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In this problem, we're going to talk about gauss's law and the electric potential.
00:06
So first, about gauss's law, we need to remember that, let's say that we have a charge distribution, whose total charge is equal to q, and that we draw, let's say that we draw a surface, a gaussian surface, around this charge.
00:29
Then gauss's law tells us that the electric flux defined as the circuit, surface integral of the electric field times the element of area is equal to the total charge enclosed by the surface divided by epsilon 0.
00:48
Also, we have the electric potential v is equal to minus the integral of the electric field.
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01:06
From the point, actually let me write this as v minus v0, this is the potential difference between our 0 and r.
01:19
Okay, so now we can go on to our exercise.
01:22
What we need to do is to consider a sphere, a non -conducting sphere that has a radius r0.
01:30
So let's say that this here is our sphere.
01:34
And this sphere has a uniform density such that the total charge of the sphere is q.
01:44
And our goal is to find what are what is the value of the potential energy first in question a when r is greater than r zero well in that case the we can use gauces law in order to find the electric field and then with that we can find the electric potential the first thing we need to notice is that when v goes to infinity we want v to go to zero so that's something that we're going to impose over our occasions.
02:28
So first, notice that since there is spherical symmetry in our system, then the electric flux is just, let's say that we draw the surface at a distance r from the center.
02:48
The electric flux is just the electric field at that point.
02:53
Notice that all throughout the surface of the electric field is the same, times the area, which is 4 pi r squared, and this is q, that's the charge enclosed by the surface, divided by epsilon 0.
03:10
So the electric field is q over 4 pi epsilon 0, r squared.
03:17
Now, v is minus the integral of the electric field.
03:28
Dr and what i'm going to do is to integrate from infinity to zero because of infinity v is equal to zero so v is equal to minus q over 4 pi epsom 0 and there is a minus here so there's an sorry, yeah, so there's an overall positive sign, so v is q over 4 pi epsilon 0r.
04:14
Notice that when r goes to infinity, v goes to infinity as we want to.
04:19
Then we want to find what is the value when r is smaller than our 0.
04:24
Here, we're going to need to impose that the electric potential must always be continuous.
04:30
So the electric potential at r, the electric potential at r is greater, greater than r0.
04:40
When calculated at, let's say, r is equal to r0 plus, this means that it's equal to i to r0.
04:49
From outside of the conducting sphere, this is equal to v of r0 minus.
05:01
Okay.
05:02
So first, i'm going to draw here our surface.
05:10
So now, we have that r0, this distance here is r0, our surface, the distance between the center and our surface is r.
05:25
So what we have is that the electric field, i'm sorry, the electric field times the area element is equal again to e times 4 pi r squared, and this is equal to the enclosed charge, i'm going to call it lowercase q, divided by epsilon 0.
05:54
Now notice that lowercase q is equal to the density times the volume of 4 pi are cubed over 3, and the row, the density is the total charge divided by 4 pi are 0 cubed over 3, by 4 pi over 3 are cubed...