1 00:00:00,180 --> 00:00:04,070 Great job, you made it through this long section about electrostatic. 2 00:00:04,590 --> 00:00:06,270 Let us summarize what we have learned. 3 00:00:07,610 --> 00:00:13,140 So in the first lecture, we have considered the Maxwell equations for electrostatic. 4 00:00:13,380 --> 00:00:20,370 So here we are only concerned with the electric fields and all of the time derivatives of these fields 5 00:00:20,370 --> 00:00:21,120 are zero. 6 00:00:21,630 --> 00:00:27,300 So this leaves us with two equations and one of them becomes zero on the right hand side. 7 00:00:28,410 --> 00:00:36,900 This in turn allowed us to establish the electrostatic potential because if you write down the rotation 8 00:00:36,900 --> 00:00:39,940 of a gradient of a scalar function, it's always zero. 9 00:00:40,440 --> 00:00:46,740 So this means it must always be possible in this case to write down the electric field in terms of a 10 00:00:46,740 --> 00:00:48,860 gradient of some scalar function. 11 00:00:49,500 --> 00:00:53,470 And this function is called the electrostatic potential. 12 00:00:54,930 --> 00:01:00,600 Now, when we use the other Maxwell equation, we can derive the song equation, which is essentially 13 00:01:00,600 --> 00:01:03,300 a second order differential equation, because this is here. 14 00:01:03,300 --> 00:01:11,250 There's a class operator that includes the sum of all the partial derivatives with respect to the coordinates 15 00:01:11,610 --> 00:01:12,710 in the second order. 16 00:01:13,380 --> 00:01:17,730 And on the right hand side, we have here the charge density of our. 17 00:01:20,140 --> 00:01:26,860 Furthermore, we have established the electrostatic potential and then also the electric field for a 18 00:01:26,860 --> 00:01:28,640 general charged distribution. 19 00:01:29,380 --> 00:01:34,870 So when you have any charged distribution, you can take these expressions to calculate fire and eat. 20 00:01:36,400 --> 00:01:42,780 So this we have used together with the integral formulation of this Maxwell equation to solve several 21 00:01:42,850 --> 00:01:44,040 several examples. 22 00:01:44,710 --> 00:01:51,760 For example, here we have calculated the electric field of a homogeneously charged sphere. 23 00:01:52,690 --> 00:01:58,750 So since this is rotational symmetric, we could immediately say that our electric field, this vector 24 00:01:58,750 --> 00:02:01,540 here, is always oriented along the radial direction. 25 00:02:02,050 --> 00:02:10,180 And we only had to consider here the absolute value of R, and the result was that inside of this sphere 26 00:02:10,600 --> 00:02:17,410 we get a linear dependence on the radius because the included charge in such an imaginary circle here 27 00:02:17,710 --> 00:02:19,970 increases when we make the radius larger. 28 00:02:20,560 --> 00:02:27,310 However, outside of this circle, the electric field decreases with the power of our to the power of 29 00:02:27,310 --> 00:02:27,940 minus two. 30 00:02:29,400 --> 00:02:33,810 So this is here the famous equation that you probably knew already before this course. 31 00:02:35,070 --> 00:02:41,170 And we could also solve another example by exactly the same method, which is the spherical capacitor. 32 00:02:41,950 --> 00:02:44,800 So here we have two metals that are charged oppositely. 33 00:02:44,800 --> 00:02:45,640 Q and minus. 34 00:02:45,640 --> 00:02:53,410 Q And we could hear calculate the electric field in all three areas and we could calculate the electrostatic 35 00:02:53,410 --> 00:02:53,960 potential. 36 00:02:54,760 --> 00:03:01,240 So for this, it's very important to remember that the electrostatic potential must be continuous here 37 00:03:01,240 --> 00:03:02,260 at these positions. 38 00:03:04,510 --> 00:03:10,380 Furthermore, we could also calculate the capacitance based on the voltage which is given by this expression. 39 00:03:10,390 --> 00:03:11,830 So we have here one of a.. 40 00:03:11,830 --> 00:03:18,040 And one of a B, which corresponds to these radii here of Metal One and metal B, uh, metal one. 41 00:03:18,060 --> 00:03:18,550 Metal two. 42 00:03:21,020 --> 00:03:26,880 Then in one of the longer sections, we have discussed the electric dipole moment. 43 00:03:27,190 --> 00:03:33,070 So this is one we have two charges, for example, here, a negative Q and positive Q that are separated 44 00:03:33,430 --> 00:03:34,460 by a vector A.. 45 00:03:34,750 --> 00:03:39,190 So then we can define this dipole moment P, which is given by this expression. 46 00:03:40,180 --> 00:03:45,850 And it took us quite a long time to calculate this expression here for the electric field. 47 00:03:46,510 --> 00:03:52,660 And you can see that the orientation of the dipole moment P really plays a role in these two terms. 48 00:03:53,580 --> 00:04:00,210 However, far away from this dipole moment, this field behaves like one of our to the power of three, 49 00:04:00,690 --> 00:04:06,000 because here we get our square, here we get our square, and here we divide by our to the power of 50 00:04:06,000 --> 00:04:06,450 five. 51 00:04:08,190 --> 00:04:14,250 Furthermore, we have considered the case where we take this dipole moment and put it into a homogeneous 52 00:04:14,520 --> 00:04:16,280 and constant electric field. 53 00:04:16,800 --> 00:04:23,330 And what we found is that this dipole moment reorients and it aligns with the electric fields. 54 00:04:24,030 --> 00:04:31,020 So this means on the center point or on the total configuration of these two charges, there's no charge. 55 00:04:31,380 --> 00:04:32,930 There is no force acting here. 56 00:04:33,630 --> 00:04:37,110 However, we have a talk that leads to this reorientation. 57 00:04:37,860 --> 00:04:42,260 And you can see this if you consider here a constant electric field that we get this talk. 58 00:04:42,270 --> 00:04:46,500 However, we do not get the force because the gradient of a constant is zero. 59 00:04:48,760 --> 00:04:54,360 In one of the last sections we have discussed and arrived the energy of electrostatic. 60 00:04:55,270 --> 00:05:00,520 So if you know the electric field, for example, here in these two examples, you can just take this 61 00:05:00,520 --> 00:05:06,670 equation that is based on the continuity equation and just integrate over the electric field squared. 62 00:05:07,810 --> 00:05:14,890 However, when you have not when you do not have the electric field, you can just take the charged 63 00:05:14,890 --> 00:05:15,640 distribution. 64 00:05:18,190 --> 00:05:24,640 And the very last section of this of this part of discourse, we also discussed boundary conditions 65 00:05:24,880 --> 00:05:32,020 and image charges and we found out that the boundary conditions are actually very important because 66 00:05:32,020 --> 00:05:37,420 if you if you have a single charge and vacuum, the electric field configuration looks like this. 67 00:05:37,420 --> 00:05:38,890 It's radial symmetric. 68 00:05:39,640 --> 00:05:45,880 However, once you bring this charge in the vicinity of a metal where you have electrons inside of it, 69 00:05:46,330 --> 00:05:48,030 these electrons will react. 70 00:05:48,400 --> 00:05:53,950 And this is because the boundary of this metal has to fulfill a boundary condition. 71 00:05:54,150 --> 00:05:59,940 It has to fulfill that the electrostatic potential is constant, for example, can be zero. 72 00:06:01,090 --> 00:06:07,420 And in order for this to be true, these field lines here must be perpendicular to this boundary. 73 00:06:07,420 --> 00:06:12,730 And this can only be the case if the electric field profile looks as is shown here. 74 00:06:13,760 --> 00:06:20,200 So the electric fields of this single charge in the vicinity of a metal behaves as if there were another 75 00:06:20,200 --> 00:06:21,930 charge inside of the metal. 76 00:06:22,390 --> 00:06:23,950 And this is called the mirror charge. 77 00:06:23,950 --> 00:06:29,110 And that emerges because the electrons react to this charge outside of the metal.