1 00:00:00,300 --> 00:00:06,000 So maybe the last exercise was a bit difficult when you tried to solve it yourself, because it was 2 00:00:06,000 --> 00:00:08,700 the very first time that you are dealing with these problems. 3 00:00:09,270 --> 00:00:15,960 But I hope that once you have seen the solution, it all made more sense to you and that you can now 4 00:00:15,960 --> 00:00:18,150 deal much better with these kinds of problems. 5 00:00:19,140 --> 00:00:21,840 So this is why here is another exercise. 6 00:00:22,020 --> 00:00:28,620 This time we will calculate the electric field and the electrostatic potential of a spherical capacitor. 7 00:00:29,970 --> 00:00:36,510 So, yeah, these are the first two tasks, and then we will also calculate the voltage between the 8 00:00:36,510 --> 00:00:40,830 two metals of the capacitor and we will calculate the capacitance. 9 00:00:41,590 --> 00:00:48,420 But like in the previous example, the only real difficult task is task number one and then maybe task 10 00:00:48,420 --> 00:00:48,990 number two. 11 00:00:49,080 --> 00:00:51,030 The other two tasks are quite easy. 12 00:00:52,080 --> 00:00:54,680 So this is what our capacitor looks like. 13 00:00:54,720 --> 00:00:59,100 We have metal, number one, and green and metal number two in blue. 14 00:00:59,940 --> 00:01:06,720 And we have here they are both spheres, basically, or surfaces of spheres. 15 00:01:07,080 --> 00:01:11,070 And they have the same middle point, same zero position. 16 00:01:12,210 --> 00:01:19,320 And to charge on both of these metals of the capacitor is equal in magnitude, but opposite insigne. 17 00:01:20,500 --> 00:01:28,780 So the main difference to the previous example is that there is no charge density inside here, so you 18 00:01:28,780 --> 00:01:34,590 really only have this charge density on this metal on the surface of the sphere. 19 00:01:35,550 --> 00:01:41,070 And now we do not only have one sphere, but we have two spheres and they have opposite charges. 20 00:01:42,410 --> 00:01:50,870 So please go ahead and post a video and really try to solve this problem yourself for the region inside 21 00:01:50,870 --> 00:01:57,590 of this green metal, then for the region between the two metals and then for the region outside of 22 00:01:57,590 --> 00:01:58,340 the blue metal. 23 00:02:03,060 --> 00:02:05,520 OK, so now I will present to you the solution. 24 00:02:07,170 --> 00:02:13,810 So the first task was to calculate the electric field of this capacitor for all three regions. 25 00:02:15,060 --> 00:02:20,190 So we begin with the region inside of the first metal and green. 26 00:02:20,400 --> 00:02:29,130 So this is for our smaller than a and here the solution is really very simple because there is no charge 27 00:02:29,370 --> 00:02:30,480 and closed. 28 00:02:31,230 --> 00:02:38,130 If we so so let's say we we consider here sphere of reference inside of this green metal. 29 00:02:38,490 --> 00:02:42,470 And so inside of the sphere of reference, there's no charge included. 30 00:02:42,750 --> 00:02:47,110 So this means he was zero and even row of R zero everywhere. 31 00:02:47,880 --> 00:02:51,960 And so this means that this integral here on the right hand side will be zero. 32 00:02:52,350 --> 00:02:57,140 And the only way this is possible is that E is equal to zero. 33 00:02:57,480 --> 00:03:00,420 So there's no electric field inside this metal. 34 00:03:00,420 --> 00:03:01,040 No one. 35 00:03:03,000 --> 00:03:07,800 Now, for the second region, we consider the region between these two metals. 36 00:03:08,610 --> 00:03:11,250 So this is for A smaller are smaller B. 37 00:03:12,610 --> 00:03:20,620 So this time, the right hand side integrates up to cue to capital cute because you can see there's 38 00:03:20,860 --> 00:03:24,430 a charge on this on the screen metal in total, we have to charge. 39 00:03:24,430 --> 00:03:28,120 Q So we have that this integral is equal to Cube. 40 00:03:29,590 --> 00:03:33,240 Now we can of course, consider again this radial symmetry. 41 00:03:33,910 --> 00:03:41,240 So this means we can pull out this E and we just have to calculate here to surface integral over the 42 00:03:41,240 --> 00:03:49,030 S, which is just the area of the surface of the sphere that we are considering and where the radius 43 00:03:49,030 --> 00:03:50,610 is somewhere between A and B.. 44 00:03:51,700 --> 00:03:56,760 So this means the surface is four PI Times Square, as we had previous examples. 45 00:03:57,250 --> 00:04:01,560 So this is the equation that we get and now we just solve for E. 46 00:04:03,070 --> 00:04:06,520 So this means each of our is equal to this prefecture. 47 00:04:06,520 --> 00:04:11,290 Times Cube divided by square times, this radio position vector. 48 00:04:12,740 --> 00:04:19,880 And maybe you remember, this is exactly the same field as we had in the previous example outside of 49 00:04:19,880 --> 00:04:27,190 the sphere, so it does not matter at all if this sphere here is filled or not. 50 00:04:27,860 --> 00:04:35,360 It's just this whole electric field just depends on the enclosed charge, but not really on the distribution 51 00:04:35,360 --> 00:04:38,780 of the charge as long as we have some radial symmetry. 52 00:04:40,650 --> 00:04:46,680 Now for the third region, we go outside of the sphere, so we have here our larger than B and here 53 00:04:46,680 --> 00:04:48,920 we have the same argument as in the first region. 54 00:04:49,380 --> 00:04:55,650 We have these two charges, plus Q and minus Q and if you consider a sphere that is even larger, the 55 00:04:55,650 --> 00:04:57,570 enclosed charge is zero. 56 00:04:57,720 --> 00:05:02,700 Those means here we get to zero and then this electric field must also be zero. 57 00:05:04,130 --> 00:05:09,140 And so this is our solution to the first task we have these electric fields and these three regions, 58 00:05:09,620 --> 00:05:13,730 and this is what the magnitude of the electric field looks like in these regions. 59 00:05:16,850 --> 00:05:22,550 The second task was to calculate the electrostatic potential, and so since we have these solutions 60 00:05:22,550 --> 00:05:28,070 here for the electric field, we just have to integrate over the electric field to get the electrostatic 61 00:05:28,070 --> 00:05:28,700 potential. 62 00:05:29,300 --> 00:05:37,280 But also we have to be careful because we have to consider the integration constant to make the electrostatic 63 00:05:37,280 --> 00:05:38,660 potential continuous. 64 00:05:40,740 --> 00:05:49,200 So what we get for the inside and the outside region is, of course, that the electrostatic potential 65 00:05:49,200 --> 00:05:57,050 is a constant number because if we integrate of a zero, we had zero plus an integration constant. 66 00:05:57,870 --> 00:06:02,130 And so as we get to your two constants, that would probably not be the same. 67 00:06:02,160 --> 00:06:10,260 So we have to determine these later and for the middle region, for R between A and B. We have to integrate 68 00:06:10,260 --> 00:06:11,040 this term here. 69 00:06:11,040 --> 00:06:16,140 So we get minus one of our and then we get to another minus signs. 70 00:06:16,150 --> 00:06:21,270 So this will be the solution and then we get another integration constant. 71 00:06:22,740 --> 00:06:29,790 And so now what we have to consider or what we have to determine are these three integration constants, 72 00:06:29,790 --> 00:06:36,150 because here we have constant here we have some constant and here we have one of our dependence and 73 00:06:36,150 --> 00:06:37,050 some constant. 74 00:06:37,620 --> 00:06:44,310 And on these on these points, A and B, these electrostatic potentials have to be continuous. 75 00:06:44,550 --> 00:06:48,000 So this is you, of course, the electrostatic potential and not the electric field. 76 00:06:49,920 --> 00:06:57,420 So what we have to do first is we just have to set one of these constants to some value. 77 00:06:57,750 --> 00:07:06,840 And I think it always makes sense to me to go to our to infinity and to make sure that the electrostatic 78 00:07:06,840 --> 00:07:07,890 potential is zero. 79 00:07:07,890 --> 00:07:12,030 In this case, it's not really something you have to do, but it's often done. 80 00:07:12,030 --> 00:07:13,320 And I think it makes sense. 81 00:07:13,680 --> 00:07:16,900 And to you, it's very easy because it's constant. 82 00:07:17,160 --> 00:07:19,890 We have constant electrostatic potential for large hours. 83 00:07:20,160 --> 00:07:22,680 So we just take this constant and set it to zero. 84 00:07:24,150 --> 00:07:32,280 So now we know what our electric and our electrostatic potential looks like at this position, B, so 85 00:07:32,280 --> 00:07:32,970 it's zero. 86 00:07:33,750 --> 00:07:42,000 And so this means when we take this solution here and set our equal to B, then we must also get zero. 87 00:07:42,690 --> 00:07:51,960 So this means that this constant must be equal to this prefecture here times Q divided by B and we need 88 00:07:51,960 --> 00:07:53,220 to have A minus sign. 89 00:07:54,330 --> 00:08:02,550 So we can write it down like this and then we have this solution, we can determine what this electrostatic 90 00:08:02,550 --> 00:08:08,580 potential looks like at the position are equal to a test would be, of course, you one over a minus 91 00:08:08,590 --> 00:08:10,310 one of a B and D prefecture. 92 00:08:10,800 --> 00:08:15,630 So this constant here for the first part must be exactly this constant. 93 00:08:16,770 --> 00:08:18,780 And so this is our whole solution. 94 00:08:19,020 --> 00:08:23,280 We have now the electrostatic potential for all three parts. 95 00:08:24,940 --> 00:08:31,990 And since we have chosen the ticket, the electricity potential constant here for this region to be 96 00:08:31,990 --> 00:08:35,740 zero, it is now very, very easy to solve the third task. 97 00:08:36,390 --> 00:08:43,150 It is that we have to calculate the voltage between the two metals and the voltage is just a difference 98 00:08:43,150 --> 00:08:44,950 in the electrostatic potential. 99 00:08:46,040 --> 00:08:54,560 And so in case you have chosen different constants here, this is absolutely fine, it will it will 100 00:08:54,560 --> 00:09:00,290 still be the correct result, but it could be different because all of these curves here could be shifted 101 00:09:00,290 --> 00:09:01,390 by some constant. 102 00:09:02,270 --> 00:09:06,770 But now for the third task, you should get exactly the same value as I get. 103 00:09:07,730 --> 00:09:09,310 So here is my solution. 104 00:09:09,620 --> 00:09:14,310 So we had inside this electrostatic potential, which is just as constant. 105 00:09:14,330 --> 00:09:15,800 So there is no our dependence. 106 00:09:16,310 --> 00:09:19,400 And outside we had also know our dependence here. 107 00:09:19,400 --> 00:09:22,000 We had that the electrostatic potential is zero. 108 00:09:22,940 --> 00:09:27,270 And so the difference in electrostatic potentials is just this value. 109 00:09:27,920 --> 00:09:34,820 So our voltage difference is kieu divided by four pipes and zero and then one over a minus one of a 110 00:09:34,820 --> 00:09:35,030 B. 111 00:09:36,330 --> 00:09:40,290 And now the solution for the fourth task is also quite straightforward. 112 00:09:40,650 --> 00:09:47,610 So here we had to calculate the capacitance of our capacitor and this capacitance is just the ratio 113 00:09:47,970 --> 00:09:51,540 of the stored charge divided by the voltage you. 114 00:09:52,710 --> 00:09:55,620 So, yeah, we of course have to charge. 115 00:09:55,630 --> 00:10:03,210 Q And we know our voltage you so we can just set it in here and now this Qs cancel out and we can make 116 00:10:03,210 --> 00:10:05,350 the rest look a bit more beautifully. 117 00:10:05,880 --> 00:10:08,250 So this is the solution that we get. 118 00:10:08,550 --> 00:10:13,680 It's for Pi Epsilon zero times B divided by B minus A.