1 00:00:00,210 --> 00:00:07,200 So now that has come to an exercise for you, we want to calculate the electric field and the electrostatic 2 00:00:07,200 --> 00:00:15,810 potential for a charged sphere so the tasks that you have to solve are calculate the electric field 3 00:00:15,810 --> 00:00:18,750 of a homogeneously charged sphere. 4 00:00:19,530 --> 00:00:26,190 And as a hint, you have to distinguish the inside region of the sphere and the outside region of the 5 00:00:26,190 --> 00:00:26,660 sphere. 6 00:00:27,930 --> 00:00:34,230 And you should use the rotational symmetry of the problem to make it a lot simpler for you to solve 7 00:00:34,230 --> 00:00:34,830 the integral. 8 00:00:35,520 --> 00:00:42,990 And you have to consider that homogeneously charged means that this row here it is charged density is 9 00:00:42,990 --> 00:00:43,660 constant. 10 00:00:43,920 --> 00:00:46,230 So it means you can pull it out of the integral. 11 00:00:47,460 --> 00:00:52,410 And then the second task is quite easy when you have determined the electric field from the electric 12 00:00:52,410 --> 00:00:54,960 field to calculate the electrostatic potential. 13 00:00:56,430 --> 00:01:03,300 So if you want to have another Hynde's, please keep on watching the video, otherwise you can now stop 14 00:01:03,300 --> 00:01:06,170 the video and go ahead and solve the problem yourself. 15 00:01:08,350 --> 00:01:14,740 So the bigger hint is that, of course, we have already solved this problem for the region outside 16 00:01:14,740 --> 00:01:15,410 of the sphere. 17 00:01:16,300 --> 00:01:25,720 So this was as a solution to Coulomb stop where we have integrated here over the charged density and 18 00:01:25,720 --> 00:01:26,650 got to that charge. 19 00:01:26,660 --> 00:01:33,970 Q And on the left hand side, we could pull out to the electric field because we have this dependence 20 00:01:34,390 --> 00:01:39,600 artist, a cemetery where the electric field is always pointing along the radio direction. 21 00:01:39,880 --> 00:01:44,840 So this vector here is parallel to the vector from the integral. 22 00:01:45,370 --> 00:01:52,030 And so this term here would just be the surface of the sphere that we are considering for our integration. 23 00:01:53,360 --> 00:01:59,750 And so this gave us the electric field, which is proportional to you, divide divided by our square. 24 00:02:01,070 --> 00:02:06,530 So the first part we have actually already solved, but still it's a nice exercise to do it yourself 25 00:02:06,530 --> 00:02:07,100 once again. 26 00:02:07,940 --> 00:02:12,140 And now what is about the region inside of the sphere? 27 00:02:13,080 --> 00:02:14,960 So this you really have to do yourself. 28 00:02:14,960 --> 00:02:20,350 So please post a video and just continue when you have tried for a few minutes. 29 00:02:20,930 --> 00:02:27,140 So since this is the first time you are doing this, please don't expect to be able to solve it just 30 00:02:27,140 --> 00:02:32,150 right away, because even though it may be easy, once you have seen the solution, sometimes it can 31 00:02:32,150 --> 00:02:37,070 be quite difficult to get the right idea and to come up with a solution. 32 00:02:37,640 --> 00:02:38,840 So please don't give up. 33 00:02:38,840 --> 00:02:42,260 Just really spend some minutes trying to solve this. 34 00:02:42,740 --> 00:02:47,270 And when you've done it or when you need more hints, please continue the video. 35 00:02:48,230 --> 00:02:50,300 And now I will present you the solution. 36 00:02:51,760 --> 00:02:57,200 So everything that we need is the Maxwell equation and its integral formulation. 37 00:02:57,220 --> 00:03:03,010 So this one here, then we need the symmetry of the problem, which allows us to write down the electric 38 00:03:03,010 --> 00:03:07,840 field as a scalar function multiplied by this vector are. 39 00:03:08,020 --> 00:03:11,540 So it means the electric field is always pointing along the radio direction. 40 00:03:12,460 --> 00:03:16,930 And so this means that this is always parallel to this D.R.C. here. 41 00:03:18,920 --> 00:03:24,760 Now, in previous light, I have already shown you what the solution looks like outside of the sphere, 42 00:03:24,770 --> 00:03:30,400 so this means this small R is larger than this capital R, which is the radius of the sphere. 43 00:03:31,160 --> 00:03:39,220 And so our right hand side of this solution is always the volume of this sphere. 44 00:03:39,230 --> 00:03:43,960 So it is four divided by three times PI times capital R to the power of three. 45 00:03:44,420 --> 00:03:50,750 And in this charge density and of course this one, we could also write down as the inclosed charge, 46 00:03:51,050 --> 00:03:52,550 but it's not really necessary. 47 00:03:53,810 --> 00:03:59,990 Now, the left hand side you can see here, I've already pulled out this E of our so we are just left 48 00:03:59,990 --> 00:04:05,550 with the surface integral of the way of the sphere that we are considering here. 49 00:04:05,570 --> 00:04:13,910 So this would be this black Dasht sphere with the radius small R and not capital R, so the surface 50 00:04:13,910 --> 00:04:17,240 of a sphere is four pi times the radius square. 51 00:04:17,300 --> 00:04:18,540 So this is what we have here. 52 00:04:19,280 --> 00:04:23,270 So as you have seen on the previous slides, this is already a solution. 53 00:04:23,270 --> 00:04:27,410 We just have to divide by Epsilon zero by four pi r square. 54 00:04:27,840 --> 00:04:34,640 And so this is the solution that we come up with so we can express it like this or we can reintroduce 55 00:04:34,640 --> 00:04:38,480 the inclose charge and then we get the result that we had previously. 56 00:04:40,530 --> 00:04:47,030 Now, if you have the solution, it is not so difficult to find a solution for the inside region. 57 00:04:47,920 --> 00:04:55,570 The main difference here is that the right hand side of this equation looks different because here the 58 00:04:55,570 --> 00:05:04,330 right hand side of this equation was always the volume of our charge, so of our charged sphere. 59 00:05:05,290 --> 00:05:11,200 So this is because this road density is only finite for this sphere. 60 00:05:11,200 --> 00:05:13,780 And in this wide region here, it would be zero. 61 00:05:13,930 --> 00:05:16,520 So it would not contribute here to this integral. 62 00:05:17,320 --> 00:05:26,080 However, now if we look on the inside, this black Dasht sphere here would be inside of this blue sphere. 63 00:05:26,530 --> 00:05:31,420 So this means now the radius of the black sphere really plays a role. 64 00:05:31,810 --> 00:05:38,160 So this means here for this integral, we do not have to consider the radius of the black out of the 65 00:05:38,230 --> 00:05:43,370 blue sphere capital R, but we have to consider the radius of the black sphere smaller. 66 00:05:44,140 --> 00:05:50,470 So this means here the left hand side is exactly the same, but the right hand side is now four pi four 67 00:05:50,470 --> 00:05:55,900 divided by three times pi times smaller to the power of three times real. 68 00:05:57,000 --> 00:06:04,110 And this means that this is now our solution, so why we had to hear a one of our square dependent's 69 00:06:04,110 --> 00:06:09,990 we have now here and our dependence, and this is, of course, because here instead of this capital, 70 00:06:09,990 --> 00:06:17,220 are we have now a small R and so these power of three and this power of two, they council and we are 71 00:06:17,220 --> 00:06:18,730 left with power of one. 72 00:06:20,010 --> 00:06:25,610 Now we can look at this electric field and you can see that on the inside we get our dependence. 73 00:06:25,980 --> 00:06:27,150 So it's a linear function. 74 00:06:27,150 --> 00:06:36,810 It increases, which is, of course, because when we go far away from the center of this sphere, we 75 00:06:36,810 --> 00:06:41,190 have more a charge generating these electric fields. 76 00:06:42,270 --> 00:06:46,730 So this means here we have this linear increase in the electric field. 77 00:06:47,070 --> 00:06:53,880 And when we are outside of the sphere, the electric field decreases with the power of one over R to 78 00:06:53,880 --> 00:06:54,720 the power of two. 79 00:06:56,220 --> 00:07:00,200 And at the position are we can also calculate the value. 80 00:07:00,600 --> 00:07:04,640 This would be divided by three epsilon zero times capital R. 81 00:07:04,950 --> 00:07:11,550 And as you can see, this is this value we get from both solutions, of course, because the electric 82 00:07:11,550 --> 00:07:12,690 field has to be steady. 83 00:07:13,410 --> 00:07:20,190 So here we would get capital R straightforwardly and here we would get R to the power of three divided 84 00:07:20,190 --> 00:07:23,640 by R to the power of two, which gives us the same value. 85 00:07:25,570 --> 00:07:34,420 Now, the second task was calculate the electrostatic potential, and so here we just have to take our 86 00:07:34,420 --> 00:07:38,390 solutions from inside and outside of the sphere for the electric field. 87 00:07:39,310 --> 00:07:43,630 And now we have to calculate the electrostatic potential by this equation here. 88 00:07:44,380 --> 00:07:47,740 So basically, we just have to integrate over the electric field. 89 00:07:48,940 --> 00:08:00,830 So here we get as a solution that we get here an exponential of this hour that is increased by one. 90 00:08:01,030 --> 00:08:07,210 So instead of R to the power of minus two, we get R to the power of minus one or we can write it down 91 00:08:07,210 --> 00:08:11,310 as one of R and then we would also get a minus sign. 92 00:08:11,320 --> 00:08:12,770 But he is also a minus sign. 93 00:08:12,790 --> 00:08:13,850 So these cancel out. 94 00:08:14,410 --> 00:08:16,770 So this is what we get for the electrostatic potential. 95 00:08:17,710 --> 00:08:25,960 And of course, you can remember that we could add here any constant to this electrostatic potential 96 00:08:26,380 --> 00:08:32,640 because we can shift our zero position of the electromagnetic potential as we wish. 97 00:08:33,340 --> 00:08:39,800 But here I have chosen it such that it is zero when we take the position vector to infinity. 98 00:08:40,090 --> 00:08:45,850 So if you go with R to infinity, this whole thing here would go to zero, which is something that, 99 00:08:46,150 --> 00:08:50,410 yeah, looks good or makes sense physically, but it's actually not that important. 100 00:08:51,070 --> 00:08:55,360 It's important that you have this term and then you would have here another constant possibly. 101 00:08:56,680 --> 00:09:01,670 Now for the inside region, we also have to integrate here. 102 00:09:01,700 --> 00:09:10,420 So what we get is we have to increase this exponent here so we get all square and then we have to divide 103 00:09:10,420 --> 00:09:11,020 by two. 104 00:09:11,960 --> 00:09:14,980 And then also we get you just minus sign from this minus sign. 105 00:09:16,060 --> 00:09:18,940 So this is what we get for the electrostatic potential. 106 00:09:19,690 --> 00:09:22,800 But also we get here another constant. 107 00:09:23,710 --> 00:09:30,820 And now this time the constant is really important because at the position are is equal to capital. 108 00:09:30,820 --> 00:09:35,120 Are these two electrostatic potentials need to match each other. 109 00:09:35,650 --> 00:09:37,700 So it must be a continuous function. 110 00:09:38,710 --> 00:09:45,130 So if we have assumed here that our constant is zero, we can now determine the constant here so that 111 00:09:45,130 --> 00:09:53,530 these to match at the position are equal capital R so you can see that this solution and we can also 112 00:09:53,530 --> 00:09:54,910 write it like this. 113 00:09:55,270 --> 00:10:02,440 And if you now put in here to Capital R, you will see that you will get here two times Capital R Square. 114 00:10:02,710 --> 00:10:11,590 So you get real divided by three Epsilon, zero Times Capital R Square and you, you get the same thing 115 00:10:11,590 --> 00:10:15,610 because you used to cancel out and we are only left with a capital R Square. 116 00:10:16,690 --> 00:10:19,650 And so this is what the electrostatic potential looks like. 117 00:10:20,080 --> 00:10:28,150 So it keeps on decreasing yappers basically because of the shape of our electric field on the previous 118 00:10:28,150 --> 00:10:28,570 slide. 119 00:10:29,200 --> 00:10:34,450 And then we have here a continuous function and to position R, which is really important. 120 00:10:35,680 --> 00:10:38,230 So I'm just saying this one more time. 121 00:10:38,740 --> 00:10:45,580 You could add here a different constant and then you would have to add here the same constant additional 122 00:10:45,580 --> 00:10:46,600 to this constant. 123 00:10:47,820 --> 00:10:54,390 So when you plot your electrostatic potential, it should look exactly like this, but this whole function 124 00:10:54,390 --> 00:10:56,310 could be shifted by any value.