21.605 ","End":"04:24.905","Text":"I have another point charge, negative q."},{"Start":"04:24.905 ","End":"04:28.250","Text":"Of course, this is the z direction."},{"Start":"04:28.250 ","End":"04:35.960","Text":"This is still the x direction and this is still the y direction."},{"Start":"04:36.290 ","End":"04:40.985","Text":"Let\u0027s begin by calculating this electric field."},{"Start":"04:40.985 ","End":"04:44.540","Text":"Let\u0027s say that I\u0027m looking at a point over here."},{"Start":"04:44.540 ","End":"04:50.825","Text":"We\u0027re located at z is equal to 0 or a point that\u0027s just above that,"},{"Start":"04:50.825 ","End":"04:53.765","Text":"but we can consider it as z equal to 0."},{"Start":"04:53.765 ","End":"04:58.600","Text":"Let\u0027s say that the distance from the origin until that point,"},{"Start":"04:58.600 ","End":"05:07.010","Text":"so this distance over here is r. On this point over here and of course,"},{"Start":"05:07.010 ","End":"05:09.920","Text":"we can place it anywhere on the x, y plane,"},{"Start":"05:09.920 ","End":"05:11.525","Text":"because that\u0027s where the surface is,"},{"Start":"05:11.525 ","End":"05:14.450","Text":"as long as it\u0027s at z is equal to 0."},{"Start":"05:14.450 ","End":"05:19.490","Text":"Our q charge is exerting an electric field on the point in"},{"Start":"05:19.490 ","End":"05:26.365","Text":"this direction and our negative q is exerting an electric field in this direction."},{"Start":"05:26.365 ","End":"05:30.620","Text":"From symmetry that we have equal and opposite charges,"},{"Start":"05:30.620 ","End":"05:33.740","Text":"so we can see that we won\u0027t have"},{"Start":"05:33.740 ","End":"05:38.160","Text":"an electric field when we add these 2 vectors up together."},{"Start":"05:38.160 ","End":"05:44.010","Text":"We won\u0027t have an electric field in the radial direction, but what we will,"},{"Start":"05:44.010 ","End":"05:49.655","Text":"is we\u0027ll have a resultant electric field in the z direction,"},{"Start":"05:49.655 ","End":"05:53.340","Text":"going in the negative z direction specifically."},{"Start":"05:53.900 ","End":"06:02.180","Text":"Now let\u0027s define this distance to our q as distance d and of course,"},{"Start":"06:02.180 ","End":"06:07.150","Text":"the distance to negative q is also d just going down here."},{"Start":"06:07.150 ","End":"06:09.580","Text":"What are we going to do is,"},{"Start":"06:09.580 ","End":"06:14.240","Text":"we\u0027re going to calculate the electric field due to charge q on"},{"Start":"06:14.240 ","End":"06:19.370","Text":"our point and we\u0027re going to find its magnitude in the z direction."},{"Start":"06:19.370 ","End":"06:22.880","Text":"Then we\u0027re just going to multiply it by 2 to give"},{"Start":"06:22.880 ","End":"06:27.390","Text":"us what we\u0027ll get also from our negative q charge."},{"Start":"06:27.390 ","End":"06:31.650","Text":"Let\u0027s rub this out over here,"},{"Start":"06:31.650 ","End":"06:34.985","Text":"we won\u0027t look at that and we\u0027re not yet looking at this."},{"Start":"06:34.985 ","End":"06:40.800","Text":"We\u0027re just looking at this electric field over here."},{"Start":"06:40.800 ","End":"06:48.785","Text":"Then we\u0027ll have going from our q down to here,"},{"Start":"06:48.785 ","End":"06:54.030","Text":"we have this vector over here and let\u0027s call that r_1."},{"Start":"06:54.850 ","End":"07:00.280","Text":"Now we can say that the magnitude of"},{"Start":"07:00.280 ","End":"07:06.045","Text":"our E field due to charge q is equal to, from Coulomb\u0027s law."},{"Start":"07:06.045 ","End":"07:11.715","Text":"We know it\u0027s kq divided by r_1^2."},{"Start":"07:11.715 ","End":"07:15.000","Text":"We don\u0027t know what r_1 is."},{"Start":"07:15.000 ","End":"07:21.750","Text":"We\u0027re going to say that this is equal to kq divided by,"},{"Start":"07:21.750 ","End":"07:25.140","Text":"we have r^2,"},{"Start":"07:25.140 ","End":"07:30.670","Text":"we\u0027re using Pythagoras plus d^2."},{"Start":"07:30.860 ","End":"07:36.910","Text":"From Pythagoras, this r_1 is the hypotenuse."},{"Start":"07:37.970 ","End":"07:45.840","Text":"What we\u0027re trying to find is the electric field in this z direction."},{"Start":"07:47.420 ","End":"07:56.160","Text":"What we can do is we can say that this angle between E and E_z is some angle Alpha."},{"Start":"07:56.240 ","End":"07:59.830","Text":"This is the same triangle as this over here,"},{"Start":"07:59.830 ","End":"08:02.360","Text":"so if this angle over here is Alpha,"},{"Start":"08:02.360 ","End":"08:07.375","Text":"then this angle over here will also be Alpha."},{"Start":"08:07.375 ","End":"08:13.220","Text":"Therefore, we can say that our E_z is going to be"},{"Start":"08:13.220 ","End":"08:19.600","Text":"equal to the magnitude our E field multiplied by,"},{"Start":"08:19.600 ","End":"08:21.750","Text":"we want it in the z direction,"},{"Start":"08:21.750 ","End":"08:25.425","Text":"so that is the adjacent side."},{"Start":"08:25.425 ","End":"08:30.780","Text":"It\u0027s going to be E multiplied by cosine of Alpha."},{"Start":"08:30.780 ","End":"08:34.330","Text":"What is cosine of Alpha?"},{"Start":"08:34.430 ","End":"08:42.540","Text":"From our trig identities we know that cosine of Alpha is adjacent over hypotenuse."},{"Start":"08:42.540 ","End":"08:46.395","Text":"We have E multiplied by cosine Alpha,"},{"Start":"08:46.395 ","End":"08:50.270","Text":"so the adjacent to the alpha is this side over here,"},{"Start":"08:50.270 ","End":"09:00.023","Text":"which is d. The hypotenuse is this side over here, which is r_1."},{"Start":"09:00.023 ","End":"09:07.840","Text":"We already saw that r_1 from Pythagoras,"},{"Start":"09:07.840 ","End":"09:15.035","Text":"it\u0027s equal to the square root of r^2 plus d^2."},{"Start":"09:15.035 ","End":"09:17.830","Text":"Then we can write the magnitude of E,"},{"Start":"09:17.830 ","End":"09:27.670","Text":"which is kq divided by r^2 plus d^2 and then we have it multiplied by d and again,"},{"Start":"09:27.670 ","End":"09:31.270","Text":"divided by the square root of r^2d^2."},{"Start":"09:31.270 ","End":"09:37.670","Text":"What we\u0027ll get in the denominator is this divided by (r^2+d^2)^3/2."},{"Start":"09:44.910 ","End":"09:50.364","Text":"Now, we know that it\u0027s pointing in the negative z direction,"},{"Start":"09:50.364 ","End":"09:55.120","Text":"so we have to add in a negative over here because we can"},{"Start":"09:55.120 ","End":"10:00.230","Text":"see from the diagram that it\u0027s pointing in the negative z direction."},{"Start":"10:00.230 ","End":"10:03.595","Text":"The next thing is that this is just the E field from"},{"Start":"10:03.595 ","End":"10:09.895","Text":"our q charge but we also have this negative q and like we said in the beginning,"},{"Start":"10:09.895 ","End":"10:14.150","Text":"we\u0027re just going to multiply this therefore by 2."},{"Start":"10:16.700 ","End":"10:21.765","Text":"Now, jus a reminder that because we\u0027re working in Cartesian coordinates,"},{"Start":"10:21.765 ","End":"10:31.810","Text":"so it\u0027s important to remember that r^2 over here is equal to x^2 plus y^2."},{"Start":"10:31.860 ","End":"10:36.865","Text":"Now, what we\u0027ve done is we\u0027ve calculated"},{"Start":"10:36.865 ","End":"10:41.890","Text":"our electric field or are jumping the electric field."},{"Start":"10:41.890 ","End":"10:45.550","Text":"In this example, it just basically means the same thing."},{"Start":"10:45.550 ","End":"10:50.440","Text":"Therefore, we can see that our Sigma I charged"},{"Start":"10:50.440 ","End":"10:56.860","Text":"density is equal to Epsilon_0 multiplied by this E_z,"},{"Start":"10:56.860 ","End":"10:59.095","Text":"z is equal to 0,"},{"Start":"10:59.095 ","End":"11:01.600","Text":"which is what we just calculated."},{"Start":"11:01.600 ","End":"11:09.580","Text":"We have negative 2kqd multiplied by Epsilon_0 and"},{"Start":"11:09.580 ","End":"11:17.950","Text":"then divided by (r^2+d^2)^3/2."},{"Start":"11:17.950 ","End":"11:22.165","Text":"The last thing that I want to show is what would happen"},{"Start":"11:22.165 ","End":"11:26.785","Text":"if we try to work out the total charge on the surface."},{"Start":"11:26.785 ","End":"11:29.020","Text":"We know that the total charge,"},{"Start":"11:29.020 ","End":"11:34.290","Text":"so let\u0027s call that q_t for q total is the integral"},{"Start":"11:34.290 ","End":"11:41.645","Text":"along the charge density per unit area multiplied by the area ds."},{"Start":"11:41.645 ","End":"11:44.050","Text":"Once we do this integral,"},{"Start":"11:44.050 ","End":"11:47.170","Text":"we\u0027ll get that this is equal to negative q,"},{"Start":"11:47.170 ","End":"11:51.590","Text":"which is exactly our image over here."},{"Start":"11:52.230 ","End":"11:56.830","Text":"Now, we\u0027ve finished with this example so what we\u0027re going to do is we\u0027re going to"},{"Start":"11:56.830 ","End":"12:01.225","Text":"look at the example of a conducting spherical shell."},{"Start":"12:01.225 ","End":"12:03.970","Text":"It\u0027s pretty much the same idea as what we\u0027ve done now"},{"Start":"12:03.970 ","End":"12:06.850","Text":"so if you feel like you completely understand this,"},{"Start":"12:06.850 ","End":"12:11.150","Text":"then feel free to skip over to the next video."},{"Start":"12:11.670 ","End":"12:16.720","Text":"Here we have our sphere and it\u0027s"},{"Start":"12:16.720 ","End":"12:20.935","Text":"grounded and so we know because this sphere is a conductor,"},{"Start":"12:20.935 ","End":"12:26.470","Text":"so we\u0027re going to only have charges along the surface."},{"Start":"12:26.470 ","End":"12:32.515","Text":"We have our charge q at distance a away from the center of our conducting sphere."},{"Start":"12:32.515 ","End":"12:34.720","Text":"Using the method of images,"},{"Start":"12:34.720 ","End":"12:36.280","Text":"we have this q tag,"},{"Start":"12:36.280 ","End":"12:40.300","Text":"a distance of b away from the center."},{"Start":"12:40.300 ","End":"12:44.365","Text":"Then what we want to do is we want to choose a point"},{"Start":"12:44.365 ","End":"12:48.820","Text":"along the surface of the sphere and we want to find our Sigma,"},{"Start":"12:48.820 ","End":"12:51.830","Text":"what our charged density is over there."},{"Start":"12:52.290 ","End":"12:58.300","Text":"Let\u0027s say that this angle between this arrow pointing"},{"Start":"12:58.300 ","End":"13:03.655","Text":"to this point on the surface and this vector over here,"},{"Start":"13:03.655 ","End":"13:07.390","Text":"where this line pointing in this direction q,"},{"Start":"13:07.390 ","End":"13:13.195","Text":"so let\u0027s say that this angle is Alpha or Theta, let\u0027s call it Theta."},{"Start":"13:13.195 ","End":"13:17.665","Text":"We\u0027ve already seen that my charge density Sigma is equal to"},{"Start":"13:17.665 ","End":"13:22.615","Text":"Epsilon_0 multiplied by my E field,"},{"Start":"13:22.615 ","End":"13:26.770","Text":"or rather multiplied by the jump"},{"Start":"13:26.770 ","End":"13:33.050","Text":"in my E field and it has to be in the direction perpendicular to my surface."},{"Start":"13:33.050 ","End":"13:34.920","Text":"Here with a sphere,"},{"Start":"13:34.920 ","End":"13:37.200","Text":"the E field that is perpendicular to"},{"Start":"13:37.200 ","End":"13:41.145","Text":"my surface is going to be the E field in the radial direction."},{"Start":"13:41.145 ","End":"13:45.415","Text":"I have Epsilon_0 multiplied by"},{"Start":"13:45.415 ","End":"13:51.700","Text":"my E field in the radial direction so at some point over here."},{"Start":"13:51.700 ","End":"13:55.400","Text":"We can see that my E field is going to be like so."},{"Start":"13:55.400 ","End":"14:01.370","Text":"I\u0027m looking for the E field somewhere just above my surface."},{"Start":"14:02.250 ","End":"14:05.515","Text":"If the radius of my sphere is r,"},{"Start":"14:05.515 ","End":"14:13.885","Text":"so somewhere at r plus minus the radial E field just before we reach the surface,"},{"Start":"14:13.885 ","End":"14:17.875","Text":"so that is at r minus."},{"Start":"14:17.875 ","End":"14:23.410","Text":"Because I know that my sphere is a conducting sphere,"},{"Start":"14:23.410 ","End":"14:27.880","Text":"so I know that I have no charges on the inside of the sphere."},{"Start":"14:27.880 ","End":"14:36.025","Text":"They are only located right at the surface at r. That means that this E_r of r minus,"},{"Start":"14:36.025 ","End":"14:38.230","Text":"which is just inside the sphere,"},{"Start":"14:38.230 ","End":"14:41.140","Text":"so there has to be no electric field over there,"},{"Start":"14:41.140 ","End":"14:46.075","Text":"because I know that the electric field within a conductor is equal to 0."},{"Start":"14:46.075 ","End":"14:51.850","Text":"Therefore, I can say that this is equal to 0."},{"Start":"14:51.850 ","End":"14:57.400","Text":"Therefore, my Sigma is going to be equal to Epsilon_0 multiplied"},{"Start":"14:57.400 ","End":"15:03.955","Text":"by the electric field and the radial direction of r plus,"},{"Start":"15:03.955 ","End":"15:09.520","Text":"where of course, have to take into account this angle Theta over here."},{"Start":"15:09.520 ","End":"15:12.790","Text":"Let\u0027s find the potential."},{"Start":"15:12.790 ","End":"15:16.345","Text":"Let\u0027s choose this point over here,"},{"Start":"15:16.345 ","End":"15:21.265","Text":"and over here, we\u0027re trying to calculate the potential."},{"Start":"15:21.265 ","End":"15:27.955","Text":"Let\u0027s define from the origin until this point over here."},{"Start":"15:27.955 ","End":"15:32.815","Text":"Let\u0027s call this the r vector then in blue,"},{"Start":"15:32.815 ","End":"15:37.120","Text":"from this charge q tag until this point,"},{"Start":"15:37.120 ","End":"15:39.970","Text":"let\u0027s call this r_1."},{"Start":"15:39.970 ","End":"15:46.990","Text":"Then let\u0027s say from a charge q until this point,"},{"Start":"15:46.990 ","End":"15:50.440","Text":"we have vector r_2."},{"Start":"15:50.440 ","End":"15:55.105","Text":"Now, what we can say is that the potential at this point"},{"Start":"15:55.105 ","End":"16:00.700","Text":"is just like we\u0027ve learned when using the method of images."},{"Start":"16:00.700 ","End":"16:04.090","Text":"The potential at this point is equal to the potential for"},{"Start":"16:04.090 ","End":"16:08.230","Text":"my charge q tag and the potential for my charge q."},{"Start":"16:08.230 ","End":"16:15.985","Text":"That\u0027s simply going to be kq tag divided by the magnitude of"},{"Start":"16:15.985 ","End":"16:26.144","Text":"r_1 plus kq divided by the magnitude of r_2."},{"Start":"16:26.144 ","End":"16:30.985","Text":"Now let\u0027s call this angle Theta."},{"Start":"16:30.985 ","End":"16:38.235","Text":"Angle Theta is always the angle between this line over here,"},{"Start":"16:38.235 ","End":"16:40.140","Text":"our a or b,"},{"Start":"16:40.140 ","End":"16:43.575","Text":"and the line that is pointing to the point that we\u0027re looking at."},{"Start":"16:43.575 ","End":"16:45.864","Text":"Here specifically, this is the point we\u0027re looking at,"},{"Start":"16:45.864 ","End":"16:51.200","Text":"so it\u0027s the angle between line our a and our i vector."},{"Start":"16:51.200 ","End":"16:55.855","Text":"Then from our law of cosines,"},{"Start":"16:55.855 ","End":"17:03.250","Text":"we get that r1 is equal to r^2 plus b^2"},{"Start":"17:03.250 ","End":"17:10.703","Text":"minus 2(rb)cosine of Theta."},{"Start":"17:10.703 ","End":"17:15.842","Text":"Of course, this is to the power of 1.5,"},{"Start":"17:15.842 ","End":"17:17.800","Text":"and r2 is the same thing."},{"Start":"17:17.800 ","End":"17:19.915","Text":"Just instead of b, we have a,"},{"Start":"17:19.915 ","End":"17:25.900","Text":"so r2 is equal to r^2 plus a^2 minus"},{"Start":"17:25.900 ","End":"17:33.590","Text":"2(ra) cosine of Theta to the power of 1.5."},{"Start":"17:33.630 ","End":"17:44.605","Text":"Now we remember that q tag is equal to negative R divided by a multiplied by q,"},{"Start":"17:44.605 ","End":"17:50.200","Text":"and that b is equal to R^2 divided by a."},{"Start":"17:50.200 ","End":"17:54.280","Text":"Now what I\u0027m going to do is I\u0027m going to substitute all of this into"},{"Start":"17:54.280 ","End":"17:59.905","Text":"these two equations and then substitute that into our potential."},{"Start":"17:59.905 ","End":"18:03.175","Text":"Once we substitute all of that in,"},{"Start":"18:03.175 ","End":"18:10.771","Text":"we\u0027ll get that our potential is equal to kq divided by."},{"Start":"18:10.771 ","End":"18:20.800","Text":"Then we\u0027ll have r^2 plus a^2 minus 2(ra)cosine of"},{"Start":"18:20.800 ","End":"18:29.830","Text":"Theta so the power of 1.5 minus kq divided by"},{"Start":"18:29.830 ","End":"18:38.590","Text":"capital R^2 plus ra divided by capital R^2"},{"Start":"18:38.590 ","End":"18:45.685","Text":"minus 2(ra)cosine of Theta"},{"Start":"18:45.685 ","End":"18:49.945","Text":"and all of this is also to the power of 1.5."},{"Start":"18:49.945 ","End":"18:54.680","Text":"This is our potential function."},{"Start":"18:54.990 ","End":"18:58.795","Text":"Now what we want to do is we want to find the E field."},{"Start":"18:58.795 ","End":"19:06.025","Text":"We can see that I potential is dependent on r lowercase r and on Theta."},{"Start":"19:06.025 ","End":"19:13.045","Text":"We know that our electric field is going to be as a function of r and Theta."},{"Start":"19:13.045 ","End":"19:16.510","Text":"However, we\u0027re trying to,"},{"Start":"19:16.510 ","End":"19:18.580","Text":"well, we\u0027ll do that in a second."},{"Start":"19:18.580 ","End":"19:21.970","Text":"It\u0027s a function of r and Theta."},{"Start":"19:21.970 ","End":"19:25.750","Text":"However, because we\u0027re trying to just find"},{"Start":"19:25.750 ","End":"19:31.210","Text":"the radial component of our electric field so we\u0027re only going to take"},{"Start":"19:31.210 ","End":"19:35.425","Text":"the derivative of our potential function with respect to"},{"Start":"19:35.425 ","End":"19:39.730","Text":"r because the Theta component is parallel to our surface,"},{"Start":"19:39.730 ","End":"19:43.615","Text":"so that doesn\u0027t help us to find our charge density Sigma."},{"Start":"19:43.615 ","End":"19:45.429","Text":"We need the component perpendicular,"},{"Start":"19:45.429 ","End":"19:48.110","Text":"which is the radial component."},{"Start":"19:48.270 ","End":"19:52.480","Text":"First of all, this is in spherical coordinates."},{"Start":"19:52.480 ","End":"19:58.360","Text":"In spherical coordinates where when we take the partial derivative with respect to r,"},{"Start":"19:58.360 ","End":"20:05.695","Text":"so it\u0027s just negative d Phi 2 and dr,"},{"Start":"20:05.695 ","End":"20:09.753","Text":"which is equal to,"},{"Start":"20:09.753 ","End":"20:12.730","Text":"if you feel like calculating this yourself,"},{"Start":"20:12.730 ","End":"20:14.065","Text":"please pause the video."},{"Start":"20:14.065 ","End":"20:24.309","Text":"But this is equal to kq divided by r^2 plus a^2 minus"},{"Start":"20:24.309 ","End":"20:31.730","Text":"2(ar)cosine of Theta to the power of"},{"Start":"20:33.420 ","End":"20:43.805","Text":"3 over 2 and then minus kq divided by"},{"Start":"20:43.805 ","End":"20:53.925","Text":"capital R^2 plus ra divided by capital R^2 minus"},{"Start":"20:53.925 ","End":"20:59.875","Text":"2(ra)cosine of Theta and"},{"Start":"20:59.875 ","End":"21:05.120","Text":"all of this is also to the power of 3 over 2."},{"Start":"21:05.640 ","End":"21:09.475","Text":"Now let\u0027s scroll down a little bit."},{"Start":"21:09.475 ","End":"21:13.495","Text":"What we want to do is we want to find the E field."},{"Start":"21:13.495 ","End":"21:16.450","Text":"I\u0027m reminding you exactly at this point,"},{"Start":"21:16.450 ","End":"21:20.740","Text":"just above our spherical surface area."},{"Start":"21:20.740 ","End":"21:25.465","Text":"We\u0027re finding the E field at r is equal to R,"},{"Start":"21:25.465 ","End":"21:29.720","Text":"and of course, as a function of Theta."},{"Start":"21:30.030 ","End":"21:33.113","Text":"Of course, I forgot to fill in the numerator,"},{"Start":"21:33.113 ","End":"21:40.390","Text":"so here we have kq multiplied by r minus a cosine of Theta divided by this and then"},{"Start":"21:40.390 ","End":"21:44.140","Text":"negative kq multiplied by r multiplied by a"},{"Start":"21:44.140 ","End":"21:48.475","Text":"divided by capital R^2 minus a cosine of Theta."},{"Start":"21:48.475 ","End":"21:52.210","Text":"Now, once we substitute in where we see lowercase r,"},{"Start":"21:52.210 ","End":"21:55.120","Text":"we substitute R to find the electric field just"},{"Start":"21:55.120 ","End":"21:59.845","Text":"above the surface of the conducting sphere."},{"Start":"21:59.845 ","End":"22:10.465","Text":"What we\u0027ll get is an electric field of kq multiplied by R^2"},{"Start":"22:10.465 ","End":"22:20.425","Text":"minus a^2 and all of this is divided by R multiplied by"},{"Start":"22:20.425 ","End":"22:24.700","Text":"R^2 plus a^2 minus"},{"Start":"22:24.700 ","End":"22:31.195","Text":"2(ra)cosine of Theta and"},{"Start":"22:31.195 ","End":"22:36.445","Text":"all of this is to the power of 3 over 2."},{"Start":"22:36.445 ","End":"22:40.000","Text":"Now if we go back to a Sigma,"},{"Start":"22:40.000 ","End":"22:46.720","Text":"we saw that Sigma is equal to Epsilon naught multiplied by this E field."},{"Start":"22:46.720 ","End":"22:50.710","Text":"Then let\u0027s just save some time,"},{"Start":"22:50.710 ","End":"22:55.255","Text":"I\u0027ll rub this out and then we can say that Sigma"},{"Start":"22:55.255 ","End":"23:00.370","Text":"is therefore equal to Epsilon naught multiplied by this."},{"Start":"23:00.370 ","End":"23:05.650","Text":"Now we have our charge density."},{"Start":"23:05.650 ","End":"23:09.715","Text":"Then again, if we want to calculate the total charge,"},{"Start":"23:09.715 ","End":"23:15.160","Text":"so we\u0027re going to be integrating along Sigma ds and of course,"},{"Start":"23:15.160 ","End":"23:18.970","Text":"we\u0027re integrating with respect to its spherical coordinates."},{"Start":"23:18.970 ","End":"23:25.195","Text":"We\u0027ll have R^2 multiplied by Sigma,"},{"Start":"23:25.195 ","End":"23:32.965","Text":"multiplied by sine of Theta d Theta d Phi."},{"Start":"23:32.965 ","End":"23:43.675","Text":"Our Theta goes from 0 until Pi and our Phi goes from 0 until 2 Pi."},{"Start":"23:43.675 ","End":"23:49.915","Text":"What will get at the end is that this is equal to q tag,"},{"Start":"23:49.915 ","End":"23:52.359","Text":"where q tag is of course,"},{"Start":"23:52.359 ","End":"23:54.955","Text":"equal to this equation over here."},{"Start":"23:54.955 ","End":"24:00.820","Text":"It\u0027s equal to negative R divided by a multiplied by q,"},{"Start":"24:00.820 ","End":"24:05.530","Text":"which is exactly the answer that we got when we were looking"},{"Start":"24:05.530 ","End":"24:11.256","Text":"at the total charge on the grounded infinite plane."},{"Start":"24:11.256 ","End":"24:13.705","Text":"So that was the exact thing we got that"},{"Start":"24:13.705 ","End":"24:19.000","Text":"the total charge was equal to the image charge that we used."},{"Start":"24:19.000 ","End":"24:22.760","Text":"That is the end of this lesson."}],"ID":14217},{"Watched":false,"Name":"Force and Energy","Duration":"8m 11s","ChapterTopicVideoID":12115,"CourseChapterTopicPlaylistID":161612,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12115.jpeg","UploadDate":"2018-06-28T03:40:32.1800000","DurationForVideoObject":"PT8M11S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.755","Text":"Hello, in this lesson,"},{"Start":"00:01.755 ","End":"00:06.960","Text":"we\u0027re going to be learning about force and energy of image charges."},{"Start":"00:06.960 ","End":"00:09.495","Text":"Here we have this system over here."},{"Start":"00:09.495 ","End":"00:15.730","Text":"Let\u0027s say that we want to find the force on this charge q over here,"},{"Start":"00:15.730 ","End":"00:19.890","Text":"which is located above this grounded plane."},{"Start":"00:20.110 ","End":"00:26.675","Text":"The force that this plane exerts on charge q is the same force that"},{"Start":"00:26.675 ","End":"00:33.270","Text":"an image charge of negative q would exert on this charge q."},{"Start":"00:34.550 ","End":"00:40.490","Text":"Let\u0027s say that charge q is a distance d away from the plane."},{"Start":"00:40.490 ","End":"00:46.385","Text":"That would mean that charge minus q is also this distance d away from the plane."},{"Start":"00:46.385 ","End":"00:53.465","Text":"The force exerted on q from minus q is equal to from Coulomb\u0027s law,"},{"Start":"00:53.465 ","End":"01:03.590","Text":"we get that it\u0027s equal to negative q^2 divided by 2d^2."},{"Start":"01:03.590 ","End":"01:07.970","Text":"Negative q^2 with a k over here."},{"Start":"01:07.970 ","End":"01:11.990","Text":"Here is q, and here we\u0027re multiplying by minus q,"},{"Start":"01:11.990 ","End":"01:19.340","Text":"and then divided by the distance between the 2 charges squared and the distance is 2d."},{"Start":"01:20.990 ","End":"01:23.690","Text":"We can see that with the force,"},{"Start":"01:23.690 ","End":"01:26.030","Text":"this is relatively straightforward."},{"Start":"01:26.030 ","End":"01:30.020","Text":"Now let\u0027s take a look at how we can calculate the energy of"},{"Start":"01:30.020 ","End":"01:35.525","Text":"this particle q when it\u0027s located above this grounded plane."},{"Start":"01:35.525 ","End":"01:40.205","Text":"Funnily enough, this is not the same as the energy of"},{"Start":"01:40.205 ","End":"01:46.550","Text":"this system of a charge q and a charge negative q at distance 2d away from one another."},{"Start":"01:46.550 ","End":"01:50.030","Text":"The energy actually works out as"},{"Start":"01:50.030 ","End":"01:55.710","Text":"half of the energy that we would have from q and minus q."},{"Start":"01:56.600 ","End":"02:03.720","Text":"We know that in this region we have an electric field."},{"Start":"02:03.800 ","End":"02:08.980","Text":"We have an electric field where z is larger than 0."},{"Start":"02:08.980 ","End":"02:14.450","Text":"However, in this region below the plane where z is smaller than 0,"},{"Start":"02:14.450 ","End":"02:17.700","Text":"we know that our electric field is equal to 0."},{"Start":"02:17.700 ","End":"02:19.760","Text":"We spoke about in another lesson."},{"Start":"02:19.760 ","End":"02:22.460","Text":"It\u0027s because the potential along the plane is equal to 0."},{"Start":"02:22.460 ","End":"02:25.265","Text":"The potential and infinity is equal to 0."},{"Start":"02:25.265 ","End":"02:28.040","Text":"This charge over here is an imaginary,"},{"Start":"02:28.040 ","End":"02:29.860","Text":"it\u0027s an image charge."},{"Start":"02:29.860 ","End":"02:34.430","Text":"Here we have a potential of 0 and at infinity we have a potential of 0,"},{"Start":"02:34.430 ","End":"02:37.835","Text":"which means that everywhere we have a potential of 0,"},{"Start":"02:37.835 ","End":"02:44.190","Text":"and therefore the electric field in this region is also going to be 0."},{"Start":"02:44.800 ","End":"02:48.095","Text":"Why is the energy 1/2?"},{"Start":"02:48.095 ","End":"02:53.360","Text":"Because in our original problem we have this grounded plane and one charge."},{"Start":"02:53.360 ","End":"02:55.775","Text":"But when we\u0027re using the method of images,"},{"Start":"02:55.775 ","End":"03:01.675","Text":"we have as if we don\u0027t have this grounded plane and we have these 2 charges."},{"Start":"03:01.675 ","End":"03:04.740","Text":"This is a positive charge."},{"Start":"03:04.740 ","End":"03:07.520","Text":"Here we\u0027ll have an electric field going out,"},{"Start":"03:07.520 ","End":"03:10.280","Text":"and here we have this negative charge,"},{"Start":"03:10.280 ","End":"03:13.225","Text":"so we have these field lines coming in."},{"Start":"03:13.225 ","End":"03:18.500","Text":"But if we look at the magnitude of the electric field in the region of z is"},{"Start":"03:18.500 ","End":"03:23.960","Text":"bigger than 0 and z is smaller than 0 the magnitude of the electric field is the same."},{"Start":"03:23.960 ","End":"03:25.340","Text":"The direction is not,"},{"Start":"03:25.340 ","End":"03:28.410","Text":"but the magnitude is."},{"Start":"03:29.560 ","End":"03:35.900","Text":"In our diagram over here where we\u0027re using the method of images,"},{"Start":"03:35.900 ","End":"03:40.730","Text":"we can see that we have twice the field lines as we have"},{"Start":"03:40.730 ","End":"03:47.130","Text":"actually going on in our original problem of our charge q and a grounded plane."},{"Start":"03:47.780 ","End":"03:53.255","Text":"If we remember our equation for energy, so that is u,"},{"Start":"03:53.255 ","End":"03:58.655","Text":"which is equal to the integral of Epsilon_0 divided by 2"},{"Start":"03:58.655 ","End":"04:06.215","Text":"multiplied by our electric field squared dv."},{"Start":"04:06.215 ","End":"04:10.895","Text":"If you remember, this is our energy density,"},{"Start":"04:10.895 ","End":"04:13.370","Text":"so Mu e. All right,"},{"Start":"04:13.370 ","End":"04:17.150","Text":"so here we can see that when we\u0027re using the method of images,"},{"Start":"04:17.150 ","End":"04:21.560","Text":"we have twice the electric field,"},{"Start":"04:21.560 ","End":"04:25.160","Text":"which means that we\u0027ll have twice the energy."},{"Start":"04:25.160 ","End":"04:26.810","Text":"That\u0027s one way of looking at it."},{"Start":"04:26.810 ","End":"04:28.685","Text":"We have twice the electric fields, so,"},{"Start":"04:28.685 ","End":"04:32.260","Text":"therefore, according to this equation will have twice the energy."},{"Start":"04:32.260 ","End":"04:37.580","Text":"Another way of looking at this is using the other equation for energy."},{"Start":"04:37.580 ","End":"04:46.955","Text":"This is where we have U is equal to the sum of 1/2 multiplied by qi Phi i,"},{"Start":"04:46.955 ","End":"04:49.205","Text":"where Phi i is, of course the potential."},{"Start":"04:49.205 ","End":"04:50.960","Text":"In the previous lesson,"},{"Start":"04:50.960 ","End":"04:59.310","Text":"we saw what the potential is for a point charge located above a grounded plane."},{"Start":"04:59.310 ","End":"05:03.110","Text":"Let\u0027s write this out. We have 1/2 multiplied by the charge,"},{"Start":"05:03.110 ","End":"05:07.715","Text":"which is q, multiplied by the potential which we saw in the previous lesson,"},{"Start":"05:07.715 ","End":"05:13.630","Text":"is equal to k multiplied by minus q divided by"},{"Start":"05:13.630 ","End":"05:22.625","Text":"2d plus we have the charge of now this is grounded plane."},{"Start":"05:22.625 ","End":"05:27.935","Text":"Now, we can work out the charge because we know that it has a charge density of Sigma."},{"Start":"05:27.935 ","End":"05:30.860","Text":"However, the definition that this is"},{"Start":"05:30.860 ","End":"05:35.720","Text":"a grounded plane means that the potential on the plane is equal to 0."},{"Start":"05:35.720 ","End":"05:39.470","Text":"Therefore, we\u0027ll have half multiplied by our charge,"},{"Start":"05:39.470 ","End":"05:41.150","Text":"which we can figure out from Sigma,"},{"Start":"05:41.150 ","End":"05:43.595","Text":"but multiplied by a potential of 0."},{"Start":"05:43.595 ","End":"05:46.750","Text":"That will give us 0."},{"Start":"05:48.440 ","End":"05:56.510","Text":"Then in total we\u0027ll get that the energy is equal to negative kq divided by 4d,"},{"Start":"05:56.510 ","End":"06:00.890","Text":"which is half of the energy that we would get if we figured out"},{"Start":"06:00.890 ","End":"06:06.155","Text":"the energy of this charge q and of this charge negative q,"},{"Start":"06:06.155 ","End":"06:08.560","Text":"and added them together."},{"Start":"06:08.560 ","End":"06:15.605","Text":"Of course, here we\u0027re going to have squared because we have this q and this q."},{"Start":"06:15.605 ","End":"06:21.200","Text":"Now another further way is that we can integrate along F. We"},{"Start":"06:21.200 ","End":"06:27.300","Text":"know that U is equal to the negative integral along Fdr."},{"Start":"06:27.300 ","End":"06:32.175","Text":"Here, let\u0027s say that this is our z-axis."},{"Start":"06:32.175 ","End":"06:35.220","Text":"D is somewhere along the z-axis."},{"Start":"06:35.220 ","End":"06:40.560","Text":"Let\u0027s say that z is equal to d. We\u0027re located at this point,"},{"Start":"06:40.560 ","End":"06:44.960","Text":"so that this is a general equation and not just for this position d,"},{"Start":"06:44.960 ","End":"06:47.359","Text":"but for any place along the z-axis."},{"Start":"06:47.359 ","End":"06:56.730","Text":"Let\u0027s say that it\u0027s d. We can say that this is the negative integral along Fdz."},{"Start":"06:57.140 ","End":"07:03.200","Text":"Then we have that this is equal to negative integral of F,"},{"Start":"07:03.200 ","End":"07:12.740","Text":"which is equal to negative kq squared divided by 2z^2 and of"},{"Start":"07:12.740 ","End":"07:17.225","Text":"course dz and we\u0027re integrating"},{"Start":"07:17.225 ","End":"07:24.640","Text":"from infinity until our point d over here where it\u0027s located."},{"Start":"07:24.640 ","End":"07:28.880","Text":"The minus and the minus will cancel out."},{"Start":"07:28.880 ","End":"07:33.799","Text":"And then we\u0027ll get kq squared divided"},{"Start":"07:33.799 ","End":"07:39.590","Text":"by 4 and then we\u0027ll have infinity."},{"Start":"07:39.590 ","End":"07:49.620","Text":"This goes to 0 when we plug in infinity minus kq^2 divided by 4d."},{"Start":"07:50.420 ","End":"07:55.340","Text":"Then in total, we\u0027ll be left with that U,"},{"Start":"07:55.340 ","End":"08:02.360","Text":"the energy is equal to negative kq^2 divided by 4d,"},{"Start":"08:02.360 ","End":"08:08.450","Text":"which is exactly what we got here and what we would have gotten over here."},{"Start":"08:08.450 ","End":"08:12.210","Text":"That is the end of this lesson."}],"ID":14218},{"Watched":false,"Name":"Grounded Sphere","Duration":"20m 50s","ChapterTopicVideoID":12116,"CourseChapterTopicPlaylistID":161612,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12116.jpeg","UploadDate":"2018-06-28T03:43:38.7130000","DurationForVideoObject":"PT20M50S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.695","Text":"Hello, in this lesson,"},{"Start":"00:01.695 ","End":"00:05.250","Text":"we\u0027re going to be learning about the case of the grounded sphere."},{"Start":"00:05.250 ","End":"00:08.715","Text":"Here we have a spherical shell which is grounded."},{"Start":"00:08.715 ","End":"00:13.755","Text":"Therefore, we know that the potential on its surface is equal to 0."},{"Start":"00:13.755 ","End":"00:17.820","Text":"Then we have a distance a from the center of the sphere,"},{"Start":"00:17.820 ","End":"00:21.165","Text":"a charged particle with a charge of q."},{"Start":"00:21.165 ","End":"00:28.775","Text":"What was discovered is that in order to calculate this charge distribution,"},{"Start":"00:28.775 ","End":"00:35.468","Text":"what we have to do is we have to place another charge somewhere within the sphere,"},{"Start":"00:35.468 ","End":"00:38.445","Text":"and this distance is called b."},{"Start":"00:38.445 ","End":"00:41.025","Text":"We\u0027ll soon find out what this distance b is."},{"Start":"00:41.025 ","End":"00:47.180","Text":"We have to place a charge in there where this charge is called q tag."},{"Start":"00:47.180 ","End":"00:51.980","Text":"This q tag is different to the charge"},{"Start":"00:51.980 ","End":"00:56.750","Text":"when we were dealing with the method of images with a grounded infinite plane."},{"Start":"00:56.750 ","End":"01:00.950","Text":"Our q tag is not going to be equal to negative q."},{"Start":"01:00.950 ","End":"01:03.960","Text":"We\u0027re soon going to see what it is equal to."},{"Start":"01:04.850 ","End":"01:10.220","Text":"What we\u0027ll see is given this b and this q tag,"},{"Start":"01:10.910 ","End":"01:14.860","Text":"so this q tag, together with this charge q,"},{"Start":"01:14.860 ","End":"01:20.050","Text":"will cancel out the potential on every single point on the sphere."},{"Start":"01:20.050 ","End":"01:25.450","Text":"If we look over here at this point right here."},{"Start":"01:25.610 ","End":"01:30.070","Text":"We\u0027re going to get potential from this charge q tag over here,"},{"Start":"01:30.070 ","End":"01:35.170","Text":"and a potential from this charge q over here."},{"Start":"01:35.170 ","End":"01:41.360","Text":"Together, there\u0027ll be a potential of 0 over here because they\u0027ll cancel each other out."},{"Start":"01:42.770 ","End":"01:45.555","Text":"What is this charge q tag?"},{"Start":"01:45.555 ","End":"01:47.190","Text":"This is something to remember."},{"Start":"01:47.190 ","End":"01:49.310","Text":"Q tag first of all we\u0027re going to have"},{"Start":"01:49.310 ","End":"01:52.655","Text":"a minus because we know of this charge q is a positive,"},{"Start":"01:52.655 ","End":"01:56.045","Text":"we need q tag to be some negative charge"},{"Start":"01:56.045 ","End":"01:59.990","Text":"in order for us to get this canceled out to potential over here."},{"Start":"01:59.990 ","End":"02:02.300","Text":"Then it\u0027s equal to r,"},{"Start":"02:02.300 ","End":"02:03.815","Text":"the radius of the sphere,"},{"Start":"02:03.815 ","End":"02:08.885","Text":"divided by a, the distance from the center of the sphere to the charge q."},{"Start":"02:08.885 ","End":"02:12.360","Text":"Then multiplied by q."},{"Start":"02:12.560 ","End":"02:15.315","Text":"What is this distance, b?"},{"Start":"02:15.315 ","End":"02:20.075","Text":"B is the distance from the center of the sphere to our charge q tag."},{"Start":"02:20.075 ","End":"02:23.795","Text":"This is equal to r divided by a."},{"Start":"02:23.795 ","End":"02:26.870","Text":"The same ratio that we have for our q tags."},{"Start":"02:26.870 ","End":"02:29.240","Text":"That might make it easier to remember."},{"Start":"02:29.240 ","End":"02:36.790","Text":"Then multiplied by R. These are two equations to memorize."},{"Start":"02:36.950 ","End":"02:40.910","Text":"There are two things to notice in this equation."},{"Start":"02:40.910 ","End":"02:43.880","Text":"Number 1, we can see that our value a over"},{"Start":"02:43.880 ","End":"02:46.970","Text":"here in the denominator for both of these equations,"},{"Start":"02:46.970 ","End":"02:49.460","Text":"is this length over here,"},{"Start":"02:49.460 ","End":"02:53.600","Text":"which is much bigger than this distance r,"},{"Start":"02:53.600 ","End":"02:55.505","Text":"which is the radius of the sphere."},{"Start":"02:55.505 ","End":"02:58.850","Text":"A is always going to be outside of the sphere,"},{"Start":"02:58.850 ","End":"03:01.175","Text":"which means that a is always going to be bigger than"},{"Start":"03:01.175 ","End":"03:04.835","Text":"r. That means that this fraction over here,"},{"Start":"03:04.835 ","End":"03:09.290","Text":"r divided by a is always going to be smaller than 1."},{"Start":"03:09.290 ","End":"03:17.640","Text":"What we can see from that is that q tag is always going to have a smaller charge than q."},{"Start":"03:17.960 ","End":"03:28.220","Text":"Also that this distance b is always going to be smaller than r. What does that mean?"},{"Start":"03:28.220 ","End":"03:33.485","Text":"That means that this distance b is always going to be located within the grounded sphere,"},{"Start":"03:33.485 ","End":"03:36.710","Text":"which makes sense, b is always"},{"Start":"03:36.710 ","End":"03:40.655","Text":"located within the sphere and a is always located out of the sphere."},{"Start":"03:40.655 ","End":"03:42.995","Text":"Now the other thing that we can do,"},{"Start":"03:42.995 ","End":"03:47.995","Text":"or that we can see is that as a becomes larger,"},{"Start":"03:47.995 ","End":"03:52.170","Text":"our point B becomes smaller."},{"Start":"03:52.170 ","End":"03:57.770","Text":"That means that as our charge q goes further and further away from the grounded sphere,"},{"Start":"03:57.770 ","End":"04:02.960","Text":"our charge q tag is going to be closer and closer to the center of the sphere."},{"Start":"04:02.960 ","End":"04:08.530","Text":"Conversely, as q moves closer to the grounded sphere."},{"Start":"04:08.530 ","End":"04:14.375","Text":"Our length b is going to increase and move closer itself to the surface of the sphere."},{"Start":"04:14.375 ","End":"04:16.670","Text":"Will move away from the center of the sphere,"},{"Start":"04:16.670 ","End":"04:18.690","Text":"but will move towards the surface."},{"Start":"04:18.690 ","End":"04:23.375","Text":"We\u0027re rarely getting this image or this mirror image that we were seeing."},{"Start":"04:23.375 ","End":"04:25.250","Text":"Just like in the grounded plane,"},{"Start":"04:25.250 ","End":"04:27.830","Text":"where if we had a particle over here,"},{"Start":"04:27.830 ","End":"04:31.385","Text":"then it\u0027s image charge would be also far away."},{"Start":"04:31.385 ","End":"04:33.560","Text":"Whereas if we had a charged particle here,"},{"Start":"04:33.560 ","End":"04:36.545","Text":"then it\u0027s image particle would be closer."},{"Start":"04:36.545 ","End":"04:39.960","Text":"The same thing with the sphere."},{"Start":"04:40.370 ","End":"04:48.340","Text":"The next thing that we can see is that as our charge q moves further away to infinity,"},{"Start":"04:48.340 ","End":"04:51.880","Text":"as a becomes infinitely large."},{"Start":"04:51.880 ","End":"04:54.935","Text":"Our charge q tag is going to become"},{"Start":"04:54.935 ","End":"04:58.415","Text":"infinitely small and it will be basically non-existent,"},{"Start":"04:58.415 ","End":"05:01.310","Text":"which is what we would expect if we had some charge,"},{"Start":"05:01.310 ","End":"05:04.985","Text":"which was infinitely far away from a grounded sphere."},{"Start":"05:04.985 ","End":"05:08.990","Text":"Then we can just work out the potential as if we have just a point charge"},{"Start":"05:08.990 ","End":"05:13.010","Text":"in space, q However,"},{"Start":"05:13.010 ","End":"05:17.050","Text":"if our q is relatively close to a grounded sphere over here of"},{"Start":"05:17.050 ","End":"05:21.340","Text":"radius r. In order to find the potential,"},{"Start":"05:21.340 ","End":"05:25.165","Text":"we\u0027re going to just calculate like we know and like we\u0027ve seen before,"},{"Start":"05:25.165 ","End":"05:31.490","Text":"the potential due to a charge q and due to a charge q tag."},{"Start":"05:31.490 ","End":"05:35.155","Text":"Now we\u0027ve seen the equations and how they work."},{"Start":"05:35.155 ","End":"05:39.940","Text":"Whoever has time and wants to see how the equations are derived,"},{"Start":"05:39.940 ","End":"05:43.975","Text":"then that\u0027s what I\u0027m going to continue with during this lesson."},{"Start":"05:43.975 ","End":"05:47.289","Text":"If you want to move on to answering some questions,"},{"Start":"05:47.289 ","End":"05:50.510","Text":"then skip to the next lesson."},{"Start":"05:51.330 ","End":"05:57.710","Text":"Let\u0027s say that this distance from the center up until here,"},{"Start":"05:57.710 ","End":"05:59.420","Text":"up until this point."},{"Start":"05:59.420 ","End":"06:04.415","Text":"Let\u0027s say that this point is some random point on the surface of the sphere."},{"Start":"06:04.415 ","End":"06:08.430","Text":"This distance over here is r."},{"Start":"06:08.920 ","End":"06:15.620","Text":"Then let\u0027s say that this distance over here in this blue line,"},{"Start":"06:15.620 ","End":"06:17.825","Text":"let\u0027s call this r_1."},{"Start":"06:17.825 ","End":"06:21.110","Text":"Then this distance over here,"},{"Start":"06:21.110 ","End":"06:24.100","Text":"let\u0027s call this r_2."},{"Start":"06:24.100 ","End":"06:29.795","Text":"Then we know that the distance from the origin to q is this distance b,"},{"Start":"06:29.795 ","End":"06:34.775","Text":"and the distance from the origin to this charge q over here,"},{"Start":"06:34.775 ","End":"06:36.350","Text":"sorry, this was q tag."},{"Start":"06:36.350 ","End":"06:39.130","Text":"Here is a distance a."},{"Start":"06:39.130 ","End":"06:44.795","Text":"What we can see is that if I connect these lines,"},{"Start":"06:44.795 ","End":"06:49.740","Text":"we can see that I have these triangles over here."},{"Start":"06:50.510 ","End":"06:53.415","Text":"We can see I have triangles."},{"Start":"06:53.415 ","End":"06:57.900","Text":"Let\u0027s call this angle over here, Theta."},{"Start":"06:57.900 ","End":"07:01.265","Text":"This over here, this angle over here is Theta."},{"Start":"07:01.265 ","End":"07:06.030","Text":"Now we\u0027re going to define all of these lengths."},{"Start":"07:06.710 ","End":"07:10.310","Text":"Using the law of cosines,"},{"Start":"07:10.310 ","End":"07:19.768","Text":"I can say that r1^2=R^2"},{"Start":"07:19.768 ","End":"07:29.570","Text":"plus b^2 minus 2Rb multiplied by cosine of the angle between the 2,"},{"Start":"07:29.570 ","End":"07:35.900","Text":"which is Theta, then I can say that r2^2."},{"Start":"07:35.900 ","End":"07:40.065","Text":"Using the same law of cosines."},{"Start":"07:40.065 ","End":"07:42.680","Text":"Now we\u0027re using a larger triangle,"},{"Start":"07:42.680 ","End":"07:44.795","Text":"but it\u0027s the same triangle."},{"Start":"07:44.795 ","End":"07:49.940","Text":"This is equal to R^2 plus A^2"},{"Start":"07:49.940 ","End":"07:57.390","Text":"minus 2Ra cosine of the angle between the 2,"},{"Start":"07:57.390 ","End":"07:59.370","Text":"which is also Theta."},{"Start":"07:59.370 ","End":"08:04.805","Text":"I want my potential at this point to be equal to 0."},{"Start":"08:04.805 ","End":"08:11.930","Text":"But it\u0027s equal to the potential from q tag plus the potential from q."},{"Start":"08:11.930 ","End":"08:18.200","Text":"This is equal to kq tag divided by r_1 plus"},{"Start":"08:18.200 ","End":"08:24.985","Text":"kq divided by r_2."},{"Start":"08:24.985 ","End":"08:29.645","Text":"Now we can divide both sides over here by k."},{"Start":"08:29.645 ","End":"08:36.830","Text":"Then let\u0027s substitute in what our r_1 and what our r_2 is equal to."},{"Start":"08:36.830 ","End":"08:38.705","Text":"What we\u0027re going to get,"},{"Start":"08:38.705 ","End":"08:40.010","Text":"let\u0027s start over here."},{"Start":"08:40.010 ","End":"08:47.555","Text":"It\u0027s a long equation is that our potential is equal to q tag divided by r_1."},{"Start":"08:47.555 ","End":"08:53.250","Text":"What we have here is the square root of all of this."},{"Start":"08:54.200 ","End":"09:02.430","Text":"What we get is q tag divided by the square root of R^2 plus b^2 minus"},{"Start":"09:02.430 ","End":"09:06.060","Text":"2Rb cosine of Theta plus q divided by"},{"Start":"09:06.060 ","End":"09:12.300","Text":"the square root of R^2 plus a^2 minus 2Ra cosine of Theta."},{"Start":"09:12.300 ","End":"09:14.055","Text":"All of this of course,"},{"Start":"09:14.055 ","End":"09:16.755","Text":"is equal to 0."},{"Start":"09:16.755 ","End":"09:24.190","Text":"Now I want to find what my q tag is equal to and what might be is equal to."},{"Start":"09:24.590 ","End":"09:29.370","Text":"From here on, we have 2 options to solve this."},{"Start":"09:29.370 ","End":"09:31.860","Text":"Number 1 is,"},{"Start":"09:31.860 ","End":"09:40.320","Text":"what we need to find is an equation that independent of this angle Theta,"},{"Start":"09:40.320 ","End":"09:44.655","Text":"what will get over here will be equal to 0."},{"Start":"09:44.655 ","End":"09:48.360","Text":"What we can say is if we take these over here with this as"},{"Start":"09:48.360 ","End":"09:52.230","Text":"equal to 0 or a point over here,"},{"Start":"09:52.230 ","End":"09:55.860","Text":"sorry, where Theta is equal to 0 and a point over here,"},{"Start":"09:55.860 ","End":"09:58.200","Text":"where Theta\u0027s equal to 90 degrees."},{"Start":"09:58.200 ","End":"10:01.005","Text":"We plug in those equations."},{"Start":"10:01.005 ","End":"10:04.595","Text":"Then what we\u0027ll have is 2 equations."},{"Start":"10:04.595 ","End":"10:09.180","Text":"Then we can solve it in order to find q tag and b."},{"Start":"10:09.850 ","End":"10:15.020","Text":"Here it actually works out that if we solve that,"},{"Start":"10:15.020 ","End":"10:18.905","Text":"then we rarely do get an equation for any Theta."},{"Start":"10:18.905 ","End":"10:23.864","Text":"Where we find that the potential at all of these points"},{"Start":"10:23.864 ","End":"10:29.805","Text":"for any Theta is going to be equal to 0 and we can solve the equation like that."},{"Start":"10:29.805 ","End":"10:33.270","Text":"The other way of doing it is by doing the algebra."},{"Start":"10:33.270 ","End":"10:38.865","Text":"We multiply out both sides and square everything."},{"Start":"10:38.865 ","End":"10:45.690","Text":"Then what we get is some expression multiplied by cosine of"},{"Start":"10:45.690 ","End":"10:53.230","Text":"Theta plus some expression which is equal to 0."},{"Start":"10:53.600 ","End":"11:01.830","Text":"Then again, we want this expression to be equal to 0 independent of Theta for any Theta."},{"Start":"11:01.830 ","End":"11:05.865","Text":"That means for any points on the sphere."},{"Start":"11:05.865 ","End":"11:07.530","Text":"In order for that to happen,"},{"Start":"11:07.530 ","End":"11:09.900","Text":"we have to have that this bracket will be equal to"},{"Start":"11:09.900 ","End":"11:13.200","Text":"0 and that this bracket will be equal to 0."},{"Start":"11:13.200 ","End":"11:15.465","Text":"Then it doesn\u0027t matter what Theta will be,"},{"Start":"11:15.465 ","End":"11:19.665","Text":"our potential will always be equal to 0."},{"Start":"11:19.665 ","End":"11:21.900","Text":"This is the second method."},{"Start":"11:21.900 ","End":"11:27.790","Text":"Now I\u0027m going to walk you through the first method and then through the second."},{"Start":"11:28.820 ","End":"11:32.580","Text":"Let\u0027s do the first method."},{"Start":"11:32.580 ","End":"11:40.710","Text":"Over here, first we\u0027ll take Theta is equal to 90 degrees or Pi divided by 2."},{"Start":"11:40.710 ","End":"11:45.840","Text":"Then cosine of Pi divided by 2 is equal to 0."},{"Start":"11:45.840 ","End":"11:49.440","Text":"We can cross off this term of negative"},{"Start":"11:49.440 ","End":"11:54.510","Text":"2Rb multiplied by 0 and negative 2Ra multiplied by 0."},{"Start":"11:54.510 ","End":"12:01.995","Text":"What we\u0027ll get after cross multiplying and taking the square root or squaring everything,"},{"Start":"12:01.995 ","End":"12:10.110","Text":"what we\u0027ll get is q tag squared multiplied by R^2 plus"},{"Start":"12:10.110 ","End":"12:21.100","Text":"a^2 will be equal to q squared multiplied by I squared plus b^2."},{"Start":"12:21.380 ","End":"12:24.795","Text":"I just did some simple algebra."},{"Start":"12:24.795 ","End":"12:29.310","Text":"You can pause the video and do it yourself to see exactly what I did."},{"Start":"12:29.310 ","End":"12:34.485","Text":"Then the next thing I\u0027m going to do is I\u0027m going to take that Theta is equal to 0."},{"Start":"12:34.485 ","End":"12:36.345","Text":"Again, if Theta is equal to 0,"},{"Start":"12:36.345 ","End":"12:39.885","Text":"then cosine of Theta is equal to 1."},{"Start":"12:39.885 ","End":"12:49.770","Text":"What we\u0027ll get in the denominators r^2 plus b^2 minus 2Rb and R^2 plus a^2 minus 2Ra."},{"Start":"12:49.770 ","End":"12:55.380","Text":"In that case, we have simply R"},{"Start":"12:55.380 ","End":"13:01.560","Text":"minus b^2 on this denominator over here."},{"Start":"13:01.560 ","End":"13:03.630","Text":"Then we take the square root,"},{"Start":"13:03.630 ","End":"13:05.865","Text":"so the square sign cancels out."},{"Start":"13:05.865 ","End":"13:08.760","Text":"For this denominator over here,"},{"Start":"13:08.760 ","End":"13:12.900","Text":"so we have R minus a^2,"},{"Start":"13:12.900 ","End":"13:15.690","Text":"and then we take the square root of all of that."},{"Start":"13:15.690 ","End":"13:17.640","Text":"That can cancel out."},{"Start":"13:17.640 ","End":"13:21.330","Text":"What we\u0027re going to get after cross-multiplying and"},{"Start":"13:21.330 ","End":"13:24.825","Text":"moving stuff over to the other side of the equation."},{"Start":"13:24.825 ","End":"13:29.280","Text":"We\u0027ll have q tag multiplied by R minus a,"},{"Start":"13:29.280 ","End":"13:37.540","Text":"which is equal to negative q multiplied by R minus b."},{"Start":"13:38.780 ","End":"13:45.570","Text":"Let\u0027s call this equation number 1 and this equation number 2."},{"Start":"13:45.570 ","End":"13:51.135","Text":"What I\u0027m going to do is I\u0027m going to take equation number 2 and I\u0027m going to square it."},{"Start":"13:51.135 ","End":"13:55.065","Text":"Then what I\u0027m going to do is I\u0027m going to take"},{"Start":"13:55.065 ","End":"14:00.555","Text":"equation number 1 and divide it by equation number 2."},{"Start":"14:00.555 ","End":"14:11.340","Text":"What we\u0027re going to get after doing that is we\u0027re going to get R^2 plus a^2 divided by"},{"Start":"14:11.340 ","End":"14:19.530","Text":"R minus a^2 is going to be equal"},{"Start":"14:19.530 ","End":"14:25.230","Text":"to R^2 plus b^2"},{"Start":"14:25.230 ","End":"14:32.020","Text":"divided by R minus b^2."},{"Start":"14:33.230 ","End":"14:37.950","Text":"Now let\u0027s scroll down to give us a bit more space."},{"Start":"14:37.950 ","End":"14:42.375","Text":"Now I\u0027m going to cross-multiply."},{"Start":"14:42.375 ","End":"14:51.840","Text":"What we\u0027re going to get is that R^2 plus a^2 multiplied by R minus b^2"},{"Start":"14:51.840 ","End":"14:57.180","Text":"is equal to R^2 plus"},{"Start":"14:57.180 ","End":"15:02.925","Text":"b^2 multiplied by R minus a^2."},{"Start":"15:02.925 ","End":"15:10.270","Text":"Now I\u0027m going to open this bracket over here that has a square sign."},{"Start":"15:10.340 ","End":"15:16.770","Text":"What we\u0027ll have is R^2 plus a^2 multiplied"},{"Start":"15:16.770 ","End":"15:24.120","Text":"by a^2 minus 2Rb plus b^2,"},{"Start":"15:24.120 ","End":"15:27.330","Text":"which is equal to R^2."},{"Start":"15:27.330 ","End":"15:28.650","Text":"I\u0027ll open up over here."},{"Start":"15:28.650 ","End":"15:34.170","Text":"We have R^2 multiplied by R minus a^2"},{"Start":"15:34.170 ","End":"15:41.265","Text":"plus b^2 multiplied by R minus a^2."},{"Start":"15:41.265 ","End":"15:45.495","Text":"I just opened up this bracket and multiplied it by this."},{"Start":"15:45.495 ","End":"15:51.540","Text":"Now I moved this equation around and put like terms with like terms, okay,"},{"Start":"15:51.540 ","End":"15:54.060","Text":"so anything that\u0027s multiplied by b^2,"},{"Start":"15:54.060 ","End":"15:55.740","Text":"I put in this brackets,"},{"Start":"15:55.740 ","End":"15:58.215","Text":"anything multiplied by b I put over here,"},{"Start":"15:58.215 ","End":"16:02.550","Text":"and anything multiplied by I squared I put in here."},{"Start":"16:02.550 ","End":"16:09.850","Text":"Now, again, I\u0027m going to play around with the algebra over here."},{"Start":"16:10.760 ","End":"16:15.960","Text":"All I did over here was I opened up these brackets."},{"Start":"16:15.960 ","End":"16:18.285","Text":"Then I subtracted everything."},{"Start":"16:18.285 ","End":"16:21.090","Text":"I was left with b^2 multiplied by 2Ra,"},{"Start":"16:21.090 ","End":"16:24.120","Text":"and then in b I did the same."},{"Start":"16:24.120 ","End":"16:28.050","Text":"I just kept it the same way and I squared again,"},{"Start":"16:28.050 ","End":"16:32.580","Text":"I opened up the brackets over here and subtract it and I was left with this."},{"Start":"16:32.580 ","End":"16:37.590","Text":"Now we can see that I can divide both sides by 2R,"},{"Start":"16:37.590 ","End":"16:43.530","Text":"2R, and 2R."},{"Start":"16:43.530 ","End":"16:47.835","Text":"Now we can see that we have a quadratic equation for b^2."},{"Start":"16:47.835 ","End":"16:54.825","Text":"We\u0027re going to plug this in to the equation for solving quadratic equations."},{"Start":"16:54.825 ","End":"17:03.180","Text":"B_12 is going to be equal to R^2 plus a^2 plus minus"},{"Start":"17:03.180 ","End":"17:09.135","Text":"the square root of R^2 plus a^2"},{"Start":"17:09.135 ","End":"17:17.260","Text":"squared minus 4a^2 R^2."},{"Start":"17:17.630 ","End":"17:22.900","Text":"All of this is divided by 2a."},{"Start":"17:22.910 ","End":"17:27.120","Text":"Then, let\u0027s just do a little side note."},{"Start":"17:27.120 ","End":"17:30.120","Text":"Let\u0027s change everything that\u0027s inside these brackets."},{"Start":"17:30.120 ","End":"17:37.620","Text":"We can see that we have R^2 plus a^2 squared minus 4a^2 R^2."},{"Start":"17:37.620 ","End":"17:42.060","Text":"We can write everything inside the square root as"},{"Start":"17:42.060 ","End":"17:49.580","Text":"simply R^2 minus a^2 squared."},{"Start":"17:49.580 ","End":"17:53.315","Text":"Then if you open up these brackets,"},{"Start":"17:53.315 ","End":"17:58.005","Text":"you\u0027ll get what we have over here inside the square root."},{"Start":"17:58.005 ","End":"18:02.770","Text":"Therefore will be left with R^2 plus"},{"Start":"18:02.770 ","End":"18:10.190","Text":"a^2 plus minus the square root of this over here,"},{"Start":"18:10.190 ","End":"18:13.910","Text":"which is R^2 minus a^2."},{"Start":"18:13.910 ","End":"18:15.650","Text":"All of this is squared."},{"Start":"18:15.650 ","End":"18:18.725","Text":"The square root can cancel out."},{"Start":"18:18.725 ","End":"18:23.995","Text":"Then all of this is divided by 2a."},{"Start":"18:23.995 ","End":"18:31.235","Text":"We can just say that this is equal to R^2 plus a^2 plus minus"},{"Start":"18:31.235 ","End":"18:39.870","Text":"R^2 minus a^2 divided by 2a."},{"Start":"18:40.880 ","End":"18:45.180","Text":"Now we get 2 options for solutions."},{"Start":"18:45.180 ","End":"18:48.140","Text":"Our b_1, our first solution is if we do plus."},{"Start":"18:48.140 ","End":"18:55.505","Text":"If R^2 plus R^2 plus a^2 minus a^2 divided by 2a,"},{"Start":"18:55.505 ","End":"19:02.330","Text":"what we get is that b_1 is equal to R^2 divided by a,"},{"Start":"19:02.330 ","End":"19:06.680","Text":"which is exactly what we got over here."},{"Start":"19:06.680 ","End":"19:10.390","Text":"Then b_2, if we take the minus option."},{"Start":"19:10.390 ","End":"19:17.325","Text":"We have R^2 minus R^2 plus a^2 plus a^2 divided by 2a."},{"Start":"19:17.325 ","End":"19:21.315","Text":"We\u0027ll get that b_2 is equal to a."},{"Start":"19:21.315 ","End":"19:23.640","Text":"This is the trivial onset."},{"Start":"19:23.640 ","End":"19:27.070","Text":"This just means that where we have q charge,"},{"Start":"19:27.070 ","End":"19:31.975","Text":"we put our negative q charge or a q tag at that exact same point."},{"Start":"19:31.975 ","End":"19:37.870","Text":"Then of course, those 2 opposite charges will cancel each other out."},{"Start":"19:37.870 ","End":"19:42.175","Text":"Then we\u0027ll get a potential of 0 on the surface of the sphere."},{"Start":"19:42.175 ","End":"19:47.200","Text":"This answer doesn\u0027t really interest us because this is something that\u0027s very trivial."},{"Start":"19:47.200 ","End":"19:53.480","Text":"But this is the answer that interests us and this is what we expected to get."},{"Start":"19:54.380 ","End":"19:58.270","Text":"Then, once we take this answer for b,"},{"Start":"19:58.270 ","End":"20:01.090","Text":"okay, where it\u0027s equal to I squared divided by a."},{"Start":"20:01.090 ","End":"20:06.000","Text":"We plug it back in to our original equations up here."},{"Start":"20:06.000 ","End":"20:08.830","Text":"We\u0027ll get our answer for q tag,"},{"Start":"20:08.830 ","End":"20:14.210","Text":"which is equal to negative R divided by a multiplied by q."},{"Start":"20:14.210 ","End":"20:16.795","Text":"This was the first method."},{"Start":"20:16.795 ","End":"20:21.190","Text":"The other method, as I said, we just multiply."},{"Start":"20:21.190 ","End":"20:25.045","Text":"We\u0027ll take our q over here to the other side of the equation."},{"Start":"20:25.045 ","End":"20:31.000","Text":"Then we multiply both sides by the respective denominators and then"},{"Start":"20:31.000 ","End":"20:37.240","Text":"we put anything that\u0027s multiplied by cosine Theta to 1 side, with brackets."},{"Start":"20:37.240 ","End":"20:42.920","Text":"Then we just set each bracket equal to 0 and then we have 2 equations with 2 unknowns."},{"Start":"20:42.920 ","End":"20:45.625","Text":"Where our 2 unknowns q tag and b."},{"Start":"20:45.625 ","End":"20:47.835","Text":"Then we can solve that."},{"Start":"20:47.835 ","End":"20:50.950","Text":"that\u0027s the end of this lesson."}],"ID":14219},{"Watched":false,"Name":"Non Grounded Sphere","Duration":"7m 51s","ChapterTopicVideoID":12117,"CourseChapterTopicPlaylistID":161612,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12117.jpeg","UploadDate":"2018-06-28T03:44:40.0270000","DurationForVideoObject":"PT7M51S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.830","Text":"Hello. In this lesson,"},{"Start":"00:01.830 ","End":"00:04.695","Text":"we\u0027re dealing with non grounded sphere."},{"Start":"00:04.695 ","End":"00:11.225","Text":"That means that the potential along the surface of the sphere is equal to V_0."},{"Start":"00:11.225 ","End":"00:16.785","Text":"What we\u0027re going to do is we\u0027re going to split the system up into 2 areas."},{"Start":"00:16.785 ","End":"00:23.505","Text":"We have Area 1 within the sphere and Area 2 outside of the sphere."},{"Start":"00:23.505 ","End":"00:26.895","Text":"Now what we want to do is we want to find"},{"Start":"00:26.895 ","End":"00:33.670","Text":"the potential everywhere due to this charge q over here."},{"Start":"00:33.850 ","End":"00:37.200","Text":"The potential in Area 1,"},{"Start":"00:37.200 ","End":"00:39.540","Text":"let\u0027s call that Phi 1,"},{"Start":"00:39.540 ","End":"00:42.430","Text":"is equal to V_0."},{"Start":"00:42.590 ","End":"00:46.360","Text":"We know that this is a conducting sphere,"},{"Start":"00:46.360 ","End":"00:50.845","Text":"which means that we have no charge in the center."},{"Start":"00:50.845 ","End":"00:59.920","Text":"Then we have charged particles causing this potential V_0 on the surface."},{"Start":"00:59.920 ","End":"01:06.820","Text":"If we take the second derivative of this potential rather,"},{"Start":"01:06.820 ","End":"01:14.600","Text":"then we\u0027ll get 0, which corresponds to having a 0 charge inside over here."},{"Start":"01:15.360 ","End":"01:21.145","Text":"We meet the Laplace rules."},{"Start":"01:21.145 ","End":"01:30.320","Text":"What we have is a constant potential or a uniform potential V_0 throughout Area 1,"},{"Start":"01:30.320 ","End":"01:32.735","Text":"so inside the sphere and on its surface,"},{"Start":"01:32.735 ","End":"01:37.830","Text":"and of course, the electric field within is equal to 0."},{"Start":"01:39.230 ","End":"01:45.895","Text":"Now, we want to find the potential in Area 2, so Phi 2."},{"Start":"01:45.895 ","End":"01:48.875","Text":"That\u0027s the potential all the way over here."},{"Start":"01:48.875 ","End":"01:55.565","Text":"A common misconception is to say that we\u0027ll take the potential due to our charge q,"},{"Start":"01:55.565 ","End":"01:58.700","Text":"and we\u0027ll add on over here,"},{"Start":"01:58.700 ","End":"02:01.605","Text":"a distance b away,"},{"Start":"02:01.605 ","End":"02:05.510","Text":"our charge q tag like we did when the sphere was grounded,"},{"Start":"02:05.510 ","End":"02:08.150","Text":"and then potential on its surface was equal to 0."},{"Start":"02:08.150 ","End":"02:10.940","Text":"We\u0027ll add on the potential due to q tag,"},{"Start":"02:10.940 ","End":"02:12.690","Text":"and then we\u0027ll add on because,"},{"Start":"02:12.690 ","End":"02:15.380","Text":"of course, this sphere isn\u0027t grounded."},{"Start":"02:15.380 ","End":"02:18.860","Text":"It also has a potential and we\u0027ll add in V_0."},{"Start":"02:19.010 ","End":"02:23.100","Text":"Now, at the beginning,"},{"Start":"02:23.100 ","End":"02:24.720","Text":"we\u0027ll say, \u0027\u0027This is great."},{"Start":"02:24.720 ","End":"02:32.205","Text":"This is our answer,\u0027\u0027 because if we take Nabla squared of Phi 2,"},{"Start":"02:32.205 ","End":"02:37.230","Text":"we\u0027ll get that this is equal to Rho divided by Epsilon naught,"},{"Start":"02:37.230 ","End":"02:40.935","Text":"which we would expect to get."},{"Start":"02:40.935 ","End":"02:43.130","Text":"We would say, great, this is the answer."},{"Start":"02:43.130 ","End":"02:46.955","Text":"However, this is not the answer. Why is this?"},{"Start":"02:46.955 ","End":"02:51.290","Text":"Because we always remember that our potential is"},{"Start":"02:51.290 ","End":"02:56.840","Text":"relative to something and our potential difference is also relative to something."},{"Start":"02:56.840 ","End":"03:00.565","Text":"We have this V_0 over here."},{"Start":"03:00.565 ","End":"03:04.490","Text":"That says that our grounded sphere located over here has"},{"Start":"03:04.490 ","End":"03:10.635","Text":"a potential V_0 relative to infinity."},{"Start":"03:10.635 ","End":"03:17.740","Text":"That means that our potential at infinity has to be equal to 0."},{"Start":"03:17.740 ","End":"03:24.530","Text":"If we look at this and we look infinitely far away over here at some point over here,"},{"Start":"03:24.530 ","End":"03:27.095","Text":"and we ask, what\u0027s the potential over here?"},{"Start":"03:27.095 ","End":"03:31.430","Text":"This potential will go to 0 and this potential will go to 0 because we know that it\u0027s"},{"Start":"03:31.430 ","End":"03:35.660","Text":"divided by R and our R would be infinity,"},{"Start":"03:35.660 ","End":"03:37.220","Text":"so these would go to 0."},{"Start":"03:37.220 ","End":"03:42.730","Text":"However, our potential would still be equal over here then to V_0."},{"Start":"03:43.820 ","End":"03:48.315","Text":"This would remain constant at infinity."},{"Start":"03:48.315 ","End":"03:51.645","Text":"This, as we know, cannot be."},{"Start":"03:51.645 ","End":"03:55.750","Text":"That means that this is not the answer."},{"Start":"03:56.000 ","End":"03:59.115","Text":"We can cross all of this out."},{"Start":"03:59.115 ","End":"04:01.960","Text":"What is the answer?"},{"Start":"04:03.860 ","End":"04:07.545","Text":"What we do is we have our charge q."},{"Start":"04:07.545 ","End":"04:11.990","Text":"We\u0027ve really do place at this distance be a charge q tag,"},{"Start":"04:11.990 ","End":"04:19.440","Text":"but we also place over here at the origin a charge q double tag."},{"Start":"04:19.450 ","End":"04:27.410","Text":"Then what we do is we say that our potential in Area 2 is"},{"Start":"04:27.410 ","End":"04:30.715","Text":"going to be equal to our potential from our charge"},{"Start":"04:30.715 ","End":"04:34.760","Text":"q plus our potential from our charge q tag,"},{"Start":"04:34.760 ","End":"04:40.445","Text":"plus our potential from our charge q double tag."},{"Start":"04:40.445 ","End":"04:44.550","Text":"Then this really is"},{"Start":"04:45.010 ","End":"04:52.500","Text":"upholding or missing the conditions that I need in Zone 2."},{"Start":"04:52.500 ","End":"04:58.535","Text":"The first thing is we\u0027re still adhering to Laplace\u0027s rule,"},{"Start":"04:58.535 ","End":"05:08.085","Text":"which means that we only have 1 charge or our original charge q in the area of Zone 2."},{"Start":"05:08.085 ","End":"05:13.890","Text":"We haven\u0027t added any more charges into this area of Zone 2."},{"Start":"05:14.030 ","End":"05:20.470","Text":"The next thing that we have is that q and q tag will cancel each other out."},{"Start":"05:20.470 ","End":"05:26.785","Text":"Of course, a potential of 0 along the surface of the sphere."},{"Start":"05:26.785 ","End":"05:29.185","Text":"That\u0027s what q and q tag do."},{"Start":"05:29.185 ","End":"05:35.745","Text":"Then q double tag causes this potential of V_0 on the surface."},{"Start":"05:35.745 ","End":"05:38.235","Text":"Q and q tag cancel out."},{"Start":"05:38.235 ","End":"05:41.410","Text":"We have a case of like when we had the grounded sphere."},{"Start":"05:41.410 ","End":"05:45.310","Text":"Then this q double tag creates this potential V_0,"},{"Start":"05:45.310 ","End":"05:51.385","Text":"which brings us back to the original problem of a non-grounded sphere."},{"Start":"05:51.385 ","End":"05:59.665","Text":"We\u0027ll see that because our q double tag and our q tag are within this Zone 1,"},{"Start":"05:59.665 ","End":"06:04.595","Text":"so we\u0027ll have the potential for point charges,"},{"Start":"06:04.595 ","End":"06:10.865","Text":"which as we know, is equal to k multiplied by the charge divided by the certain radius."},{"Start":"06:10.865 ","End":"06:15.860","Text":"Which means that when we take the potential at infinity,"},{"Start":"06:15.860 ","End":"06:19.910","Text":"it will be equal to 0 because each 1 of the radiuses or"},{"Start":"06:19.910 ","End":"06:23.870","Text":"the denominator for the potential for q, q tag,"},{"Start":"06:23.870 ","End":"06:27.245","Text":"and q double tag will approach infinity,"},{"Start":"06:27.245 ","End":"06:30.590","Text":"which means that the potentials for q, q tag,"},{"Start":"06:30.590 ","End":"06:33.560","Text":"and q double tag will approach 0,"},{"Start":"06:33.560 ","End":"06:40.770","Text":"which means that our potential in Zone 2 will approach 0 at infinity."},{"Start":"06:42.170 ","End":"06:48.350","Text":"We already know what q tag is equal to and what B is equal to from previous lessons,"},{"Start":"06:48.350 ","End":"06:54.145","Text":"so all we have to do is we have to find out what q double tag is equal to."},{"Start":"06:54.145 ","End":"07:00.200","Text":"We know that the potential of q double tag is equal to V_0,"},{"Start":"07:00.200 ","End":"07:04.715","Text":"because that is what gives us this potential V_0 over here."},{"Start":"07:04.715 ","End":"07:14.280","Text":"We also know that this is equal to kq double tag divided by R, the radius."},{"Start":"07:14.280 ","End":"07:19.370","Text":"Therefore, all we have to do is we have to isolate out q double tag."},{"Start":"07:19.370 ","End":"07:23.420","Text":"We\u0027ll get that q double tag is simply equal to"},{"Start":"07:23.420 ","End":"07:30.680","Text":"V_0R divided by k. That\u0027s it."},{"Start":"07:30.680 ","End":"07:33.830","Text":"We\u0027ve already seen what q tag has to be,"},{"Start":"07:33.830 ","End":"07:35.525","Text":"we know what q is,"},{"Start":"07:35.525 ","End":"07:39.650","Text":"and then in order to solve a problem with a non-grounded sphere,"},{"Start":"07:39.650 ","End":"07:42.065","Text":"we just have to add on this q double tag,"},{"Start":"07:42.065 ","End":"07:46.610","Text":"which is simply equal to the potential on the sphere multiplied"},{"Start":"07:46.610 ","End":"07:52.380","Text":"by R divided by k. That\u0027s the end of the lesson."}],"ID":14220},{"Watched":false,"Name":"Charge in a Sphere","Duration":"7m 40s","ChapterTopicVideoID":12118,"CourseChapterTopicPlaylistID":161612,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12118.jpeg","UploadDate":"2018-06-28T03:45:41.3230000","DurationForVideoObject":"PT7M40S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.815","Text":"Hello. In this lesson,"},{"Start":"00:01.815 ","End":"00:07.350","Text":"we\u0027re going to be dealing with the case where we have a charge within"},{"Start":"00:07.350 ","End":"00:14.955","Text":"a grounded conducting sphere of radius R. This charge q was placed inside the sphere,"},{"Start":"00:14.955 ","End":"00:19.500","Text":"a distance of r_1 away from the center of the sphere."},{"Start":"00:19.500 ","End":"00:24.960","Text":"Our question is, what is the potential?"},{"Start":"00:24.960 ","End":"00:30.050","Text":"What we\u0027re going to do is we\u0027re going to divide this into 2 different zones."},{"Start":"00:30.050 ","End":"00:31.430","Text":"We have Zone 1,"},{"Start":"00:31.430 ","End":"00:33.500","Text":"which is the area inside the sphere,"},{"Start":"00:33.500 ","End":"00:34.700","Text":"and Zone 2,"},{"Start":"00:34.700 ","End":"00:38.160","Text":"which is all the area outside the sphere."},{"Start":"00:38.160 ","End":"00:43.361","Text":"First of all, we know the potential on the surface of the sphere"},{"Start":"00:43.361 ","End":"00:50.730","Text":"at r is equal to R is equal to 0 because this sphere is grounded."},{"Start":"00:51.140 ","End":"00:57.860","Text":"We know that the potential at r is equal to infinity,"},{"Start":"00:57.860 ","End":"01:00.770","Text":"so infinitely far away from this grounded sphere,"},{"Start":"01:00.770 ","End":"01:04.250","Text":"the potential is also going to be equal to 0."},{"Start":"01:04.250 ","End":"01:08.235","Text":"We\u0027ve seen that before. Because we can see"},{"Start":"01:08.235 ","End":"01:13.175","Text":"that in between the surface of the sphere and infinity,"},{"Start":"01:13.175 ","End":"01:14.720","Text":"we have no charges."},{"Start":"01:14.720 ","End":"01:18.395","Text":"Our only charge is this charge Q inside the sphere."},{"Start":"01:18.395 ","End":"01:21.635","Text":"In Area 2, there\u0027s no charges."},{"Start":"01:21.635 ","End":"01:30.450","Text":"We can say therefore that the potential in Region 2 is equal therefore to 0."},{"Start":"01:30.450 ","End":"01:35.165","Text":"Now, all we have to do is we have to find the potential in Region 1."},{"Start":"01:35.165 ","End":"01:39.240","Text":"This is what we\u0027re trying to find now."},{"Start":"01:40.610 ","End":"01:43.100","Text":"The way we\u0027re going to solve this,"},{"Start":"01:43.100 ","End":"01:44.945","Text":"where we have a charge in a sphere,"},{"Start":"01:44.945 ","End":"01:49.970","Text":"is the same way that we solved previous questions where we saw that we had"},{"Start":"01:49.970 ","End":"01:56.446","Text":"some charge located outside a grounded sphere,"},{"Start":"01:56.446 ","End":"02:03.140","Text":"a distance of r_2 away from the center of the grounded sphere."},{"Start":"02:03.140 ","End":"02:08.240","Text":"Then we put an image charge of q tag inside the sphere."},{"Start":"02:08.240 ","End":"02:10.865","Text":"This is the exact same question."},{"Start":"02:10.865 ","End":"02:12.395","Text":"Let\u0027s call the charge,"},{"Start":"02:12.395 ","End":"02:15.880","Text":"our original charge inside the sphere as q_1."},{"Start":"02:15.880 ","End":"02:19.355","Text":"Our image charge, which is in this example,"},{"Start":"02:19.355 ","End":"02:21.725","Text":"located outside the sphere,"},{"Start":"02:21.725 ","End":"02:24.313","Text":"let\u0027s say located over here,"},{"Start":"02:24.313 ","End":"02:28.020","Text":"let\u0027s call this q_2."},{"Start":"02:28.160 ","End":"02:31.985","Text":"In previous lessons, we saw that q_1,"},{"Start":"02:31.985 ","End":"02:34.250","Text":"which in the previous examples,"},{"Start":"02:34.250 ","End":"02:38.360","Text":"when q_2 was our charge and q_1 was our image charge,"},{"Start":"02:38.360 ","End":"02:40.850","Text":"we saw that the equation for q_1,"},{"Start":"02:40.850 ","End":"02:45.455","Text":"in order to have a potential of 0 along every point on the sphere,"},{"Start":"02:45.455 ","End":"02:51.110","Text":"so q_1 would be equal to negative R divided by a,"},{"Start":"02:51.110 ","End":"02:54.800","Text":"which was the distance of q or in this example here,"},{"Start":"02:54.800 ","End":"02:57.575","Text":"q_2 away from the center of the sphere."},{"Start":"02:57.575 ","End":"03:02.550","Text":"Multiplied by the charge of q over here, it\u0027s q_2."},{"Start":"03:02.550 ","End":"03:09.245","Text":"That is going to be equal to negative R divided by the distance over here,"},{"Start":"03:09.245 ","End":"03:10.910","Text":"of q_2 from the center,"},{"Start":"03:10.910 ","End":"03:16.270","Text":"which is r_2, multiplied by q_2."},{"Start":"03:16.790 ","End":"03:19.480","Text":"Then we saw that b,"},{"Start":"03:19.480 ","End":"03:22.460","Text":"which is the distance between the center of"},{"Start":"03:22.460 ","End":"03:26.780","Text":"the sphere and the charge located inside the sphere,"},{"Start":"03:26.780 ","End":"03:30.920","Text":"is equal to R squared divided by a,"},{"Start":"03:30.920 ","End":"03:35.555","Text":"where a is the distance from the center of the sphere to the charge outside the sphere."},{"Start":"03:35.555 ","End":"03:37.430","Text":"Therefore, in our example,"},{"Start":"03:37.430 ","End":"03:40.570","Text":"b is equal to r_1."},{"Start":"03:40.570 ","End":"03:45.530","Text":"That\u0027s the distance from the center of sphere into the charge inside the sphere,"},{"Start":"03:45.530 ","End":"03:50.390","Text":"which is equal to R squared divided by a,"},{"Start":"03:50.390 ","End":"03:53.970","Text":"which in this example is r_2."},{"Start":"03:54.800 ","End":"03:59.040","Text":"We already know what q_1 and r_1 is equal to."},{"Start":"03:59.040 ","End":"04:00.935","Text":"That was given to us in the question."},{"Start":"04:00.935 ","End":"04:03.995","Text":"But what we\u0027re trying to find is the image."},{"Start":"04:03.995 ","End":"04:07.955","Text":"In this question, when our original charge was in the sphere,"},{"Start":"04:07.955 ","End":"04:11.385","Text":"our image charge is this charge q_2."},{"Start":"04:11.385 ","End":"04:15.700","Text":"You want to isolate out q_2 and r_2."},{"Start":"04:15.950 ","End":"04:19.095","Text":"Let\u0027s deal with that."},{"Start":"04:19.095 ","End":"04:23.955","Text":"Then what we\u0027re going to get is that q_2 is equal to"},{"Start":"04:23.955 ","End":"04:30.660","Text":"negative r_2 divided by R multiplied by q_1."},{"Start":"04:30.660 ","End":"04:33.510","Text":"But we don\u0027t know what r_2 is."},{"Start":"04:33.510 ","End":"04:36.300","Text":"We\u0027re trying to find that. Let\u0027s isolate out over here."},{"Start":"04:36.300 ","End":"04:42.465","Text":"R_2 is equal to R squared divided by r_1."},{"Start":"04:42.465 ","End":"04:46.100","Text":"Then if we plug this in over here,"},{"Start":"04:46.100 ","End":"04:55.390","Text":"we\u0027ll get that q_2 is equal to R squared divided by r_1R multiplied by q_1."},{"Start":"04:55.390 ","End":"04:58.810","Text":"Then this R can cancel out with this R and a negative here."},{"Start":"04:58.810 ","End":"05:09.360","Text":"Then what we\u0027re left with is negative R divided by r_1 multiplied by q_1."},{"Start":"05:09.360 ","End":"05:15.890","Text":"That\u0027s all great, but now I want to show you something cool. All right."},{"Start":"05:15.890 ","End":"05:22.170","Text":"Remember our original charge was this q_1."},{"Start":"05:22.170 ","End":"05:27.600","Text":"Usually, we called the original charge q."},{"Start":"05:27.600 ","End":"05:32.100","Text":"Now q in this example is q_1."},{"Start":"05:32.100 ","End":"05:39.730","Text":"Usually we called the distance between the center of the sphere to the original charge a."},{"Start":"05:39.730 ","End":"05:43.545","Text":"A in this example is r_1."},{"Start":"05:43.545 ","End":"05:49.235","Text":"A is the distance between the center of the sphere to the original charge."},{"Start":"05:49.235 ","End":"05:52.025","Text":"Then we had our q tag,"},{"Start":"05:52.025 ","End":"05:53.810","Text":"which is our image charge,"},{"Start":"05:53.810 ","End":"05:57.270","Text":"which over here in this example was q_2."},{"Start":"05:57.270 ","End":"06:02.660","Text":"The distance between the center of the sphere to our image charge, we called b."},{"Start":"06:02.660 ","End":"06:06.270","Text":"In this example, that\u0027s r_2."},{"Start":"06:07.100 ","End":"06:14.315","Text":"Now, if we substitute all of this into the equations,"},{"Start":"06:14.315 ","End":"06:17.990","Text":"what we\u0027ll have is q_2 over here,"},{"Start":"06:17.990 ","End":"06:25.905","Text":"so q_2, which we know is equal to q tag,"},{"Start":"06:25.905 ","End":"06:31.870","Text":"which is equal to negative R divided by r_1."},{"Start":"06:31.870 ","End":"06:37.810","Text":"So r_1 was equal to a multiplied by q_1,"},{"Start":"06:37.810 ","End":"06:43.170","Text":"where q_1 was equal to our original charge q."},{"Start":"06:43.170 ","End":"06:47.010","Text":"Then we had, for our equation b."},{"Start":"06:47.010 ","End":"06:49.830","Text":"Here, b is r_2."},{"Start":"06:49.830 ","End":"06:53.130","Text":"So r_2 is equal to b,"},{"Start":"06:53.130 ","End":"07:01.060","Text":"which is equal to R squared divided by r_1 where r_1 is equal to a."},{"Start":"07:01.910 ","End":"07:06.130","Text":"What we can see over here is that we\u0027re left with"},{"Start":"07:06.130 ","End":"07:10.700","Text":"the exact same equations to solve this question if"},{"Start":"07:10.700 ","End":"07:14.630","Text":"we had the example where our original charge is outside"},{"Start":"07:14.630 ","End":"07:19.035","Text":"the sphere and our image charge has to be in the sphere."},{"Start":"07:19.035 ","End":"07:23.465","Text":"It doesn\u0027t matter if the original charge is outside or inside the sphere,"},{"Start":"07:23.465 ","End":"07:28.940","Text":"and therefore the image charge is inside or outside the sphere."},{"Start":"07:28.940 ","End":"07:33.185","Text":"These same equations that we saw a few lessons ago"},{"Start":"07:33.185 ","End":"07:38.120","Text":"are still the equations to use in order to solve a question like this."},{"Start":"07:38.120 ","End":"07:41.370","Text":"That\u0027s the end of this lesson."}],"ID":14221},{"Watched":false,"Name":"Exercise 3","Duration":"5m 26s","ChapterTopicVideoID":12119,"CourseChapterTopicPlaylistID":161612,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12119.jpeg","UploadDate":"2018-06-28T03:46:26.0230000","DurationForVideoObject":"PT5M26S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.140","Text":"Hello. In this lesson we\u0027re going to be answering the following question."},{"Start":"00:04.140 ","End":"00:05.760","Text":"A point charge, q,"},{"Start":"00:05.760 ","End":"00:10.005","Text":"is placed inside a grounded half sphere of radius,"},{"Start":"00:10.005 ","End":"00:15.240","Text":"R. The charge is located at height a above the center of the sphere."},{"Start":"00:15.240 ","End":"00:17.370","Text":"What is the potential?"},{"Start":"00:17.370 ","End":"00:25.895","Text":"We know that the potential on the surface where r=R=0 because it\u0027s grounded,"},{"Start":"00:25.895 ","End":"00:32.955","Text":"and we know that the potential at infinity is also equal to 0."},{"Start":"00:32.955 ","End":"00:42.660","Text":"Therefore, if we call this Region 1 and everywhere outside this sphere,"},{"Start":"00:42.660 ","End":"00:47.120","Text":"Region 2, we can say that the potential in Region 2 is equal,"},{"Start":"00:47.120 ","End":"00:48.560","Text":"therefore, to 0."},{"Start":"00:48.560 ","End":"00:53.520","Text":"So now we\u0027re trying to find the potential in Region 1."},{"Start":"00:54.410 ","End":"00:58.875","Text":"Now we have 2 problems to solve."},{"Start":"00:58.875 ","End":"01:07.010","Text":"We need to find some system of image charges such that the potential on this base of"},{"Start":"01:07.010 ","End":"01:11.330","Text":"the spherical sphere will be equal to 0 and that"},{"Start":"01:11.330 ","End":"01:17.030","Text":"the potential along this half sphere over here will also be equal to 0,"},{"Start":"01:17.030 ","End":"01:19.975","Text":"so we have 2 regions to deal with."},{"Start":"01:19.975 ","End":"01:25.250","Text":"If I take a charge over here,"},{"Start":"01:25.250 ","End":"01:30.560","Text":"a distance of a away from the center in the other direction,"},{"Start":"01:30.560 ","End":"01:33.785","Text":"and I put here a charge of negative q,"},{"Start":"01:33.785 ","End":"01:41.405","Text":"so I know that the potential due to negative q and due to q along this plane over here,"},{"Start":"01:41.405 ","End":"01:43.700","Text":"this base for the half sphere,"},{"Start":"01:43.700 ","End":"01:46.880","Text":"the potential over here will be equal to 0."},{"Start":"01:46.880 ","End":"01:52.740","Text":"But I know that the potential on the half sphere still isn\u0027t equal to 0."},{"Start":"01:52.740 ","End":"01:58.315","Text":"So now I have to find some other image charge in order to do that."},{"Start":"01:58.315 ","End":"02:05.900","Text":"The next thing I can do is I can place a charge over here,"},{"Start":"02:05.900 ","End":"02:14.483","Text":"and this is q\u0027, and it\u0027s a distance b away from the center of the sphere."},{"Start":"02:14.483 ","End":"02:18.030","Text":"Then let\u0027s just write this down."},{"Start":"02:18.080 ","End":"02:29.218","Text":"I know that q\u0027 from my equations is equal to negative r divided by a multiplied by q,"},{"Start":"02:29.218 ","End":"02:33.260","Text":"and b, so this distance over here,"},{"Start":"02:33.260 ","End":"02:38.430","Text":"is equal to R^2 divided by a."},{"Start":"02:39.430 ","End":"02:42.350","Text":"So far so good,"},{"Start":"02:42.350 ","End":"02:44.000","Text":"but it\u0027s not perfect."},{"Start":"02:44.000 ","End":"02:52.290","Text":"What I can see is that these 2 charges cause a potential of 0 on this plane,"},{"Start":"02:52.290 ","End":"02:54.140","Text":"but now I\u0027ve added in this q\u0027,"},{"Start":"02:54.140 ","End":"02:59.450","Text":"and its charge on this plane isn\u0027t yet canceled out."},{"Start":"02:59.450 ","End":"03:01.190","Text":"This charge and this charge,"},{"Start":"03:01.190 ","End":"03:03.200","Text":"q and q\u0027,"},{"Start":"03:03.200 ","End":"03:06.665","Text":"cancel out the potential on this half sphere,"},{"Start":"03:06.665 ","End":"03:13.645","Text":"but the potential of negative q isn\u0027t yet canceled out on this sphere."},{"Start":"03:13.645 ","End":"03:19.670","Text":"So what can I do? If I take a distance, b,"},{"Start":"03:19.670 ","End":"03:21.380","Text":"but in this direction,"},{"Start":"03:21.380 ","End":"03:27.040","Text":"and here I place a charge of negative q\u0027."},{"Start":"03:27.040 ","End":"03:33.120","Text":"Now somehow everything works out."},{"Start":"03:33.120 ","End":"03:36.875","Text":"So let\u0027s take a look at what exactly is happening over here."},{"Start":"03:36.875 ","End":"03:41.420","Text":"If I complete this sphere to,"},{"Start":"03:41.420 ","End":"03:44.415","Text":"so it will look something like so,"},{"Start":"03:44.415 ","End":"03:47.925","Text":"so we can see that my q and my negative q,"},{"Start":"03:47.925 ","End":"03:50.599","Text":"together with my q\u0027 and my negative q\u0027,"},{"Start":"03:50.599 ","End":"03:55.445","Text":"are all equidistant from this plane over here,"},{"Start":"03:55.445 ","End":"03:57.315","Text":"and so they\u0027ll cancel each other out,"},{"Start":"03:57.315 ","End":"04:00.470","Text":"q with negative q and q\u0027 with negative q\u0027,"},{"Start":"04:00.470 ","End":"04:07.350","Text":"such that the potential along this plane over here will be equal to 0."},{"Start":"04:07.520 ","End":"04:13.700","Text":"Then we can see that the potential along"},{"Start":"04:13.700 ","End":"04:20.575","Text":"this half sphere over here is going to be equal to 0 because of our q and our q\u0027."},{"Start":"04:20.575 ","End":"04:25.040","Text":"Then we can see that the potential along this bottom half will be equal to"},{"Start":"04:25.040 ","End":"04:30.660","Text":"0 because of negative q and negative q\u0027."},{"Start":"04:31.930 ","End":"04:36.275","Text":"We were playing by the rules of the method of images,"},{"Start":"04:36.275 ","End":"04:38.900","Text":"so the region where we were finding the potential is"},{"Start":"04:38.900 ","End":"04:44.925","Text":"this enclosed half sphere where we have 1 charge,"},{"Start":"04:44.925 ","End":"04:50.180","Text":"and we didn\u0027t change the number of charges in this region."},{"Start":"04:50.180 ","End":"04:53.635","Text":"We still have the original charge only inside there."},{"Start":"04:53.635 ","End":"04:59.873","Text":"The only charges that we added were outside this region."},{"Start":"04:59.873 ","End":"05:06.500","Text":"Our boundary conditions, which is that the potential on the surface of this half sphere,"},{"Start":"05:06.500 ","End":"05:09.545","Text":"so on the base and along the cap,"},{"Start":"05:09.545 ","End":"05:11.270","Text":"will be equal to 0,"},{"Start":"05:11.270 ","End":"05:14.960","Text":"which according to this system, we actually get."},{"Start":"05:14.960 ","End":"05:16.970","Text":"Then to find the potential,"},{"Start":"05:16.970 ","End":"05:24.275","Text":"we just have to sum up the potential at a certain point due to these 4 charges."},{"Start":"05:24.275 ","End":"05:27.390","Text":"That\u0027s the end of this lesson."}],"ID":14222},{"Watched":false,"Name":"Exercise 4","Duration":"19m 16s","ChapterTopicVideoID":12120,"CourseChapterTopicPlaylistID":161612,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12120.jpeg","UploadDate":"2018-06-28T03:49:19.7500000","DurationForVideoObject":"PT19M16S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.995","Text":"Hello. In this lesson,"},{"Start":"00:01.995 ","End":"00:04.710","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.710 ","End":"00:08.445","Text":"A charge q is placed along the x-axis,"},{"Start":"00:08.445 ","End":"00:15.600","Text":"a distance of 2a from the center of a grounded spherical shell of radius a."},{"Start":"00:15.600 ","End":"00:17.610","Text":"Points A, B,"},{"Start":"00:17.610 ","End":"00:21.300","Text":"and C are also along the x-axis,"},{"Start":"00:21.300 ","End":"00:25.800","Text":"such that x_A is at a divided by 2,"},{"Start":"00:25.800 ","End":"00:28.020","Text":"x_B is at a,"},{"Start":"00:28.020 ","End":"00:32.175","Text":"and x_C is at 3a divided by 2."},{"Start":"00:32.175 ","End":"00:34.065","Text":"Question number 1 is,"},{"Start":"00:34.065 ","End":"00:39.370","Text":"what is the potential at points A, B, and C?"},{"Start":"00:40.450 ","End":"00:47.195","Text":"What we\u0027re going to do is we\u0027re going to split the area up into zones."},{"Start":"00:47.195 ","End":"00:50.355","Text":"Here we have Zone 1,"},{"Start":"00:50.355 ","End":"00:53.910","Text":"and here we have Zone 2."},{"Start":"00:53.910 ","End":"00:57.980","Text":"Zone 1 includes everything inside the grounded sphere,"},{"Start":"00:57.980 ","End":"01:01.755","Text":"and Zone 2 is everything outside."},{"Start":"01:01.755 ","End":"01:06.140","Text":"First of all, let\u0027s find the potential in Zone 1."},{"Start":"01:06.140 ","End":"01:10.265","Text":"We know that we have a grounded spherical shell."},{"Start":"01:10.265 ","End":"01:16.295","Text":"Then we can see that the net charge over here inside"},{"Start":"01:16.295 ","End":"01:23.645","Text":"the spherical shell is equal to 0 and the only charge as this external charge q."},{"Start":"01:23.645 ","End":"01:28.010","Text":"We know that the potential in Zone 1 is equal to 0 because of that,"},{"Start":"01:28.010 ","End":"01:31.770","Text":"because it\u0027s a grounded conducting spherical shell."},{"Start":"01:31.790 ","End":"01:37.955","Text":"That means that every point inside Zone 1 is also going to be equal to 0."},{"Start":"01:37.955 ","End":"01:44.960","Text":"Therefore, the potential at point A will also be equal to 0 and we know that"},{"Start":"01:44.960 ","End":"01:53.850","Text":"the potential on the surface of some grounded conductor is also equal to 0."},{"Start":"01:54.160 ","End":"01:59.435","Text":"That we know. Therefore, we can also say that the potential at point B,"},{"Start":"01:59.435 ","End":"02:02.540","Text":"which is located exactly on the surface of"},{"Start":"02:02.540 ","End":"02:09.070","Text":"the spherical shell so that is also going to be equal to 0."},{"Start":"02:09.080 ","End":"02:14.930","Text":"Now what we want to do is we want to find the potential outside the spherical shell."},{"Start":"02:14.930 ","End":"02:18.925","Text":"Then point C is located outside the spherical shell"},{"Start":"02:18.925 ","End":"02:22.680","Text":"and then we\u0027ll just sub in where C is,"},{"Start":"02:22.680 ","End":"02:26.070","Text":"so use position and we\u0027ll find the potential there."},{"Start":"02:26.890 ","End":"02:29.570","Text":"As we know, in order to do this,"},{"Start":"02:29.570 ","End":"02:34.490","Text":"what we\u0027re going to have to do is we\u0027re going to have to use our method of images."},{"Start":"02:34.490 ","End":"02:43.055","Text":"That means that we\u0027re placing some kind of charge q tag somewhere inside the sphere and"},{"Start":"02:43.055 ","End":"02:51.330","Text":"then we\u0027re going to work out the potential in Area 2 due to charge q and q tag."},{"Start":"02:51.330 ","End":"02:54.890","Text":"First of all, let\u0027s find out what q tag is equal to."},{"Start":"02:54.890 ","End":"02:58.340","Text":"We know that the equation for q tag is equal to"},{"Start":"02:58.340 ","End":"03:03.200","Text":"negative the radius of the grounded sphere divided by a,"},{"Start":"03:03.200 ","End":"03:07.355","Text":"the distance from the center of the sphere until r charge q,"},{"Start":"03:07.355 ","End":"03:10.160","Text":"multiplied by the charge q."},{"Start":"03:10.160 ","End":"03:15.980","Text":"That means that in our example q tag is equal to negative the radius of the sphere,"},{"Start":"03:15.980 ","End":"03:18.580","Text":"which is a,"},{"Start":"03:20.000 ","End":"03:26.628","Text":"divided by the distance to I charge q so that is 2a,"},{"Start":"03:26.628 ","End":"03:28.785","Text":"so divided by 2a,"},{"Start":"03:28.785 ","End":"03:31.800","Text":"multiplied by its charge q."},{"Start":"03:31.800 ","End":"03:38.620","Text":"What we\u0027re going to get is that q tag is equal to negative 1/2 of q."},{"Start":"03:38.620 ","End":"03:43.160","Text":"Then the position of q tag so is equal to"},{"Start":"03:43.160 ","End":"03:47.975","Text":"b and the equation is equal to R^2 divided by a,"},{"Start":"03:47.975 ","End":"03:51.755","Text":"so is the radius of the grounded sphere squared,"},{"Start":"03:51.755 ","End":"03:54.850","Text":"divided by the distance to our charge q."},{"Start":"03:54.850 ","End":"03:57.045","Text":"Therefore, in our example,"},{"Start":"03:57.045 ","End":"03:59.730","Text":"b is equal to, here,"},{"Start":"03:59.730 ","End":"04:04.415","Text":"it\u0027s a^2 divided by the distance to q,"},{"Start":"04:04.415 ","End":"04:07.115","Text":"which is divided by 2a."},{"Start":"04:07.115 ","End":"04:13.260","Text":"Then we\u0027ll get that b is equal to 1/2a."},{"Start":"04:13.260 ","End":"04:19.590","Text":"So 1/2a is the position of point a."},{"Start":"04:19.590 ","End":"04:29.790","Text":"What we can do is we can rub this out and we can place q tag on this point a,"},{"Start":"04:29.790 ","End":"04:32.130","Text":"because it\u0027s at 1/2a,"},{"Start":"04:32.130 ","End":"04:34.990","Text":"which is exactly point a."},{"Start":"04:35.870 ","End":"04:43.460","Text":"Now we want to find the potential at C. Let\u0027s carry this on over here."},{"Start":"04:43.460 ","End":"04:47.360","Text":"The potential at C is just equal to"},{"Start":"04:47.360 ","End":"04:53.470","Text":"the potential at C due to q plus the potential at C due to q tag."},{"Start":"04:53.470 ","End":"04:57.860","Text":"This distance between c and q,"},{"Start":"04:57.860 ","End":"05:03.745","Text":"so q is at 2a away from the center minus the position of C,"},{"Start":"05:03.745 ","End":"05:09.275","Text":"which is 3a divided by 2 away from the center."},{"Start":"05:09.275 ","End":"05:16.530","Text":"This is going to be equal to kq divided by 2a minus 3 over"},{"Start":"05:16.530 ","End":"05:24.900","Text":"2a is simply equal to 1/2a plus the potential due to q tag."},{"Start":"05:24.900 ","End":"05:31.805","Text":"This distance over here between q tag and C is the same distance between A and C."},{"Start":"05:31.805 ","End":"05:40.440","Text":"That means it\u0027s 3 divided by 2a minus a over 2."},{"Start":"05:40.440 ","End":"05:47.865","Text":"That is simply equal to kq tag divided by a."},{"Start":"05:47.865 ","End":"05:55.250","Text":"Then once we add all of this up and when we sub in q tag,"},{"Start":"05:55.250 ","End":"06:01.920","Text":"we\u0027re going to get that this is equal to 3kq divided by 2a."},{"Start":"06:03.290 ","End":"06:06.840","Text":"Now we found the potential at A, B,"},{"Start":"06:06.840 ","End":"06:11.595","Text":"and C. Now let\u0027s move on to question number 2."},{"Start":"06:11.595 ","End":"06:18.420","Text":"Question number 2 is what is the surface charge distribution at B?"},{"Start":"06:18.520 ","End":"06:24.920","Text":"The surface charge distribution at point B is equal"},{"Start":"06:24.920 ","End":"06:31.205","Text":"to Epsilon_0 multiplied by the jump in the electric field,"},{"Start":"06:31.205 ","End":"06:36.270","Text":"which is perpendicular to our conducting surface."},{"Start":"06:36.350 ","End":"06:40.925","Text":"That means that in order to calculate this jump in the field,"},{"Start":"06:40.925 ","End":"06:44.450","Text":"I\u0027m going to measure the electric field at 2 points very,"},{"Start":"06:44.450 ","End":"06:49.205","Text":"very close to the surface of our grounded spherical shell."},{"Start":"06:49.205 ","End":"06:52.550","Text":"That means calculating the electric field at"},{"Start":"06:52.550 ","End":"06:56.750","Text":"this point just outside of the conducting spherical shell,"},{"Start":"06:56.750 ","End":"07:00.140","Text":"and at this point just inside of"},{"Start":"07:00.140 ","End":"07:06.330","Text":"the conducting spherical shell and then I\u0027ll find the difference in the electric field."},{"Start":"07:07.070 ","End":"07:14.140","Text":"Let\u0027s call this point point number 1 and this point, point number point 2."},{"Start":"07:14.330 ","End":"07:21.185","Text":"Let\u0027s see what the electric field at point number 1 perpendicular,"},{"Start":"07:21.185 ","End":"07:23.420","Text":"of course, is equal to."},{"Start":"07:23.420 ","End":"07:27.880","Text":"We know that our charge q tag doesn\u0027t really exist."},{"Start":"07:27.880 ","End":"07:31.010","Text":"We just use it for the method of images to find"},{"Start":"07:31.010 ","End":"07:35.160","Text":"the potential outside of the charge spherical shell."},{"Start":"07:35.600 ","End":"07:38.300","Text":"This is a theoretical charge,"},{"Start":"07:38.300 ","End":"07:39.380","Text":"but in our actual problem,"},{"Start":"07:39.380 ","End":"07:41.480","Text":"it doesn\u0027t exist, as we know,"},{"Start":"07:41.480 ","End":"07:44.495","Text":"it isn\u0027t really inside the spherical shell."},{"Start":"07:44.495 ","End":"07:51.890","Text":"As we know, the total charge inside the spherical shell is equal to 0,"},{"Start":"07:51.890 ","End":"07:58.805","Text":"which means that the electric field inside the spherical shell is equal to 0."},{"Start":"07:58.805 ","End":"08:03.005","Text":"Now let\u0027s find the electric field at 0.2,"},{"Start":"08:03.005 ","End":"08:06.305","Text":"and of course it is also perpendicular."},{"Start":"08:06.305 ","End":"08:08.780","Text":"The electric field at 0.2,"},{"Start":"08:08.780 ","End":"08:11.225","Text":"we\u0027re going to calculate via knowing"},{"Start":"08:11.225 ","End":"08:16.565","Text":"the potential in this region outside the spherical shell."},{"Start":"08:16.565 ","End":"08:20.855","Text":"That we know from this equation over here,"},{"Start":"08:20.855 ","End":"08:26.960","Text":"we have our charge q and we have a charge q tag."},{"Start":"08:26.960 ","End":"08:32.330","Text":"Here, I\u0027m reminding you we can use q tag because in order to"},{"Start":"08:32.330 ","End":"08:34.100","Text":"find the potential and therefore"},{"Start":"08:34.100 ","End":"08:37.250","Text":"the electric field outside of a grounded spherical shell,"},{"Start":"08:37.250 ","End":"08:41.190","Text":"we have to use this theoretical q tag."},{"Start":"08:41.230 ","End":"08:44.945","Text":"This is simply going to be equal to,"},{"Start":"08:44.945 ","End":"08:49.645","Text":"so we have kq divided by the radius squared."},{"Start":"08:49.645 ","End":"08:55.210","Text":"The radius squared is the distance between our q and our point B."},{"Start":"08:55.210 ","End":"08:59.120","Text":"We know that point 2 where we\u0027re measuring the electric field is very,"},{"Start":"08:59.120 ","End":"09:00.320","Text":"very close to point B."},{"Start":"09:00.320 ","End":"09:04.445","Text":"It\u0027s just on the outside of our shell."},{"Start":"09:04.445 ","End":"09:13.340","Text":"That is going to be a distance of a and it\u0027s the radius squared and of course,"},{"Start":"09:13.340 ","End":"09:16.465","Text":"the electric field is a vector."},{"Start":"09:16.465 ","End":"09:21.230","Text":"The direction of the electric field due to point q."},{"Start":"09:21.230 ","End":"09:25.580","Text":"So q, we\u0027re assuming is a positive charge."},{"Start":"09:25.580 ","End":"09:28.670","Text":"If it isn\u0027t, once we add in the negative,"},{"Start":"09:28.670 ","End":"09:31.295","Text":"the directions will soil each other out."},{"Start":"09:31.295 ","End":"09:33.400","Text":"If q is a positive charge,"},{"Start":"09:33.400 ","End":"09:37.190","Text":"then the electric field is coming out of q and we\u0027re"},{"Start":"09:37.190 ","End":"09:41.615","Text":"measuring the electric field from q to our point B,"},{"Start":"09:41.615 ","End":"09:45.915","Text":"which means that it\u0027s going in this direction and this leftwards direction,"},{"Start":"09:45.915 ","End":"09:52.511","Text":"which means that it\u0027s going in the negative x direction."},{"Start":"09:52.511 ","End":"09:57.680","Text":"Now we add on the electric field due to our q tag."},{"Start":"09:57.680 ","End":"10:01.175","Text":"We have k multiplied by q tag."},{"Start":"10:01.175 ","End":"10:02.435","Text":"Let\u0027s already sub in."},{"Start":"10:02.435 ","End":"10:07.280","Text":"So q tag is equal to negative half q multiplied by"},{"Start":"10:07.280 ","End":"10:13.715","Text":"negative half q divided by the distance between q tag and B."},{"Start":"10:13.715 ","End":"10:16.355","Text":"So B is at a,"},{"Start":"10:16.355 ","End":"10:22.722","Text":"q tag is at 1/2a divided by 1/2a^2."},{"Start":"10:22.722 ","End":"10:27.695","Text":"Then, so q tag,"},{"Start":"10:27.695 ","End":"10:29.770","Text":"we can see is a negative charge."},{"Start":"10:29.770 ","End":"10:38.330","Text":"We have a minus so the electric field lines are going into the charge q tag,"},{"Start":"10:38.330 ","End":"10:40.440","Text":"which is at point a."},{"Start":"10:41.680 ","End":"10:45.485","Text":"That\u0027s going to be in the negative x direction."},{"Start":"10:45.485 ","End":"10:51.215","Text":"But we\u0027re going from A to B because that\u0027s the direction of R. We have negative"},{"Start":"10:51.215 ","End":"10:58.230","Text":"multiplied by a negative so this is just going to be in the x direction."},{"Start":"10:58.610 ","End":"11:06.855","Text":"I\u0027ll just explain that the electric field lines are going in the negative x direction,"},{"Start":"11:06.855 ","End":"11:12.110","Text":"from B to A, because the electric field lines are going in to q tag."},{"Start":"11:12.110 ","End":"11:15.445","Text":"However, our r vector,"},{"Start":"11:15.445 ","End":"11:19.115","Text":"let\u0027s just draw it, is in this direction,"},{"Start":"11:19.115 ","End":"11:22.040","Text":"which is in the negative direction to"},{"Start":"11:22.040 ","End":"11:24.260","Text":"the electric field lines and"},{"Start":"11:24.260 ","End":"11:27.170","Text":"the electric field lines are going in the negative x direction."},{"Start":"11:27.170 ","End":"11:30.200","Text":"We have negative x multiplied by"},{"Start":"11:30.200 ","End":"11:33.860","Text":"the negative vector because it\u0027s in the opposite direction to r,"},{"Start":"11:33.860 ","End":"11:36.785","Text":"which means we get a plus over here."},{"Start":"11:36.785 ","End":"11:41.090","Text":"Then we can just add all of this together."},{"Start":"11:41.090 ","End":"11:50.405","Text":"What we\u0027re going to have is we\u0027re going to have negative kq divided by a squared,"},{"Start":"11:50.405 ","End":"11:56.880","Text":"and then we have multiplied by 1 plus 2."},{"Start":"11:57.750 ","End":"12:00.805","Text":"Once you get the Algebra over here,"},{"Start":"12:00.805 ","End":"12:02.890","Text":"and you make a common denominator,"},{"Start":"12:02.890 ","End":"12:12.565","Text":"so what we\u0027re going to get is that this is equal to negative 3kq divided by a^2."},{"Start":"12:12.565 ","End":"12:14.800","Text":"This, of course, is in the x direction,"},{"Start":"12:14.800 ","End":"12:18.320","Text":"and so is this."},{"Start":"12:19.050 ","End":"12:24.655","Text":"Now let\u0027s substitute all of this into our equation for Sigma_B,"},{"Start":"12:24.655 ","End":"12:27.220","Text":"the surface charge distribution."},{"Start":"12:27.220 ","End":"12:33.925","Text":"This is equal to Epsilon_0 multiplied by the jump in the perpendicular electric field,"},{"Start":"12:33.925 ","End":"12:38.605","Text":"so we have here E_2 minus E_1."},{"Start":"12:38.605 ","End":"12:44.200","Text":"We have negative 3kq divided by a^2."},{"Start":"12:44.200 ","End":"12:48.715","Text":"Notice we shouldn\u0027t put in"},{"Start":"12:48.715 ","End":"12:56.950","Text":"the direction because Sigma_B is not a vector, it\u0027s a quantity."},{"Start":"12:56.950 ","End":"12:58.915","Text":"Then minus 0,"},{"Start":"12:58.915 ","End":"13:00.370","Text":"so we just get this."},{"Start":"13:00.370 ","End":"13:03.440","Text":"This is the answer to Question 2."},{"Start":"13:04.620 ","End":"13:08.455","Text":"Now let\u0027s answer question number 3."},{"Start":"13:08.455 ","End":"13:11.410","Text":"Calculate the force acting on q."},{"Start":"13:11.410 ","End":"13:15.415","Text":"We\u0027re just going to use our classic Coulomb\u0027s law."},{"Start":"13:15.415 ","End":"13:18.985","Text":"It\u0027s the force acting on q."},{"Start":"13:18.985 ","End":"13:22.330","Text":"Let\u0027s see f on q."},{"Start":"13:22.330 ","End":"13:26.800","Text":"That is just going to be equal to k multiplied by"},{"Start":"13:26.800 ","End":"13:33.860","Text":"q_1 multiplied by q_2 divided by the distance between the 2^2 squared."},{"Start":"13:33.930 ","End":"13:42.280","Text":"In our example, we have kq_1 as just q. Q_2 is q tag,"},{"Start":"13:42.280 ","End":"13:46.510","Text":"so q tag is negative 1/2q,"},{"Start":"13:46.510 ","End":"13:52.090","Text":"and then divided by the distance between the 2^2."},{"Start":"13:52.090 ","End":"13:59.320","Text":"Q is located at this distance over here,"},{"Start":"13:59.320 ","End":"14:03.459","Text":"which is 2_a, and q tag is located,"},{"Start":"14:03.459 ","End":"14:05.500","Text":"obviously this distance over here,"},{"Start":"14:05.500 ","End":"14:08.290","Text":"which is 1/2 a."},{"Start":"14:08.290 ","End":"14:15.860","Text":"What we\u0027re going to have is 2_a minus 1/2 a^2."},{"Start":"14:17.310 ","End":"14:24.625","Text":"Of course we can add in over here that this is going to be in the x-direction."},{"Start":"14:24.625 ","End":"14:29.170","Text":"Specifically, here we can see it\u0027s going to be in some way in the x-direction,"},{"Start":"14:29.170 ","End":"14:31.820","Text":"either the positive or the negative."},{"Start":"14:31.860 ","End":"14:38.395","Text":"We can see that we have a negative value over here."},{"Start":"14:38.395 ","End":"14:41.515","Text":"We can put this over like this,"},{"Start":"14:41.515 ","End":"14:44.995","Text":"and then we can carry this through."},{"Start":"14:44.995 ","End":"14:50.155","Text":"We\u0027ll get that F is equal to simply,"},{"Start":"14:50.155 ","End":"14:52.270","Text":"once we solve all of this,"},{"Start":"14:52.270 ","End":"15:01.735","Text":"this will be equal to 2_kq^2 divided by 9_a^2,"},{"Start":"15:01.735 ","End":"15:03.955","Text":"and we have the negative over here,"},{"Start":"15:03.955 ","End":"15:06.460","Text":"so this is in the negative x-direction,"},{"Start":"15:06.460 ","End":"15:11.900","Text":"which means that it\u0027s an attractive force between q and q tag."},{"Start":"15:13.680 ","End":"15:18.415","Text":"Now let\u0027s move on to question number 4."},{"Start":"15:18.415 ","End":"15:19.930","Text":"I\u0027ll do it over here."},{"Start":"15:19.930 ","End":"15:24.010","Text":"What is the energy required to build the system?"},{"Start":"15:24.010 ","End":"15:29.335","Text":"Here, our equation for energy is equal to 1/2"},{"Start":"15:29.335 ","End":"15:34.825","Text":"of the sum of q_i divided by Phi,"},{"Start":"15:34.825 ","End":"15:43.250","Text":"where q_i is each individual charge multiplied by the potential at that point."},{"Start":"15:43.650 ","End":"15:50.005","Text":"Let\u0027s do this. We have 1/2 multiplied by q_i,"},{"Start":"15:50.005 ","End":"15:51.355","Text":"so our first q,"},{"Start":"15:51.355 ","End":"15:53.290","Text":"let\u0027s start with just q."},{"Start":"15:53.290 ","End":"15:55.195","Text":"We have our charge q,"},{"Start":"15:55.195 ","End":"16:05.365","Text":"and the potential on charge q is going to be due to the potential that we had in area 2."},{"Start":"16:05.365 ","End":"16:08.725","Text":"Q is located in Zone 2."},{"Start":"16:08.725 ","End":"16:13.525","Text":"That means the potential in Zone 2 is acting on Q."},{"Start":"16:13.525 ","End":"16:16.465","Text":"What is the potential in Zone 2?"},{"Start":"16:16.465 ","End":"16:22.390","Text":"The potential in Zone 2 is equal to kq divided"},{"Start":"16:22.390 ","End":"16:29.665","Text":"by r_1 plus kq tag divided by r_2."},{"Start":"16:29.665 ","End":"16:38.755","Text":"Q can\u0027t cause a potential on q itself because they are located at the exact same point,"},{"Start":"16:38.755 ","End":"16:41.020","Text":"so that r is equal to 0,"},{"Start":"16:41.020 ","End":"16:43.090","Text":"and then we get an infinity thing over here,"},{"Start":"16:43.090 ","End":"16:44.845","Text":"and it\u0027s a bit of a problem."},{"Start":"16:44.845 ","End":"16:51.110","Text":"The only charge causing a potential at q is q tag."},{"Start":"16:51.720 ","End":"16:57.730","Text":"That is simply going to be equal to k multiplied by q tag,"},{"Start":"16:57.730 ","End":"17:00.505","Text":"so that is negative 1/2q,"},{"Start":"17:00.505 ","End":"17:04.915","Text":"divided by the distance between the two."},{"Start":"17:04.915 ","End":"17:09.190","Text":"We\u0027ve already seen that that is simply equal"},{"Start":"17:09.190 ","End":"17:17.320","Text":"to 3 over 2 multiplied by a."},{"Start":"17:17.320 ","End":"17:25.735","Text":"Then our next charge is naught q tag."},{"Start":"17:25.735 ","End":"17:29.455","Text":"Here it gets complicated, please concentrate."},{"Start":"17:29.455 ","End":"17:32.560","Text":"q tag is what we put in,"},{"Start":"17:32.560 ","End":"17:35.395","Text":"in order to use the method of images."},{"Start":"17:35.395 ","End":"17:38.410","Text":"We know that q tag is simply"},{"Start":"17:38.410 ","End":"17:44.590","Text":"a mathematical tool in order to calculate this potential in Zone 2."},{"Start":"17:44.590 ","End":"17:47.500","Text":"This is a theoretical charge,"},{"Start":"17:47.500 ","End":"17:49.285","Text":"it doesn\u0027t really exist,"},{"Start":"17:49.285 ","End":"17:50.950","Text":"and it\u0027s a mathematical tool."},{"Start":"17:50.950 ","End":"17:56.530","Text":"We only use q tag in equations when we\u0027re dealing with something that"},{"Start":"17:56.530 ","End":"18:03.620","Text":"includes the potential in this area outside of the grounded sphere."},{"Start":"18:03.960 ","End":"18:07.195","Text":"But q tag doesn\u0027t really exist."},{"Start":"18:07.195 ","End":"18:14.240","Text":"What does exist is the tiny little charges on the surface of the spherical shell."},{"Start":"18:14.970 ","End":"18:18.115","Text":"Let\u0027s call that q Tilde."},{"Start":"18:18.115 ","End":"18:24.805","Text":"q Tilde, all of the tiny little charges all along the surface."},{"Start":"18:24.805 ","End":"18:29.875","Text":"However, we know that this is a grounded spherical shell,"},{"Start":"18:29.875 ","End":"18:35.455","Text":"so when we go to multiply all of these charges on the surface of their spherical shell,"},{"Start":"18:35.455 ","End":"18:39.730","Text":"multiplied by the potential acting on them,"},{"Start":"18:39.730 ","End":"18:46.464","Text":"we know that the potential on the surface over here is equal to 0."},{"Start":"18:46.464 ","End":"18:50.425","Text":"Then this is going to be simply multiplied by 0,"},{"Start":"18:50.425 ","End":"18:55.075","Text":"and then it doesn\u0027t really affect our calculation."},{"Start":"18:55.075 ","End":"18:58.945","Text":"All that we\u0027ll get is a half multiplied by this."},{"Start":"18:58.945 ","End":"19:07.900","Text":"What we\u0027re going to get is negative kq^2 divided by 3_a."},{"Start":"19:07.900 ","End":"19:10.870","Text":"This is the answer to question number 4,"},{"Start":"19:10.870 ","End":"19:13.090","Text":"the energy required to build the system,"},{"Start":"19:13.090 ","End":"19:16.760","Text":"and that is the end of this lesson."}],"ID":14223},{"Watched":false,"Name":"Two Parallel Infinite Cylinders","Duration":"24m 8s","ChapterTopicVideoID":12121,"CourseChapterTopicPlaylistID":161612,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12121.jpeg","UploadDate":"2018-06-28T03:54:59.4870000","DurationForVideoObject":"PT24M8S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.725","Text":"Hello. In this lesson we have 2 parallel infinite cylinders,"},{"Start":"00:04.725 ","End":"00:08.055","Text":"and we\u0027re being told that they are conductors."},{"Start":"00:08.055 ","End":"00:13.350","Text":"We can see that they\u0027re a distance D away from one another."},{"Start":"00:13.350 ","End":"00:17.925","Text":"We\u0027re being told that there is a potential difference of V_0 between"},{"Start":"00:17.925 ","End":"00:24.225","Text":"the cylinders and we\u0027re being asked to find the potential outside of the cylinders."},{"Start":"00:24.225 ","End":"00:26.190","Text":"What we\u0027ll do is,"},{"Start":"00:26.190 ","End":"00:31.054","Text":"we\u0027ll say that this place right in the center between the 2 cylinders,"},{"Start":"00:31.054 ","End":"00:35.525","Text":"so we\u0027ll say that the potential over here is equal to 0."},{"Start":"00:35.525 ","End":"00:41.060","Text":"Therefore, we can say that the potential or the voltage,"},{"Start":"00:41.060 ","End":"00:47.900","Text":"or the potential difference on this cylinder is equal to V_0 divided by 2."},{"Start":"00:47.900 ","End":"00:51.770","Text":"On this cylinder, the voltage or the potential difference is equal"},{"Start":"00:51.770 ","End":"00:56.320","Text":"to negative V_0 divided by 2."},{"Start":"00:57.110 ","End":"01:04.550","Text":"First of all, we can notice that if we have a problem where we have 2 infinite wires,"},{"Start":"01:04.550 ","End":"01:10.460","Text":"that have charged density Lambda on this one a negative Lambda on this one."},{"Start":"01:10.460 ","End":"01:17.525","Text":"Then what we do is we find surfaces of uniform potential,"},{"Start":"01:17.525 ","End":"01:23.965","Text":"which is just cylindrical surfaces that go around the wire."},{"Start":"01:23.965 ","End":"01:28.760","Text":"Here we have uniform potential along this surface."},{"Start":"01:28.760 ","End":"01:31.400","Text":"If we can do that for a wire,"},{"Start":"01:31.400 ","End":"01:36.070","Text":"then we can do that for these cylinders."},{"Start":"01:36.070 ","End":"01:38.490","Text":"What we\u0027re going to do is,"},{"Start":"01:38.490 ","End":"01:44.200","Text":"we\u0027re going to put 2 infinitely long wires inside these cylinders."},{"Start":"01:44.200 ","End":"01:46.680","Text":"One will be over here."},{"Start":"01:46.680 ","End":"01:52.335","Text":"This is an infinitely long wire of a charge density Lambda."},{"Start":"01:52.335 ","End":"01:55.600","Text":"This distance over here is a,"},{"Start":"01:55.600 ","End":"01:59.510","Text":"and the other one we\u0027re going to place over here."},{"Start":"01:59.510 ","End":"02:02.270","Text":"Because this has the negative potential difference,"},{"Start":"02:02.270 ","End":"02:05.900","Text":"so we\u0027ll say that this is of charge negative Lambda"},{"Start":"02:05.900 ","End":"02:11.020","Text":"and that this distance over here is also a."},{"Start":"02:11.020 ","End":"02:13.280","Text":"Now what we want to do is,"},{"Start":"02:13.280 ","End":"02:15.740","Text":"we want to find out what a is,"},{"Start":"02:15.740 ","End":"02:19.910","Text":"so what this distance is and what Lambda is,"},{"Start":"02:19.910 ","End":"02:26.510","Text":"so that we can know what charged tile or what charge wire we have"},{"Start":"02:26.510 ","End":"02:32.570","Text":"to put inside the cylinder such that we\u0027ll get this potential difference over here,"},{"Start":"02:32.570 ","End":"02:34.850","Text":"V_0 divided by 2, and of here,"},{"Start":"02:34.850 ","End":"02:38.310","Text":"negative V_0 divided by 2."},{"Start":"02:38.640 ","End":"02:44.310","Text":"A negative Lambda over here are our image charges."},{"Start":"02:44.310 ","End":"02:50.275","Text":"Then once we find this electric field from these 2,"},{"Start":"02:50.275 ","End":"02:53.840","Text":"so it will be the same electric field and"},{"Start":"02:53.840 ","End":"02:58.795","Text":"potential that we would have outside of the 2 cylinders."},{"Start":"02:58.795 ","End":"03:02.000","Text":"Now what we\u0027re going to do is we\u0027re going to choose"},{"Start":"03:02.000 ","End":"03:05.600","Text":"an arbitrary point on the surface of one of the cylinders."},{"Start":"03:05.600 ","End":"03:07.310","Text":"Let\u0027s say over here."},{"Start":"03:07.310 ","End":"03:11.735","Text":"We\u0027re going to calculate the potential at this point."},{"Start":"03:11.735 ","End":"03:16.020","Text":"We can say that the angle over here is Theta."},{"Start":"03:16.370 ","End":"03:18.750","Text":"The next thing that we have to do,"},{"Start":"03:18.750 ","End":"03:20.415","Text":"is we have to calibrate this."},{"Start":"03:20.415 ","End":"03:24.220","Text":"We have to find out what IC is equal to."},{"Start":"03:24.220 ","End":"03:30.650","Text":"What we know is that our potential over here is equal to 0."},{"Start":"03:30.650 ","End":"03:36.230","Text":"Let\u0027s call this distance over here for the meantime,"},{"Start":"03:36.230 ","End":"03:40.705","Text":"S. We\u0027ll see what it is later if we need it."},{"Start":"03:40.705 ","End":"03:47.083","Text":"Then our potential at the origin,"},{"Start":"03:47.083 ","End":"03:54.375","Text":"at point 0 is equal to the potential from this wire with a charge density Lambda."},{"Start":"03:54.375 ","End":"03:59.570","Text":"That we got as equal to negative Lambda divided"},{"Start":"03:59.570 ","End":"04:05.885","Text":"by 2 Pi Epsilon naught multiplied by ln of r,"},{"Start":"04:05.885 ","End":"04:08.375","Text":"which here is S plus,"},{"Start":"04:08.375 ","End":"04:10.760","Text":"let\u0027s call it in the meantime C_1."},{"Start":"04:10.760 ","End":"04:17.000","Text":"This C. Then we\u0027re adding plus"},{"Start":"04:17.000 ","End":"04:23.735","Text":"the potential that we get from this Y over here of charge density negative Lambda."},{"Start":"04:23.735 ","End":"04:30.280","Text":"Then we have negative of negative Lambda divided by"},{"Start":"04:30.280 ","End":"04:37.604","Text":"2 Pi Epsilon naught and because these are equidistant,"},{"Start":"04:37.604 ","End":"04:43.150","Text":"we can say that this distance over here is also S. Again,"},{"Start":"04:43.150 ","End":"04:46.770","Text":"ln of S plus C_2."},{"Start":"04:46.770 ","End":"04:50.660","Text":"Of course the potential over here is equal to 0."},{"Start":"04:50.660 ","End":"04:55.845","Text":"What we see is that we have negative Lambda divided by 2 Pi Epsilon naught"},{"Start":"04:55.845 ","End":"05:01.880","Text":"ln S. We have plus and then a minus and a minus is plus,"},{"Start":"05:01.880 ","End":"05:08.510","Text":"so plus Lambda divided by 2 Pi Epsilon naught ln S. Those will cancel out,"},{"Start":"05:08.510 ","End":"05:15.080","Text":"and therefore what we\u0027ll get is that C_1 plus C_2 is equal to 0."},{"Start":"05:15.080 ","End":"05:20.060","Text":"Therefore, we can say that C_1 is equal to C_2,"},{"Start":"05:20.060 ","End":"05:22.500","Text":"which is equal to 0."},{"Start":"05:24.230 ","End":"05:30.200","Text":"In that case, what we can do is we can say that the total potential due to"},{"Start":"05:30.200 ","End":"05:38.760","Text":"these 2 charged wires is equal to the potential from this positively charged one."},{"Start":"05:38.760 ","End":"05:49.205","Text":"That is equal to negative Lambda divided by 2 Pi Epsilon naught ln of r,"},{"Start":"05:49.205 ","End":"05:50.615","Text":"and then it\u0027s the positive one."},{"Start":"05:50.615 ","End":"05:53.120","Text":"Let\u0027s call it r plus."},{"Start":"05:53.120 ","End":"06:02.140","Text":"Then we\u0027re adding this negative Lambda."},{"Start":"06:02.140 ","End":"06:04.425","Text":"We can just add it,"},{"Start":"06:04.425 ","End":"06:07.305","Text":"plus Lambda divided by"},{"Start":"06:07.305 ","End":"06:13.775","Text":"2 Pi Epsilon naught of ln of the radius from this negatively charged ones."},{"Start":"06:13.775 ","End":"06:15.965","Text":"Let\u0027s call this r minus."},{"Start":"06:15.965 ","End":"06:22.500","Text":"All in all, we can write this as Lambda divided by"},{"Start":"06:22.500 ","End":"06:28.000","Text":"2 Pi Epsilon naught multiplied by"},{"Start":"06:28.000 ","End":"06:34.515","Text":"ln of r minus divided by r plus."},{"Start":"06:34.515 ","End":"06:36.450","Text":"This is the potential."},{"Start":"06:36.450 ","End":"06:40.625","Text":"Now all I have to do is I have to check that this"},{"Start":"06:40.625 ","End":"06:46.925","Text":"potential abides by my conditions in the question."},{"Start":"06:46.925 ","End":"06:51.650","Text":"Namely that, well, we\u0027ve seen that the potential here is equal to 0,"},{"Start":"06:51.650 ","End":"06:57.770","Text":"but also that my potential difference all around my cylinder over here,"},{"Start":"06:57.770 ","End":"07:01.100","Text":"let\u0027s say, is equal to V_0 divided by 2."},{"Start":"07:01.100 ","End":"07:05.670","Text":"On this cylinder to negative V_0 divided by 2."},{"Start":"07:05.880 ","End":"07:09.910","Text":"Okay. We have this point over here."},{"Start":"07:09.910 ","End":"07:12.505","Text":"Let\u0027s call this point P,"},{"Start":"07:12.505 ","End":"07:15.730","Text":"and we want to work out the potential over there."},{"Start":"07:15.730 ","End":"07:18.700","Text":"We have in blue,"},{"Start":"07:18.700 ","End":"07:22.090","Text":"from our Lambda all the way to point P,"},{"Start":"07:22.090 ","End":"07:25.885","Text":"this is our r1 vector."},{"Start":"07:25.885 ","End":"07:30.610","Text":"For my negative Lambda all the way to P,"},{"Start":"07:30.610 ","End":"07:34.495","Text":"we have our r2 vector."},{"Start":"07:34.495 ","End":"07:39.400","Text":"I want to work out the potential over here."},{"Start":"07:39.400 ","End":"07:44.480","Text":"I\u0027m going to rub out everything that we don\u0027t need to give us a little bit more space."},{"Start":"07:44.910 ","End":"07:54.670","Text":"Now what I need is that my potential at this point P has to equal to V_0 divided by 2."},{"Start":"07:54.670 ","End":"08:01.660","Text":"Let\u0027s write, my potential at point P has to be equal to my equation over here,"},{"Start":"08:01.660 ","End":"08:10.435","Text":"which is Lambda divided by 2Pi Epsilon naught multiplied by ln."},{"Start":"08:10.435 ","End":"08:15.850","Text":"Then I have my r minus divided by my r plus."},{"Start":"08:15.850 ","End":"08:22.040","Text":"This has to be equal to V_0 divided by 2."},{"Start":"08:22.140 ","End":"08:28.150","Text":"What we want to do is we want to isolate out this fraction of r minus divided by r plus."},{"Start":"08:28.150 ","End":"08:30.895","Text":"We\u0027re going to multiply both sides by 2,"},{"Start":"08:30.895 ","End":"08:37.220","Text":"and we\u0027re going to divide both sides by Lambda divided by Pi Epsilon naught."},{"Start":"08:38.160 ","End":"08:47.065","Text":"Then we\u0027re going to raise everything by E. What we\u0027re going to get therefore,"},{"Start":"08:47.065 ","End":"08:55.570","Text":"is that r minus divided by r plus is equal to e to"},{"Start":"08:55.570 ","End":"09:04.555","Text":"the power of Pi Epsilon Naught multiplied by V_0 divided by Lambda."},{"Start":"09:04.555 ","End":"09:08.425","Text":"All of this we can consider as a constant right now,"},{"Start":"09:08.425 ","End":"09:12.890","Text":"that is independent of our angle over here, Theta."},{"Start":"09:14.100 ","End":"09:17.620","Text":"Because we\u0027re considering this a constant for now,"},{"Start":"09:17.620 ","End":"09:21.670","Text":"let\u0027s call all of this C. Then we\u0027ll get that"},{"Start":"09:21.670 ","End":"09:28.940","Text":"r minus is equal to C multiplied by r plus."},{"Start":"09:29.730 ","End":"09:39.099","Text":"Now let\u0027s call this point M and this point N. Then using my law of cosines,"},{"Start":"09:39.099 ","End":"09:43.790","Text":"I can say that my triangle O_1PM."},{"Start":"09:46.860 ","End":"09:50.485","Text":"Through that using my law of cosines,"},{"Start":"09:50.485 ","End":"09:56.620","Text":"so I can say that r plus squared is equal to"},{"Start":"09:56.620 ","End":"10:04.700","Text":"r^2 plus a^2 minus 2aR,"},{"Start":"10:05.490 ","End":"10:14.390","Text":"multiplied by cosine of Theta and through my other triangle."},{"Start":"10:14.580 ","End":"10:23.020","Text":"That is triangle O_1 PN to this point over here."},{"Start":"10:23.020 ","End":"10:31.287","Text":"Then I\u0027ll get that my r minus squared is equal to R^2 plus,"},{"Start":"10:31.287 ","End":"10:37.825","Text":"this distance over here is D minus this distance over here,"},{"Start":"10:37.825 ","End":"10:47.000","Text":"a plus D minus a^2 minus 2."},{"Start":"10:47.550 ","End":"10:56.750","Text":"Then again, D minus aR multiplied by cosine of Theta."},{"Start":"10:57.600 ","End":"11:02.500","Text":"Now let\u0027s raise this to the power of 2."},{"Start":"11:02.500 ","End":"11:05.110","Text":"Of course, C is a constant,"},{"Start":"11:05.110 ","End":"11:08.035","Text":"so C^2 is also a constant."},{"Start":"11:08.035 ","End":"11:10.690","Text":"Let\u0027s scroll down."},{"Start":"11:10.690 ","End":"11:12.790","Text":"We\u0027re plugging in here,"},{"Start":"11:12.790 ","End":"11:17.530","Text":"r minus squared, so that r minus squared, let\u0027s write it here."},{"Start":"11:17.530 ","End":"11:24.490","Text":"Is equal to R^2 plus D minus a^2 minus"},{"Start":"11:24.490 ","End":"11:34.270","Text":"2D minus aR cosine of Theta is equal to C^2."},{"Start":"11:34.270 ","End":"11:38.230","Text":"This constant multiplied by r plus squared,"},{"Start":"11:38.230 ","End":"11:48.590","Text":"which is R^2 plus a^2 minus 2ar cosine of Theta."},{"Start":"11:48.810 ","End":"11:52.540","Text":"Now we\u0027re going to move everything to 1 side and"},{"Start":"11:52.540 ","End":"11:56.020","Text":"take out cosine Theta as a common multiple."},{"Start":"11:56.020 ","End":"12:05.260","Text":"What we\u0027re going to get is that R^2 plus D minus a^2 plus,"},{"Start":"12:07.680 ","End":"12:15.670","Text":"or rather minus C^2 multiplied by R^2 plus a^2"},{"Start":"12:15.670 ","End":"12:23.830","Text":"plus C^2 multiplied by R multiplied by 2,"},{"Start":"12:23.830 ","End":"12:32.845","Text":"multiplied by a minus 2R D minus a."},{"Start":"12:32.845 ","End":"12:40.870","Text":"All of this multiplied by cosine of Theta is equal to 0."},{"Start":"12:40.870 ","End":"12:42.940","Text":"I just moved everything to 1 side."},{"Start":"12:42.940 ","End":"12:44.965","Text":"I opened up these brackets over here."},{"Start":"12:44.965 ","End":"12:46.615","Text":"I made a cosine Theta,"},{"Start":"12:46.615 ","End":"12:53.870","Text":"a common multiple of what\u0027s inside this bracket over here."},{"Start":"12:56.010 ","End":"13:00.700","Text":"Now what I need to do is I need to ensure"},{"Start":"13:00.700 ","End":"13:05.845","Text":"that my potential along any point along the cylinder."},{"Start":"13:05.845 ","End":"13:08.545","Text":"That means for any value of Theta,"},{"Start":"13:08.545 ","End":"13:12.850","Text":"that this equation will be equal to 0."},{"Start":"13:12.850 ","End":"13:18.280","Text":"What I have to do is I have to find a way for this to be equal to"},{"Start":"13:18.280 ","End":"13:26.690","Text":"0 and for this to be equal to 0."},{"Start":"13:27.450 ","End":"13:30.760","Text":"This to be equal to 0 and this to be equal to"},{"Start":"13:30.760 ","End":"13:36.530","Text":"0 independent of what value of Theta I plug-in over here."},{"Start":"13:37.290 ","End":"13:45.490","Text":"Now what we get is 2 equations with 2 unknowns."},{"Start":"13:45.490 ","End":"13:53.931","Text":"We have on the 1 hand that R^2 plus D minus a^2"},{"Start":"13:53.931 ","End":"14:02.380","Text":"minus C^2 multiplied by R^2 plus a^2 is equal to 0."},{"Start":"14:02.380 ","End":"14:09.505","Text":"The other equation is that 2C^2 Ra"},{"Start":"14:09.505 ","End":"14:17.800","Text":"minus 2R multiplied by D minus a is equal to 0."},{"Start":"14:17.800 ","End":"14:20.455","Text":"These are 2 equations."},{"Start":"14:20.455 ","End":"14:24.170","Text":"What you can do is either you can use a computer to solve this,"},{"Start":"14:24.170 ","End":"14:27.890","Text":"because we\u0027re going to get equations where we have a to the power of"},{"Start":"14:27.890 ","End":"14:32.660","Text":"3 or we can use a trick."},{"Start":"14:32.660 ","End":"14:37.400","Text":"The trick that we\u0027re going to use is this knowledge that we needed this over here,"},{"Start":"14:37.400 ","End":"14:45.420","Text":"this fraction of R minus divided by R plus to be equal to a constant."},{"Start":"14:47.100 ","End":"14:53.870","Text":"What this means, if this over here is a constant,"},{"Start":"14:53.870 ","End":"15:00.020","Text":"that means that what we\u0027ll have over here is that this over here is a constant."},{"Start":"15:00.020 ","End":"15:07.114","Text":"That means that for any point that we would choose along the surface of the cylinder,"},{"Start":"15:07.114 ","End":"15:12.635","Text":"the ratio of r minus divided by r plus will be a constant value,"},{"Start":"15:12.635 ","End":"15:16.490","Text":"which means that our potential at these points will always be,"},{"Start":"15:16.490 ","End":"15:19.285","Text":"a constant, will be the same."},{"Start":"15:19.285 ","End":"15:25.955","Text":"What we\u0027re going to do is we\u0027re going to use the trick, of similar triangles."},{"Start":"15:25.955 ","End":"15:32.052","Text":"What I need to make sure that triangle,"},{"Start":"15:32.052 ","End":"15:35.005","Text":"and it\u0027s important how I\u0027m writing it the order."},{"Start":"15:35.005 ","End":"15:43.280","Text":"The triangle of 0_1 PM so 0_1 PM,"},{"Start":"15:43.280 ","End":"15:48.830","Text":"this triangle has to be equal to or similar to"},{"Start":"15:48.830 ","End":"15:57.650","Text":"triangle O_1NP so O_1NP."},{"Start":"16:00.270 ","End":"16:07.315","Text":"What do we get if we have this is That therefore,"},{"Start":"16:07.315 ","End":"16:12.580","Text":"our r plus divided by r minus has to be"},{"Start":"16:12.580 ","End":"16:18.445","Text":"equal to R divided by D minus a,"},{"Start":"16:18.445 ","End":"16:28.280","Text":"which is also equal to a divided by R. We need these relationships between these sides."},{"Start":"16:28.920 ","End":"16:34.015","Text":"Here is our r plus."},{"Start":"16:34.015 ","End":"16:38.095","Text":"It has to be the same as this r minus."},{"Start":"16:38.095 ","End":"16:42.250","Text":"The relationship between r plus and r minus is"},{"Start":"16:42.250 ","End":"16:48.710","Text":"the same relationship between r and this over here, D minus a."},{"Start":"16:48.780 ","End":"16:59.185","Text":"It\u0027s the same relationship as a divided by R."},{"Start":"16:59.185 ","End":"17:03.175","Text":"Now because we want r minus divided by r plus,"},{"Start":"17:03.175 ","End":"17:05.439","Text":"we\u0027re going to take the reciprocal of this,"},{"Start":"17:05.439 ","End":"17:08.120","Text":"so I\u0027m going to flip it around."},{"Start":"17:08.910 ","End":"17:14.140","Text":"This is the condition that we need to meet in order to have"},{"Start":"17:14.140 ","End":"17:20.005","Text":"these similar triangles and therefore in order to answer these two equations."},{"Start":"17:20.005 ","End":"17:25.105","Text":"What we\u0027re going to do is we\u0027re trying to simplify this a little bit."},{"Start":"17:25.105 ","End":"17:27.850","Text":"What we\u0027ll get, therefore,"},{"Start":"17:27.850 ","End":"17:30.940","Text":"is that D minus a,"},{"Start":"17:30.940 ","End":"17:37.090","Text":"K multiplied by a is equal to R^2."},{"Start":"17:37.090 ","End":"17:40.270","Text":"Therefore, let\u0027s get a quadratic equation."},{"Start":"17:40.270 ","End":"17:45.750","Text":"We get that a^2 minus Da"},{"Start":"17:45.750 ","End":"17:51.595","Text":"plus R^2 is equal to 0."},{"Start":"17:51.595 ","End":"17:56.030","Text":"Now we\u0027re going to solve this quadratic equation for a."},{"Start":"17:56.310 ","End":"18:03.010","Text":"In that case, we\u0027ll get that a is equal to D plus"},{"Start":"18:03.010 ","End":"18:11.610","Text":"minus the square root of D^2 minus 4R^2,"},{"Start":"18:11.610 ","End":"18:16.035","Text":"and all of this is divided by 2."},{"Start":"18:16.035 ","End":"18:21.959","Text":"Then what we have is D divided by 2 plus,"},{"Start":"18:21.959 ","End":"18:26.280","Text":"minus the square root of D divided"},{"Start":"18:26.280 ","End":"18:33.665","Text":"by 2^2 minus R^2."},{"Start":"18:33.665 ","End":"18:37.390","Text":"Now what we want to do is we want to show that we can get rid"},{"Start":"18:37.390 ","End":"18:41.560","Text":"of this option for D plus the square root."},{"Start":"18:41.560 ","End":"18:45.370","Text":"We\u0027re trying to show that the only viable answer"},{"Start":"18:45.370 ","End":"18:49.880","Text":"here is D divided by 2 minus the square root."},{"Start":"18:50.190 ","End":"18:59.350","Text":"What we want to say is that a has to be smaller than D divided by 2. Why is this?"},{"Start":"18:59.350 ","End":"19:02.410","Text":"Let\u0027s go back to our diagram to explain it."},{"Start":"19:02.410 ","End":"19:05.710","Text":"a is the distance over here."},{"Start":"19:05.710 ","End":"19:07.435","Text":"Let\u0027s say in this cylinder,"},{"Start":"19:07.435 ","End":"19:14.000","Text":"between the center of the cylinder to where we put our charged wire over here."},{"Start":"19:15.570 ","End":"19:22.795","Text":"The distance up until this midpoint is D divided by 2."},{"Start":"19:22.795 ","End":"19:26.500","Text":"If a is bigger than D divided by 2,"},{"Start":"19:26.500 ","End":"19:29.125","Text":"then it\u0027s outside of the cylinder."},{"Start":"19:29.125 ","End":"19:33.565","Text":"Even if the cylinders were side-by-side,"},{"Start":"19:33.565 ","End":"19:39.580","Text":"so not overlapping, but we had 2 cylinders side-by-side."},{"Start":"19:39.580 ","End":"19:42.856","Text":"Here\u0027s the center of one and here\u0027s the center of the other,"},{"Start":"19:42.856 ","End":"19:49.000","Text":"this distance over here will be D. If D is bigger than D divided by 2,"},{"Start":"19:49.000 ","End":"19:53.830","Text":"then our wire will be going already into the second cylinder,"},{"Start":"19:53.830 ","End":"19:56.620","Text":"which as we know, we don\u0027t want."},{"Start":"19:56.620 ","End":"19:59.635","Text":"I hope that that diagram was clear."},{"Start":"19:59.635 ","End":"20:03.265","Text":"That\u0027s why a has to always be before this point."},{"Start":"20:03.265 ","End":"20:09.355","Text":"Of course, a also has to be smaller than R because we\u0027re putting this wire inside"},{"Start":"20:09.355 ","End":"20:12.610","Text":"the cylinder of radius R. But it\u0027s"},{"Start":"20:12.610 ","End":"20:16.525","Text":"enough for what we need right now to say that a has to,"},{"Start":"20:16.525 ","End":"20:21.535","Text":"by definition be smaller for sure than D divided by 2."},{"Start":"20:21.535 ","End":"20:23.695","Text":"Now we can go down here."},{"Start":"20:23.695 ","End":"20:29.065","Text":"If we look, if we take D divided by 2 plus whatever is in the square root,"},{"Start":"20:29.065 ","End":"20:34.660","Text":"then D divided by 2 plus something is going to be bigger than D divided by 2."},{"Start":"20:34.660 ","End":"20:38.830","Text":"Therefore, we can only use the option of D divided by"},{"Start":"20:38.830 ","End":"20:46.570","Text":"2 minus this thing in the square root as then it satisfies this."},{"Start":"20:46.570 ","End":"20:53.500","Text":"Therefore, we can say that a is equal to D divided"},{"Start":"20:53.500 ","End":"21:00.400","Text":"by 2 minus the square root of D divided by 2."},{"Start":"21:00.400 ","End":"21:05.815","Text":"This is a D^2 minus R^2."},{"Start":"21:05.815 ","End":"21:11.935","Text":"We found our a, which as I remind you if we go up here,"},{"Start":"21:11.935 ","End":"21:14.305","Text":"we were trying to find a and Lambda."},{"Start":"21:14.305 ","End":"21:16.450","Text":"We\u0027ve already found our a."},{"Start":"21:16.450 ","End":"21:18.280","Text":"Now let\u0027s find Lambda."},{"Start":"21:18.280 ","End":"21:24.483","Text":"We have this equation over here that R minus divided by R plus is equal to C,"},{"Start":"21:24.483 ","End":"21:28.030","Text":"and within this constant C,"},{"Start":"21:28.030 ","End":"21:30.800","Text":"we have our Lambda."},{"Start":"21:31.260 ","End":"21:39.475","Text":"We already know what our R minus divided by R plus has to be equal to."},{"Start":"21:39.475 ","End":"21:41.455","Text":"We saw from over here,"},{"Start":"21:41.455 ","End":"21:44.740","Text":"R minus divided by R plus has to be equal to D minus a"},{"Start":"21:44.740 ","End":"21:51.670","Text":"divided by R. My C which is equal to"},{"Start":"21:51.670 ","End":"21:58.150","Text":"e to the power of Pi Epsilon naught V naught divided"},{"Start":"21:58.150 ","End":"22:04.930","Text":"by Lambda is also equal to my R minus divided by R plus."},{"Start":"22:04.930 ","End":"22:12.530","Text":"This over here, which we saw is equal to D minus a divided by R,"},{"Start":"22:12.530 ","End":"22:15.160","Text":"and I already know what my a is equal to."},{"Start":"22:15.160 ","End":"22:16.780","Text":"It\u0027s equal to this."},{"Start":"22:16.780 ","End":"22:20.980","Text":"Then we\u0027re just going to use this equation over here with this,"},{"Start":"22:20.980 ","End":"22:25.960","Text":"and therefore we\u0027ll get that"},{"Start":"22:25.960 ","End":"22:29.545","Text":"Lambda is equal to once we plug in what"},{"Start":"22:29.545 ","End":"22:33.625","Text":"a is and we\u0027ve rearranged all of this to isolate out Lambda,"},{"Start":"22:33.625 ","End":"22:43.750","Text":"we\u0027ll get that Lambda is equal to Pi Epsilon naught multiplied by V naught divided"},{"Start":"22:43.750 ","End":"22:52.090","Text":"by ln of D divided by 2R plus"},{"Start":"22:52.090 ","End":"23:02.780","Text":"the square root of D divided by (2R)^2 minus 1."},{"Start":"23:05.850 ","End":"23:07.990","Text":"These are our answers."},{"Start":"23:07.990 ","End":"23:10.660","Text":"We have a which is equal to this and Lambda which is"},{"Start":"23:10.660 ","End":"23:13.375","Text":"equal to this and then we can substitute it"},{"Start":"23:13.375 ","End":"23:19.675","Text":"in to these equations and see that we really get that these equations are equal to 0."},{"Start":"23:19.675 ","End":"23:23.710","Text":"Then once we have those values,"},{"Start":"23:23.710 ","End":"23:27.250","Text":"we can see that we\u0027ll get that the potential is"},{"Start":"23:27.250 ","End":"23:31.540","Text":"equal to what we have over here where we found out Lambda,"},{"Start":"23:31.540 ","End":"23:38.960","Text":"and we know that R minus divided by R plus is equal to this over here."},{"Start":"23:39.030 ","End":"23:42.415","Text":"Then we know what the potential of"},{"Start":"23:42.415 ","End":"23:47.350","Text":"two conducting cylinders is equal to when there\u0027s a distance D between the two,"},{"Start":"23:47.350 ","End":"23:52.675","Text":"and there\u0027s a potential difference of V_0 between the two of them."},{"Start":"23:52.675 ","End":"23:56.635","Text":"The potential is as if we would put"},{"Start":"23:56.635 ","End":"24:02.017","Text":"two charged wires a distance a from the center of each cylinder,"},{"Start":"24:02.017 ","End":"24:06.040","Text":"and we would work out the potential of that scenario."},{"Start":"24:06.040 ","End":"24:09.470","Text":"That is the end of this lesson."}],"ID":14224},{"Watched":false,"Name":"Exercise 5","Duration":"23m 18s","ChapterTopicVideoID":12122,"CourseChapterTopicPlaylistID":161612,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12122.jpeg","UploadDate":"2018-06-28T04:08:06.8870000","DurationForVideoObject":"PT23M18S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.425","Text":"Hello, in this lesson we\u0027re going to be answering the following question."},{"Start":"00:04.425 ","End":"00:11.265","Text":"A charge Q is placed a distance D away from 2 grounded infinite planes."},{"Start":"00:11.265 ","End":"00:15.945","Text":"Calculate the potential in the region between the 2 planes."},{"Start":"00:15.945 ","End":"00:19.530","Text":"We\u0027re calculating the potential which also means"},{"Start":"00:19.530 ","End":"00:24.345","Text":"the potential difference or the voltage in this region over here."},{"Start":"00:24.345 ","End":"00:29.705","Text":"We have 2 grounded infinite planes which means that the potential along"},{"Start":"00:29.705 ","End":"00:35.960","Text":"this plane is equal to 0 and the potential along this plane is equal to 0."},{"Start":"00:35.960 ","End":"00:40.265","Text":"What we\u0027re going to do is we\u0027re going to use the method of images"},{"Start":"00:40.265 ","End":"00:44.660","Text":"in order to calculate the potential over here in this region."},{"Start":"00:44.660 ","End":"00:47.510","Text":"We\u0027ve seen how to do this."},{"Start":"00:47.510 ","End":"00:56.055","Text":"We\u0027re going to put on each side of the 2 planes image charges."},{"Start":"00:56.055 ","End":"01:03.485","Text":"A distance D away will have a charge over here which is negative q and a distance D"},{"Start":"01:03.485 ","End":"01:12.310","Text":"away from over here will have another charge over here, negative q."},{"Start":"01:12.470 ","End":"01:14.600","Text":"We\u0027re playing by the rules."},{"Start":"01:14.600 ","End":"01:19.970","Text":"We\u0027ve only added charges in this region over here and this region over"},{"Start":"01:19.970 ","End":"01:26.150","Text":"here which is not in the area we\u0027re trying to find out the potential, so that\u0027s fine."},{"Start":"01:26.150 ","End":"01:28.580","Text":"We can see that this charge q and"},{"Start":"01:28.580 ","End":"01:32.240","Text":"this charge negative q will cancel out the potential along"},{"Start":"01:32.240 ","End":"01:36.155","Text":"this infinite plane and this charge q and this charge"},{"Start":"01:36.155 ","End":"01:41.040","Text":"negative q will cancel out the potential along this infinite plane."},{"Start":"01:41.040 ","End":"01:47.000","Text":"That\u0027s great. However, now we have a negative q charge over here and"},{"Start":"01:47.000 ","End":"01:53.030","Text":"a negative q charge over here and they don\u0027t cancel each other out."},{"Start":"01:53.030 ","End":"01:58.870","Text":"We\u0027re going to have to add in some other image charges."},{"Start":"01:59.660 ","End":"02:03.830","Text":"What we\u0027re going to do is something a bit weird."},{"Start":"02:03.830 ","End":"02:11.040","Text":"We can see that the distance between each 2 charges is the distance of d plus d, so 2d."},{"Start":"02:11.270 ","End":"02:13.755","Text":"Let\u0027s just draw that."},{"Start":"02:13.755 ","End":"02:18.635","Text":"Here we have d and another d and then here"},{"Start":"02:18.635 ","End":"02:23.870","Text":"we\u0027ll put a charge q plus and the same thing over here."},{"Start":"02:23.870 ","End":"02:26.970","Text":"Here we have a distance d plus d,"},{"Start":"02:26.970 ","End":"02:32.400","Text":"so that\u0027s 2d and over here we again put a charge q."},{"Start":"02:32.750 ","End":"02:37.675","Text":"Now we can see again the potential on each plane is equal to 0."},{"Start":"02:37.675 ","End":"02:42.030","Text":"We\u0027ve only added charges to the areas where we can."},{"Start":"02:42.030 ","End":"02:48.880","Text":"Then we can see that the negatives cancel with the positive qs and that\u0027s great."},{"Start":"02:48.880 ","End":"02:53.014","Text":"However, again, we have this charge over here plus"},{"Start":"02:53.014 ","End":"02:58.085","Text":"q and this charge over here plus q that don\u0027t cancel each other out."},{"Start":"02:58.085 ","End":"03:05.890","Text":"Then we\u0027re going to add again a distance of 2d away from each one."},{"Start":"03:05.890 ","End":"03:09.400","Text":"This is again 2d and we\u0027re going to have"},{"Start":"03:09.400 ","End":"03:14.415","Text":"a negative q over here and a negative q over here."},{"Start":"03:14.415 ","End":"03:16.310","Text":"Then again we have the same problem."},{"Start":"03:16.310 ","End":"03:18.050","Text":"All of the charges cancel out."},{"Start":"03:18.050 ","End":"03:21.245","Text":"All of the potential along the plane cancels out."},{"Start":"03:21.245 ","End":"03:27.110","Text":"But again, we have this negative q and negative q which don\u0027t cancel each other out."},{"Start":"03:27.110 ","End":"03:31.340","Text":"What we\u0027re going to do is we\u0027re going to keep adding at a distance"},{"Start":"03:31.340 ","End":"03:35.855","Text":"of 2d away and we\u0027re going to keep adding alternate charges."},{"Start":"03:35.855 ","End":"03:37.835","Text":"First it was negative,"},{"Start":"03:37.835 ","End":"03:39.630","Text":"then positive q,"},{"Start":"03:39.630 ","End":"03:44.110","Text":"then negative q, then again a positive q on each side."},{"Start":"03:44.110 ","End":"03:47.765","Text":"We\u0027re going to keep doing this until we\u0027ve added"},{"Start":"03:47.765 ","End":"03:51.590","Text":"an infinite amount of charges. What does that mean?"},{"Start":"03:51.590 ","End":"03:54.905","Text":"That infinitely far away from"},{"Start":"03:54.905 ","End":"03:59.765","Text":"each one of these infinite planes we\u0027re going to have some charge,"},{"Start":"03:59.765 ","End":"04:01.806","Text":"either plus q or minus q,"},{"Start":"04:01.806 ","End":"04:04.115","Text":"which is infinitely far away."},{"Start":"04:04.115 ","End":"04:06.530","Text":"Which means that its effect,"},{"Start":"04:06.530 ","End":"04:10.280","Text":"so I\u0027m reminding you that the potential is given"},{"Start":"04:10.280 ","End":"04:19.410","Text":"as kq divided by r. If r is approaching infinity,"},{"Start":"04:19.730 ","End":"04:24.230","Text":"so that will mean that therefore"},{"Start":"04:24.230 ","End":"04:30.980","Text":"our potential will be approaching 0 over here."},{"Start":"04:30.980 ","End":"04:33.830","Text":"Its effect is going to approach 0."},{"Start":"04:33.830 ","End":"04:35.210","Text":"The same over here."},{"Start":"04:35.210 ","End":"04:38.120","Text":"If we have some charge either plus q or negative q,"},{"Start":"04:38.120 ","End":"04:39.680","Text":"whichever is leftover,"},{"Start":"04:39.680 ","End":"04:42.320","Text":"once it\u0027s infinitely far away"},{"Start":"04:42.320 ","End":"04:46.865","Text":"its effect on the potential over here is going to be negligible."},{"Start":"04:46.865 ","End":"04:49.960","Text":"It\u0027s going to be approaching 0."},{"Start":"04:49.960 ","End":"04:55.725","Text":"Each charge, if we have let\u0027s say plus q leftover on this side,"},{"Start":"04:55.725 ","End":"04:58.110","Text":"and plus q leftover on this side,"},{"Start":"04:58.110 ","End":"05:03.750","Text":"and they\u0027re both infinitely far away from this infinite plane,"},{"Start":"05:03.750 ","End":"05:05.130","Text":"and this infinite plane,"},{"Start":"05:05.130 ","End":"05:10.340","Text":"that means they are also by default infinitely far away from one another."},{"Start":"05:10.340 ","End":"05:13.280","Text":"Which means that their effects on one another is"},{"Start":"05:13.280 ","End":"05:17.000","Text":"also negligible and approaching 0 which means that we can"},{"Start":"05:17.000 ","End":"05:20.997","Text":"consider it as if they cancel each other out"},{"Start":"05:20.997 ","End":"05:22.790","Text":"because they\u0027re so far away from each"},{"Start":"05:22.790 ","End":"05:26.850","Text":"other that they don\u0027t have an effect on one another."},{"Start":"05:27.830 ","End":"05:37.970","Text":"Now I\u0027m going to rub this out and let us begin calculating the potential in this region."},{"Start":"05:37.970 ","End":"05:41.585","Text":"Let\u0027s take some random point."},{"Start":"05:41.585 ","End":"05:46.190","Text":"Let\u0027s say it\u0027s located over here and the vector from point"},{"Start":"05:46.190 ","End":"05:51.545","Text":"q to this point is some general vector, x, y,"},{"Start":"05:51.545 ","End":"05:56.765","Text":"z. I\u0027m reminding you that now that we\u0027ve found"},{"Start":"05:56.765 ","End":"06:02.600","Text":"the parallel problem to if we have 2 infinite ground on planes,"},{"Start":"06:02.600 ","End":"06:07.400","Text":"so now we\u0027re calculating the potential due to all of these image charges that"},{"Start":"06:07.400 ","End":"06:12.970","Text":"we\u0027ve added and we\u0027re imagining that these grounded planes don\u0027t exist."},{"Start":"06:14.000 ","End":"06:19.965","Text":"Let\u0027s show that this is some infinite line of charges."},{"Start":"06:19.965 ","End":"06:27.095","Text":"Over here some distance away we have our charge q and this is the nth charge."},{"Start":"06:27.095 ","End":"06:32.060","Text":"Let\u0027s call it q_n where n over here is equal to 0."},{"Start":"06:32.060 ","End":"06:34.340","Text":"Our original charge q,"},{"Start":"06:34.340 ","End":"06:38.180","Text":"the real charge that we have is at n is equal to"},{"Start":"06:38.180 ","End":"06:42.320","Text":"0 and then here we have n is equal to 1 and"},{"Start":"06:42.320 ","End":"06:45.980","Text":"so on and so on until we get to n. We\u0027re going to find"},{"Start":"06:45.980 ","End":"06:50.330","Text":"the potential at this random point by,"},{"Start":"06:50.330 ","End":"06:56.220","Text":"first of all, calculating the potential due to 1 of these image charges."},{"Start":"06:56.720 ","End":"07:01.890","Text":"If we want to know what our charge is for q_n,"},{"Start":"07:01.890 ","End":"07:07.860","Text":"so it\u0027s simply going to be equal to our charge q multiplied by negative 1 to"},{"Start":"07:07.860 ","End":"07:13.260","Text":"the power of n. Let\u0027s say if n is equal to 1,"},{"Start":"07:13.260 ","End":"07:16.665","Text":"so we\u0027ll have that q_1 is equal to"},{"Start":"07:16.665 ","End":"07:21.285","Text":"q multiplied by negative 1 to the power of 1 which is just negative 1."},{"Start":"07:21.285 ","End":"07:26.880","Text":"Then we\u0027ll have that q at position 1 is equal to negative q."},{"Start":"07:26.880 ","End":"07:33.220","Text":"Here we have n is equal to 1 and that is in fact equal to negative q."},{"Start":"07:35.490 ","End":"07:38.080","Text":"Now we can see,"},{"Start":"07:38.080 ","End":"07:45.940","Text":"so if we take the potential due to this charge q_n."},{"Start":"07:45.940 ","End":"07:48.550","Text":"Let\u0027s call that v_n."},{"Start":"07:48.550 ","End":"07:57.250","Text":"That is going to be equal to k multiplied by our charge q_n divided by"},{"Start":"07:57.250 ","End":"08:07.280","Text":"the radius between q_n and out point where we\u0027re calculating the potential."},{"Start":"08:08.100 ","End":"08:13.270","Text":"This vector over here is r_n."},{"Start":"08:13.270 ","End":"08:18.940","Text":"This is going to be divided by r_n vector and of course,"},{"Start":"08:18.940 ","End":"08:22.460","Text":"we\u0027re taking the magnitude right now."},{"Start":"08:23.130 ","End":"08:26.995","Text":"What on earth is r_n equal to?"},{"Start":"08:26.995 ","End":"08:34.780","Text":"Let\u0027s call this vector from our original point or original charge q to this point."},{"Start":"08:34.780 ","End":"08:38.710","Text":"Let\u0027s call this just the regular r vector."},{"Start":"08:38.710 ","End":"08:42.865","Text":"Then we have another vector."},{"Start":"08:42.865 ","End":"08:45.205","Text":"Let\u0027s draw it in red."},{"Start":"08:45.205 ","End":"08:48.235","Text":"That is going from our original charge,"},{"Start":"08:48.235 ","End":"08:53.515","Text":"q at n is equal to 0 up until I charge q_n."},{"Start":"08:53.515 ","End":"08:58.070","Text":"Let\u0027s call this vector r_q."},{"Start":"08:59.040 ","End":"09:01.945","Text":"This is r_q."},{"Start":"09:01.945 ","End":"09:06.415","Text":"What is r_q equal to?"},{"Start":"09:06.415 ","End":"09:11.845","Text":"We can see that it is fast of all going in this x direction."},{"Start":"09:11.845 ","End":"09:16.540","Text":"Its y and z components are equal to 0."},{"Start":"09:16.540 ","End":"09:18.580","Text":"Its x component,"},{"Start":"09:18.580 ","End":"09:24.160","Text":"we can see that with every jump in the value of n, we\u0027re moving 2d."},{"Start":"09:24.160 ","End":"09:30.640","Text":"If we jump from 0-1n, we\u0027re jumping 2d."},{"Start":"09:30.640 ","End":"09:34.915","Text":"If we go from 0 to n is equal to 2,"},{"Start":"09:34.915 ","End":"09:37.000","Text":"then we\u0027ve jumped 4d."},{"Start":"09:37.000 ","End":"09:44.570","Text":"What we have is that this is equal to 2d multiplied by n. Then of course 0,"},{"Start":"09:44.570 ","End":"09:48.500","Text":"0 in the y, z direction."},{"Start":"09:48.500 ","End":"09:53.145","Text":"Then what is r_n equal to?"},{"Start":"09:53.145 ","End":"09:58.300","Text":"R_n is equal to our radius r minus r_q."},{"Start":"10:00.050 ","End":"10:06.345","Text":"We have r minus r_q. We can see that."},{"Start":"10:06.345 ","End":"10:11.505","Text":"From q_n, how do we get to this point over here?"},{"Start":"10:11.505 ","End":"10:13.395","Text":"We go down r_q."},{"Start":"10:13.395 ","End":"10:17.985","Text":"We\u0027re going in the negative direction to r_q and then we\u0027re going up r,"},{"Start":"10:17.985 ","End":"10:22.770","Text":"so in the positive direction of our r. Then we get to this point over here."},{"Start":"10:22.770 ","End":"10:27.430","Text":"Of course, that is the same as just going along r_n."},{"Start":"10:28.260 ","End":"10:31.435","Text":"I just moved all of this over here,"},{"Start":"10:31.435 ","End":"10:35.510","Text":"so r_n is going to be equal to r minus r_q."},{"Start":"10:35.670 ","End":"10:46.120","Text":"That is x minus 2dn in the x component."},{"Start":"10:46.120 ","End":"10:54.695","Text":"Then we have y minus"},{"Start":"10:54.695 ","End":"11:03.130","Text":"0 and then z minus 0."},{"Start":"11:03.130 ","End":"11:07.630","Text":"Now we can substitute this into the equation."},{"Start":"11:07.630 ","End":"11:11.215","Text":"Let\u0027s do this."},{"Start":"11:11.215 ","End":"11:19.413","Text":"Then we\u0027ll get that v_n is equal to k multiplied by q_n."},{"Start":"11:19.413 ","End":"11:25.880","Text":"Q_n, we already said was equal to q multiplied by negative 1^n divided by r_n."},{"Start":"11:29.070 ","End":"11:34.360","Text":"Let\u0027s just write this over here."},{"Start":"11:34.360 ","End":"11:36.520","Text":"The magnitude of i_n,"},{"Start":"11:36.520 ","End":"11:38.545","Text":"as we know as due to Pythagoras."},{"Start":"11:38.545 ","End":"11:44.110","Text":"We\u0027re taking the square root of x minus"},{"Start":"11:44.110 ","End":"11:51.520","Text":"2dn^2 plus y^2 plus z^2."},{"Start":"11:51.520 ","End":"11:58.126","Text":"The square root. What we\u0027re going to have over here is simply x"},{"Start":"11:58.126 ","End":"12:03.206","Text":"minus 2dn^2"},{"Start":"12:03.206 ","End":"12:09.430","Text":"plus y^2 plus z^2^1.5."},{"Start":"12:09.430 ","End":"12:14.680","Text":"Then if we want to find the total potential over here,"},{"Start":"12:14.680 ","End":"12:19.420","Text":"then what we have to do is we have to sum up on all of the charges."},{"Start":"12:19.420 ","End":"12:21.783","Text":"This is the potential over here,"},{"Start":"12:21.783 ","End":"12:25.155","Text":"just due to this q_n."},{"Start":"12:25.155 ","End":"12:27.915","Text":"But we have lots of other charges."},{"Start":"12:27.915 ","End":"12:33.935","Text":"We can see that v_t is simply going to be equal to."},{"Start":"12:33.935 ","End":"12:43.090","Text":"What we\u0027re going to do is we\u0027re going to sum on all of the charges. What does that mean?"},{"Start":"12:43.090 ","End":"12:46.600","Text":"We\u0027re going from negative infinity in this direction,"},{"Start":"12:46.600 ","End":"12:49.960","Text":"all of the charges in this direction until infinity,"},{"Start":"12:49.960 ","End":"12:58.660","Text":"which has all of the charges in this direction over here multiplied by this v_n."},{"Start":"12:58.660 ","End":"13:02.815","Text":"We\u0027re summing up all of this, so, therefore,"},{"Start":"13:02.815 ","End":"13:10.960","Text":"that is going to be equal to the sum from negative infinity up until infinity of v_n,"},{"Start":"13:10.960 ","End":"13:12.400","Text":"which is simply equal"},{"Start":"13:12.400 ","End":"13:21.460","Text":"to negative 1^n multiplied by k_q divided by what we had over here,"},{"Start":"13:21.460 ","End":"13:25.806","Text":"which is equal to x minus"},{"Start":"13:25.806 ","End":"13:32.818","Text":"2dn^2 plus y^2 plus z^2,"},{"Start":"13:32.818 ","End":"13:33.850","Text":"and all of this to the power of 1/2."},{"Start":"13:33.850 ","End":"13:37.944","Text":"What we have in front of us is an alternating series."},{"Start":"13:37.944 ","End":"13:44.230","Text":"The first term could be negative k_q divided by all of this."},{"Start":"13:44.230 ","End":"13:48.175","Text":"Then the next term is k_q divided by all of this."},{"Start":"13:48.175 ","End":"13:51.550","Text":"Then again negative and then again plus."},{"Start":"13:51.550 ","End":"13:54.895","Text":"We have an alternating series."},{"Start":"13:54.895 ","End":"14:01.075","Text":"As we know, alternating series always converge to some value."},{"Start":"14:01.075 ","End":"14:03.730","Text":"This isn\u0027t a diverging series,"},{"Start":"14:03.730 ","End":"14:05.485","Text":"we have a converging series."},{"Start":"14:05.485 ","End":"14:11.480","Text":"We\u0027re going to get some value for the potential in this region."},{"Start":"14:12.450 ","End":"14:15.475","Text":"Whilst we\u0027re on this topic,"},{"Start":"14:15.475 ","End":"14:22.550","Text":"let\u0027s imagine that this question has a question too to it, a part B."},{"Start":"14:23.550 ","End":"14:28.225","Text":"This would have been the answer to question number 1."},{"Start":"14:28.225 ","End":"14:30.340","Text":"Let\u0027s imagine the question number 2 is,"},{"Start":"14:30.340 ","End":"14:34.760","Text":"what is the energy required to build the system?"},{"Start":"14:35.880 ","End":"14:43.060","Text":"Let\u0027s scroll down. We\u0027re calculating the energy required to build the system."},{"Start":"14:43.060 ","End":"14:45.895","Text":"The equation is equal to,"},{"Start":"14:45.895 ","End":"14:53.361","Text":"so u is equal to 1/2 multiplied by the sum of q_i,"},{"Start":"14:53.361 ","End":"14:59.740","Text":"each charge, multiplied by the potential at that point, Phi."},{"Start":"14:59.740 ","End":"15:04.360","Text":"Let\u0027s see. All of these image charges that we\u0027ve put,"},{"Start":"15:04.360 ","End":"15:06.640","Text":"we have to remember that this is"},{"Start":"15:06.640 ","End":"15:12.260","Text":"just a mathematical tool in order to calculate the potential in this region."},{"Start":"15:12.260 ","End":"15:14.840","Text":"All of these charges are theoretical."},{"Start":"15:14.840 ","End":"15:17.045","Text":"They don\u0027t really exist."},{"Start":"15:17.045 ","End":"15:20.630","Text":"We use it in order to calculate the potential here."},{"Start":"15:20.630 ","End":"15:22.976","Text":"But of course,"},{"Start":"15:22.976 ","End":"15:27.720","Text":"they don\u0027t exist, so they\u0027re not going to be included in this q_i."},{"Start":"15:27.960 ","End":"15:33.895","Text":"Let\u0027s see what I mean. What I\u0027m going to have is half multiplied by,"},{"Start":"15:33.895 ","End":"15:36.625","Text":"let\u0027s start with our first charge."},{"Start":"15:36.625 ","End":"15:42.145","Text":"As this charge, our original one that we have are only actual charge in the system."},{"Start":"15:42.145 ","End":"15:49.000","Text":"It is equal to q multiplied by the potential at this point."},{"Start":"15:49.000 ","End":"15:55.150","Text":"The potential at this point is due to the potential from all of these charges,"},{"Start":"15:55.150 ","End":"15:57.010","Text":"all of our image charges."},{"Start":"15:57.010 ","End":"16:05.845","Text":"But remember that the potential due to charge q on charge q is 0."},{"Start":"16:05.845 ","End":"16:12.760","Text":"Charge q isn\u0027t causing a potential on itself. What does that mean?"},{"Start":"16:12.760 ","End":"16:14.785","Text":"Here our V_t,"},{"Start":"16:14.785 ","End":"16:16.855","Text":"our total potential over here,"},{"Start":"16:16.855 ","End":"16:23.770","Text":"at some point in this region includes a charge at n=0."},{"Start":"16:23.770 ","End":"16:29.140","Text":"What we\u0027re going to do is we\u0027re going to subtract that from this equation."},{"Start":"16:29.140 ","End":"16:35.520","Text":"What we could either do is sum from negative infinity to infinity of all of this,"},{"Start":"16:35.520 ","End":"16:40.280","Text":"and then subtract at the end -1^n=0."},{"Start":"16:42.740 ","End":"16:49.300","Text":"That would just be 1*kq divided by all of this."},{"Start":"16:49.300 ","End":"16:57.160","Text":"We could do that or we could just write it out in a slightly faster way."},{"Start":"16:57.160 ","End":"17:04.970","Text":"If we sum from negative infinity up until negative 1,"},{"Start":"17:07.230 ","End":"17:16.660","Text":"then we add on summing from one up until infinity that\u0027s the same as multiplying all of"},{"Start":"17:16.660 ","End":"17:25.540","Text":"this by 2 multiplied by the sum of just from 1 up until infinity of all of this."},{"Start":"17:25.540 ","End":"17:36.890","Text":"Negative 1 to the power of n kq divided by this denominator over here."},{"Start":"17:38.040 ","End":"17:40.405","Text":"If we just do that,"},{"Start":"17:40.405 ","End":"17:44.470","Text":"we take the side of the series that doesn\u0027t include"},{"Start":"17:44.470 ","End":"17:48.730","Text":"0 but the positive n. All of these charges."},{"Start":"17:48.730 ","End":"17:50.440","Text":"We\u0027re going from 1 to infinity,"},{"Start":"17:50.440 ","End":"17:53.860","Text":"but then we want to include all of these charges on the other side over"},{"Start":"17:53.860 ","End":"17:59.095","Text":"here so that we just multiply it by 2, and then it\u0027s the same."},{"Start":"17:59.095 ","End":"18:02.530","Text":"What we have is V_t just without taking"},{"Start":"18:02.530 ","End":"18:06.160","Text":"into account the potential of our original charge q,"},{"Start":"18:06.160 ","End":"18:11.060","Text":"because it doesn\u0027t exert a potential on itself."},{"Start":"18:12.750 ","End":"18:17.920","Text":"Now we know that along these charged infinite planes,"},{"Start":"18:17.920 ","End":"18:22.045","Text":"we have point charges all along."},{"Start":"18:22.045 ","End":"18:28.045","Text":"We have plus our next q is going to be this q tilde of"},{"Start":"18:28.045 ","End":"18:30.460","Text":"all the point charges along each of"},{"Start":"18:30.460 ","End":"18:35.845","Text":"these infinite planes multiplied by the potential acting on these charges."},{"Start":"18:35.845 ","End":"18:38.874","Text":"These infinite planes are grounded."},{"Start":"18:38.874 ","End":"18:41.740","Text":"We know and we saw at the beginning of the lesson that"},{"Start":"18:41.740 ","End":"18:44.935","Text":"the potential along this plane is equal to 0."},{"Start":"18:44.935 ","End":"18:49.120","Text":"So all of these charges are located at a potential 0."},{"Start":"18:49.120 ","End":"18:51.355","Text":"We just multiply it by 0."},{"Start":"18:51.355 ","End":"18:55.210","Text":"We just didn\u0027t have to include that at all."},{"Start":"18:55.210 ","End":"19:03.560","Text":"Now what we need to do is we\u0027re trying to find the energy to build the system."},{"Start":"19:04.320 ","End":"19:08.855","Text":"Which is simply equal to this over here."},{"Start":"19:08.855 ","End":"19:10.405","Text":"Now, if they would ask us though,"},{"Start":"19:10.405 ","End":"19:15.835","Text":"what is the energy required to bring this charge q from infinity?"},{"Start":"19:15.835 ","End":"19:19.160","Text":"Let\u0027s add in that question."},{"Start":"19:19.590 ","End":"19:21.790","Text":"If Question 3 was,"},{"Start":"19:21.790 ","End":"19:26.185","Text":"what is the energy required to bring charge q from infinity?"},{"Start":"19:26.185 ","End":"19:29.500","Text":"Here we answered Question 2."},{"Start":"19:29.500 ","End":"19:32.750","Text":"Now let\u0027s answer Question 3."},{"Start":"19:32.970 ","End":"19:35.920","Text":"In order to answer Question 3,"},{"Start":"19:35.920 ","End":"19:42.685","Text":"that means the energy to bring q from infinity to this point at the origin."},{"Start":"19:42.685 ","End":"19:44.380","Text":"Where this is at the origin,"},{"Start":"19:44.380 ","End":"19:49.690","Text":"this is at point x=0,"},{"Start":"19:49.690 ","End":"19:53.050","Text":"y=0, and z=0."},{"Start":"19:53.050 ","End":"19:56.020","Text":"That is this point over here."},{"Start":"19:56.020 ","End":"20:01.015","Text":"That means we just substitute in 0 into each of these points."},{"Start":"20:01.015 ","End":"20:04.419","Text":"We\u0027re bringing q from infinity to the origin,"},{"Start":"20:04.419 ","End":"20:08.290","Text":"where the origin is at x=0,"},{"Start":"20:08.290 ","End":"20:10.750","Text":"y=0, and z=0."},{"Start":"20:10.750 ","End":"20:15.715","Text":"Then we could say in this case that u is going to be equal to,"},{"Start":"20:15.715 ","End":"20:17.170","Text":"what I can do is I can take all of"},{"Start":"20:17.170 ","End":"20:21.190","Text":"the constants out from the summation because they\u0027re constants,"},{"Start":"20:21.190 ","End":"20:28.770","Text":"we\u0027re only summing on -1^n divided by anything in the denominator which is"},{"Start":"20:28.770 ","End":"20:36.955","Text":"multiplied by n. What we\u0027ll have is that this is equal to q multiplied by kq."},{"Start":"20:36.955 ","End":"20:41.335","Text":"We can just write that this is equal to kq^2."},{"Start":"20:41.335 ","End":"20:49.420","Text":"Then we\u0027re summing from 1 until infinity on negative 1^n divided by,"},{"Start":"20:49.420 ","End":"20:51.490","Text":"then we have x which is 0."},{"Start":"20:51.490 ","End":"20:54.220","Text":"We have negative 2(dn)^2."},{"Start":"20:54.220 ","End":"21:04.430","Text":"That\u0027s just 4d^2n^2 plus 0^2 plus 0^2 to the power of a half."},{"Start":"21:05.370 ","End":"21:10.360","Text":"Now what we can do is we can just simplify this."},{"Start":"21:10.360 ","End":"21:13.375","Text":"We can take out the square root."},{"Start":"21:13.375 ","End":"21:18.430","Text":"Then what we\u0027re just going to have is simply 2dn."},{"Start":"21:18.430 ","End":"21:23.840","Text":"Because the square root of 4d^2n^2 is just 2dn."},{"Start":"21:24.630 ","End":"21:29.200","Text":"Then what we can have is we can see that this is,"},{"Start":"21:29.200 ","End":"21:30.400","Text":"if we take out the constants,"},{"Start":"21:30.400 ","End":"21:34.820","Text":"this is going to be equal to kq2 divided by 2d."},{"Start":"21:38.580 ","End":"21:43.420","Text":"All of this is multiplied by the sum from 1 until"},{"Start":"21:43.420 ","End":"21:48.700","Text":"infinity of negative 1 to the power of n divided"},{"Start":"21:48.700 ","End":"21:53.650","Text":"by n. Then what we can"},{"Start":"21:53.650 ","End":"21:59.140","Text":"see is we\u0027re summing up from 1 until infinity of this series over here,"},{"Start":"21:59.140 ","End":"22:02.815","Text":"which is simply equal to this side of the series."},{"Start":"22:02.815 ","End":"22:06.340","Text":"The series from 1 until infinity of negative 1 to the power"},{"Start":"22:06.340 ","End":"22:10.435","Text":"of n divided by n is equal to negative ln 2."},{"Start":"22:10.435 ","End":"22:15.295","Text":"We\u0027ll just multiply it negative ln 2 by this over here."},{"Start":"22:15.295 ","End":"22:18.560","Text":"What we\u0027ll have is negative kq^2 divided by 2d multiplied by ln of 2."},{"Start":"22:26.970 ","End":"22:30.925","Text":"That\u0027s what this series converges to."},{"Start":"22:30.925 ","End":"22:35.605","Text":"This is the energy required to bring our charge q"},{"Start":"22:35.605 ","End":"22:41.365","Text":"from infinity and up until this point over here at the origin."},{"Start":"22:41.365 ","End":"22:45.223","Text":"Let\u0027s go back to the question,"},{"Start":"22:45.223 ","End":"22:51.220","Text":"if you\u0027re ever asked what is the energy required to bring a charge q from infinity,"},{"Start":"22:51.220 ","End":"22:55.795","Text":"then first you calculate the energy required to build the system."},{"Start":"22:55.795 ","End":"23:00.880","Text":"Then you just substitute in the position of the charge,"},{"Start":"23:00.880 ","End":"23:06.085","Text":"which in this case is at the origin at 000."},{"Start":"23:06.085 ","End":"23:10.690","Text":"Then you just substitute that into the energy required to build the system."},{"Start":"23:10.690 ","End":"23:16.300","Text":"Then you have your answer for the energy required to bring the charge from infinity."},{"Start":"23:16.300 ","End":"23:19.430","Text":"That\u0027s the end of this lesson."}],"ID":14225},{"Watched":false,"Name":"Exercise 6","Duration":"11m 13s","ChapterTopicVideoID":12123,"CourseChapterTopicPlaylistID":161612,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/12123.jpeg","UploadDate":"2018-06-28T04:13:46.1230000","DurationForVideoObject":"PT11M13S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.025","Text":"Hello. In this lesson,"},{"Start":"00:02.025 ","End":"00:04.605","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.605 ","End":"00:10.995","Text":"Charge q_1 is located a distance a above an infinite grounded plane."},{"Start":"00:10.995 ","End":"00:15.795","Text":"Charge q_2 is located a distance b below the plane."},{"Start":"00:15.795 ","End":"00:19.575","Text":"What is the electric field and potential throughout?"},{"Start":"00:19.575 ","End":"00:24.960","Text":"We\u0027ve seen in previous questions how to solve via the method of images if"},{"Start":"00:24.960 ","End":"00:30.120","Text":"we have an infinite grounded plane with 1 charge above it."},{"Start":"00:30.120 ","End":"00:32.325","Text":"Let\u0027s say over here q_1."},{"Start":"00:32.325 ","End":"00:35.365","Text":"We know how to solve this type of question."},{"Start":"00:35.365 ","End":"00:40.100","Text":"What we can do is we can split up this question with"},{"Start":"00:40.100 ","End":"00:45.035","Text":"a charge above and below into two separate questions."},{"Start":"00:45.035 ","End":"00:49.800","Text":"Then we can superimpose the onsets."},{"Start":"00:49.800 ","End":"00:51.240","Text":"Why can we do this?"},{"Start":"00:51.240 ","End":"00:55.055","Text":"We know that q_1 and q_2 act independently"},{"Start":"00:55.055 ","End":"01:00.035","Text":"of one another because there\u0027s this grounded plane between the 2."},{"Start":"01:00.035 ","End":"01:06.670","Text":"We know that q_1 doesn\u0027t affect the potential or the electric fields below the plane."},{"Start":"01:06.670 ","End":"01:12.365","Text":"We know that q_2 doesn\u0027t affect the potential and the electric field above the plane."},{"Start":"01:12.365 ","End":"01:14.695","Text":"Because the plane is grounded."},{"Start":"01:14.695 ","End":"01:17.410","Text":"That means that these 2 charges are acting"},{"Start":"01:17.410 ","End":"01:22.700","Text":"independently and so we can work this out as 2 separate questions."},{"Start":"01:22.840 ","End":"01:29.135","Text":"Another way at looking why we can split this up into"},{"Start":"01:29.135 ","End":"01:36.795","Text":"the following problems is because we\u0027re dealing with an infinite plane which is grounded."},{"Start":"01:36.795 ","End":"01:45.580","Text":"Even though the plane might be extremely thin it\u0027s still is going to have some width."},{"Start":"01:45.580 ","End":"01:50.195","Text":"Let\u0027s draw it like so where we can see the width,"},{"Start":"01:50.195 ","End":"01:52.715","Text":"imagine that these lines are straight."},{"Start":"01:52.715 ","End":"01:56.960","Text":"We know that the infinite plane is also a conductor."},{"Start":"01:56.960 ","End":"02:00.390","Text":"The conductor is grounded."},{"Start":"02:00.390 ","End":"02:04.565","Text":"We know that the charge inside over here,"},{"Start":"02:04.565 ","End":"02:07.890","Text":"the net charge, is going to be equal to 0."},{"Start":"02:08.230 ","End":"02:14.200","Text":"We always know that the charge inside a conductor is equal to 0."},{"Start":"02:14.200 ","End":"02:21.350","Text":"That means that all of the charges are located along the surface of the plane."},{"Start":"02:21.350 ","End":"02:24.650","Text":"Either above or below."},{"Start":"02:24.650 ","End":"02:27.830","Text":"We know that because the plane is grounded,"},{"Start":"02:27.830 ","End":"02:30.040","Text":"that the potential, therefore,"},{"Start":"02:30.040 ","End":"02:32.740","Text":"has to be equal to 0."},{"Start":"02:32.740 ","End":"02:35.705","Text":"In order for the potential to be equal to 0,"},{"Start":"02:35.705 ","End":"02:43.490","Text":"we know that all of the charges over here cancel out with our charge q_2,"},{"Start":"02:43.490 ","End":"02:49.590","Text":"and all of the charges over here on top cancel out with our charge q_1."},{"Start":"02:51.680 ","End":"02:57.935","Text":"Therefore, we can see that we\u0027re dealing with 2 separate problems."},{"Start":"02:57.935 ","End":"03:02.720","Text":"Because the surface of our plane or the surfaces of"},{"Start":"03:02.720 ","End":"03:10.595","Text":"our plane cancel out each 1 of the other charges relative to the alternate charge."},{"Start":"03:10.595 ","End":"03:12.610","Text":"If we\u0027re looking at q_2,"},{"Start":"03:12.610 ","End":"03:15.350","Text":"q_1 is canceled out by the charges on"},{"Start":"03:15.350 ","End":"03:18.515","Text":"the surface of the plane and if we\u0027re looking at q_1,"},{"Start":"03:18.515 ","End":"03:25.550","Text":"q_2 is canceled out by the charges at the bottom surface of the plane."},{"Start":"03:25.550 ","End":"03:30.420","Text":"Now we can solve the following 2 questions."},{"Start":"03:31.580 ","End":"03:37.140","Text":"Let\u0027s call this area over here, the up,"},{"Start":"03:37.140 ","End":"03:40.235","Text":"because we\u0027re up and this over here,"},{"Start":"03:40.235 ","End":"03:43.190","Text":"we can call this area down."},{"Start":"03:43.190 ","End":"03:47.085","Text":"Let\u0027s say we want to calculate E_up."},{"Start":"03:47.085 ","End":"03:50.340","Text":"Let\u0027s work this out over here."},{"Start":"03:50.340 ","End":"03:57.440","Text":"E_up is simply going to be equal to the electric field due to this system."},{"Start":"03:57.440 ","End":"04:00.695","Text":"We already know that in order to answer a question like this,"},{"Start":"04:00.695 ","End":"04:03.780","Text":"we\u0027re going to have to have the method of images."},{"Start":"04:03.780 ","End":"04:07.205","Text":"A distance a below the ground field,"},{"Start":"04:07.205 ","End":"04:11.330","Text":"we\u0027re going to have this charge negative q_1."},{"Start":"04:11.330 ","End":"04:14.390","Text":"Then this is the parallel system."},{"Start":"04:14.390 ","End":"04:16.835","Text":"Then we can ignore this grounded plane."},{"Start":"04:16.835 ","End":"04:24.735","Text":"Then let\u0027s say we want to calculate the electric field over here at this point."},{"Start":"04:24.735 ","End":"04:30.200","Text":"E_up is going to be equal to the electric field due to our charge q_1."},{"Start":"04:30.200 ","End":"04:39.070","Text":"That\u0027s simply going to be equal to kq_1 divided by the vector from q_1 to this point."},{"Start":"04:39.800 ","End":"04:45.865","Text":"Let\u0027s call this r plus vector"},{"Start":"04:45.865 ","End":"04:52.940","Text":"divided by the magnitude of r plus vector squared."},{"Start":"04:52.940 ","End":"05:01.160","Text":"Then we\u0027re going to add on the electric field at this point due to our negative q_1."},{"Start":"05:01.160 ","End":"05:04.550","Text":"Let\u0027s call this r minus vector."},{"Start":"05:04.550 ","End":"05:08.465","Text":"Then we\u0027ll have plus negative kq_1."},{"Start":"05:08.465 ","End":"05:11.580","Text":"We\u0027ll just have plus, negative kq_1."},{"Start":"05:11.920 ","End":"05:16.940","Text":"Negative comes from this being negative q_1 divided by"},{"Start":"05:16.940 ","End":"05:23.420","Text":"the magnitude of our r minus vector."},{"Start":"05:24.350 ","End":"05:28.910","Text":"Of course, the electric field is a vector."},{"Start":"05:28.910 ","End":"05:34.445","Text":"We can say that all of this is going in the radial direction."},{"Start":"05:34.445 ","End":"05:37.595","Text":"Then similarly our E_down,"},{"Start":"05:37.595 ","End":"05:42.185","Text":"so again, we\u0027re using the method of images so a distance b above the plane."},{"Start":"05:42.185 ","End":"05:44.240","Text":"We\u0027ll put our image charge,"},{"Start":"05:44.240 ","End":"05:47.060","Text":"so that will be negative q_2."},{"Start":"05:47.060 ","End":"05:52.860","Text":"Then what we\u0027ll have is that E_down is equal 2,"},{"Start":"05:52.860 ","End":"05:56.580","Text":"so then we have kq_2."},{"Start":"05:56.580 ","End":"06:01.235","Text":"Let\u0027s say we want to measure the electric field over here."},{"Start":"06:01.235 ","End":"06:06.735","Text":"What we can say is that we have this vector over here."},{"Start":"06:06.735 ","End":"06:13.030","Text":"Let\u0027s call this r tag plus."},{"Start":"06:13.030 ","End":"06:15.365","Text":"Then from over here,"},{"Start":"06:15.365 ","End":"06:18.620","Text":"because it\u0027s on top of the plane,"},{"Start":"06:18.620 ","End":"06:21.575","Text":"so we\u0027ll call it plus and here because it\u0027s below the plane,"},{"Start":"06:21.575 ","End":"06:24.455","Text":"so we\u0027ll call this r tag minus."},{"Start":"06:24.455 ","End":"06:27.635","Text":"Then just like before,"},{"Start":"06:27.635 ","End":"06:34.455","Text":"kq_2 divided by r tag minus magnitude,"},{"Start":"06:34.455 ","End":"06:37.890","Text":"and of course this is squared as well."},{"Start":"06:37.890 ","End":"06:40.245","Text":"Don\u0027t forget the squared."},{"Start":"06:40.245 ","End":"06:47.220","Text":"Then we have minus kq_2 divided"},{"Start":"06:47.220 ","End":"06:53.930","Text":"by r tag plus the magnitude and of course squared."},{"Start":"06:53.930 ","End":"06:58.530","Text":"All of this is also in the radial direction."},{"Start":"07:00.080 ","End":"07:07.955","Text":"The total electric field is a superposition of E_up and E_down and that\u0027s just that."},{"Start":"07:07.955 ","End":"07:11.875","Text":"Of course, the potential throughout is,"},{"Start":"07:11.875 ","End":"07:15.230","Text":"obviously, you can get the potential from the electric field."},{"Start":"07:15.230 ","End":"07:19.665","Text":"You just take off the squared at the bottom."},{"Start":"07:19.665 ","End":"07:23.070","Text":"That is it. That\u0027s the answer to question number 1."},{"Start":"07:23.070 ","End":"07:26.450","Text":"Question number 2, what is the charge distribution"},{"Start":"07:26.450 ","End":"07:31.200","Text":"on the plane and what is the total charge on the plane?"},{"Start":"07:32.150 ","End":"07:42.410","Text":"The charged distribution is given by Sigma and it is equal to kq multiplied by"},{"Start":"07:42.410 ","End":"07:46.249","Text":"Epsilon naught multiplied by 2"},{"Start":"07:46.249 ","End":"07:52.005","Text":"multiplied by the distance between the charge and the plane,"},{"Start":"07:52.005 ","End":"07:54.225","Text":"the distance d,"},{"Start":"07:54.225 ","End":"07:59.680","Text":"divided by (r^2"},{"Start":"07:59.680 ","End":"08:07.440","Text":"plus d^2)^3/2."},{"Start":"08:07.440 ","End":"08:08.955","Text":"What is r^2?"},{"Start":"08:08.955 ","End":"08:15.140","Text":"The charge distribution is dependent on how far away we are from the origin."},{"Start":"08:15.140 ","End":"08:18.035","Text":"Here we can see is the origin."},{"Start":"08:18.035 ","End":"08:25.980","Text":"This is the closest point on the plane to our charge q."},{"Start":"08:25.990 ","End":"08:29.194","Text":"If we\u0027re located at this point,"},{"Start":"08:29.194 ","End":"08:33.200","Text":"so this distance over here is our r,"},{"Start":"08:33.200 ","End":"08:38.740","Text":"the distance from the origin to the point that we\u0027re measuring the charge distribution."},{"Start":"08:38.740 ","End":"08:48.020","Text":"In our question, so our Sigma_1 corresponding to q_1 is going to be equal to k multiplied"},{"Start":"08:48.020 ","End":"08:52.640","Text":"by q_1 multiplied by Epsilon naught multiplied"},{"Start":"08:52.640 ","End":"08:57.575","Text":"by 2 times the distance of q_1 from the plane,"},{"Start":"08:57.575 ","End":"09:06.790","Text":"which is a divided by r^2 multiplied,"},{"Start":"09:06.790 ","End":"09:11.600","Text":"sorry, plus the distance between the charge and the plane squared."},{"Start":"09:11.600 ","End":"09:15.750","Text":"That\u0027s a^2 to the power of 3/2."},{"Start":"09:17.210 ","End":"09:20.190","Text":"Of course I forgot the minus."},{"Start":"09:20.190 ","End":"09:22.830","Text":"There\u0027s a minus over here."},{"Start":"09:22.830 ","End":"09:27.990","Text":"Sigma_2, the charge distribution at the bottom"},{"Start":"09:27.990 ","End":"09:32.570","Text":"due to q_2 is going to be equal to negative and"},{"Start":"09:32.570 ","End":"09:34.700","Text":"then k multiplied by"},{"Start":"09:34.700 ","End":"09:41.765","Text":"q_2 Epsilon naught multiplied by 2 times the distance between q_2 and the plane,"},{"Start":"09:41.765 ","End":"09:47.595","Text":"which is b, divided by this bar."},{"Start":"09:47.595 ","End":"09:50.585","Text":"If we\u0027re measuring this point over here, it\u0027s the same r,"},{"Start":"09:50.585 ","End":"09:56.690","Text":"so divided by r^2 plus the distance between q_2 and the plane"},{"Start":"09:56.690 ","End":"10:04.260","Text":", which is b^2^3/2."},{"Start":"10:04.940 ","End":"10:16.290","Text":"Then the total charge distribution is simply going to be equal to Sigma_1 plus Sigma_2."},{"Start":"10:18.010 ","End":"10:26.980","Text":"That is simply going to be equal to negative k Epsilon naught multiplied by 2."},{"Start":"10:26.980 ","End":"10:33.575","Text":"All of this is going to be multiplied by q_1a divided by r^2 plus a^2^3/2"},{"Start":"10:33.575 ","End":"10:41.480","Text":"plus q_2b"},{"Start":"10:41.480 ","End":"10:49.470","Text":"divided by r^2 plus b^2^3/2."},{"Start":"10:49.470 ","End":"10:51.305","Text":"That is the end of the question."},{"Start":"10:51.305 ","End":"10:55.715","Text":"What\u0027s important to take from this lesson is that if you have"},{"Start":"10:55.715 ","End":"11:00.830","Text":"some infinite plane and a charge above and a charge below and the plane"},{"Start":"11:00.830 ","End":"11:05.000","Text":"is grounded then you can split up this type of question into"},{"Start":"11:05.000 ","End":"11:10.950","Text":"two separate questions and then superimpose the answers."},{"Start":"11:10.950 ","End":"11:14.320","Text":"That\u0027s it, that\u0027s the end of this lesson."}],"ID":14226}],"Thumbnail":null,"ID":161612}]

[{"ID":161612,"Videos":[14213,14214,14215,14216,14217,14218,14219,14220,14221,14222,14223,14224,14225,14226]}];

[14213,14214,14215,14216,14217,14218,14219,14220,14221,14222,14223,14224,14225,14226];

1

3

Get unlimited access to **1500 subjects** including **personalized courses**

Start your free trial
We couldn't find any results for

Upload your syllabus now and our team will create a customized course especially for you!

Alert

and we will create a personalized course (just for you) in less than **48 hours...**