1

00:00:00,180 --> 00:00:03,570

None of us come to more general charged distribution's.



2

00:00:04,590 --> 00:00:11,640

So in the previous lecture, we have determined the electrostatic potential for a sphere, a four point



3

00:00:11,640 --> 00:00:12,150

charge.



4

00:00:13,020 --> 00:00:17,640

So we have considered that our charge is positioned at the position of zero.



5

00:00:17,670 --> 00:00:20,460

So in this case, our one is equal to zero.



6

00:00:20,910 --> 00:00:24,740

And we found out that the electrostatic potential is just as constant.



7

00:00:24,750 --> 00:00:27,960

Your time times to charge divided by hour.



8

00:00:28,860 --> 00:00:35,640

So in general, if our charge is positioned at some other position, like one in this case, we can



9

00:00:35,640 --> 00:00:36,900

write it down like this.



10

00:00:38,740 --> 00:00:45,760

Now, of course, if we have multiple point charges, we can just add up their individual electrostatic



11

00:00:45,760 --> 00:00:46,460

potentials.



12

00:00:47,320 --> 00:00:53,580

So this is because this equation is including here does the plant's operator.



13

00:00:53,890 --> 00:00:57,850

So that's a sum of second-order partial derivatives.



14

00:00:58,310 --> 00:01:02,240

So the solution can be considered the sum of individual solutions.



15

00:01:02,560 --> 00:01:08,530

So if we have multiple poison charges, we can write down our electrostatic potential just like this.



16

00:01:09,490 --> 00:01:16,210

So, for example, consider we have a charged Q1 at this position or one, and we have another charge,



17

00:01:16,210 --> 00:01:18,370

Q2 at the position or two.



18

00:01:18,910 --> 00:01:22,000

Then we would just have two terms here and here.



19

00:01:22,000 --> 00:01:26,980

We would have Q1 and our one and a plus Q2 and our two.



20

00:01:29,100 --> 00:01:36,470

So what this means is that the electrostatic potential is additive, and if we can write it down as



21

00:01:36,470 --> 00:01:42,750

a sum, we can also write it down as an integral, because maybe you remember this.



22

00:01:42,750 --> 00:01:49,110

If we make our our space smaller and smaller, just some here becomes an integral.



23

00:01:49,830 --> 00:01:56,670

So this means we can write down for a general charged distribution row of R that the electrostatic potential



24

00:01:56,670 --> 00:01:58,170

is just as integral here.



25

00:02:00,170 --> 00:02:05,450

So you see, for in terms of the structure, it looks very much the same as here, it's just that we



26

00:02:05,450 --> 00:02:13,880

have to some here, this is now an integral and here we have expressed our cue in terms of this real



27

00:02:13,880 --> 00:02:14,420

times.



28

00:02:14,420 --> 00:02:20,510

These are so if you remember the charges, just as integral overall times D.R.



29

00:02:22,140 --> 00:02:29,820

So now we have all our tools to consider more difficult examples, and in the next section we will start



30

00:02:30,150 --> 00:02:35,820

by calculating the electric field and the electrostatic potential of an electric dipole.