00:01
Okay, so we've got a sphere of radius a and volume charge distribution c over r squared coulombs per meter cubed, and we want to find the electric field in and out of the sphere respectively.
00:20
So to find the electric field in, we're first going to take our gauss sphere to be of radius r, or our gaussian surface to be a shell of radius r, which is less than a.
00:32
So for e in, we're going to be doing gauss's law.
00:41
So the integral of the electric field dot -producted with the normal to the area of the gaussian sphere is given by the enclosed charge divided by epsilon naught.
00:55
Now the enclosed charge in this sphere of radius r, it's inside the sphere, and so we're just going to times the volume of this sphere by the charge density.
01:08
So that's going to be charge density is c over r squared.
01:12
We've got the volume is going to be four thirds of pi r cubed, and then we've got our one over epsilon naught there.
01:22
And tidying that up quickly, we're going to get this r squared is going to cancel with the two of the r factors there.
01:31
So we're going to get four c pi r over three epsilon naught.
01:39
As for this integral, it's just going to be the electric field points in the same direction as the normal to the area vector.
01:45
So this is just going to be equal to the electric field times the surface area of the gaussian sphere, which is four pi r squared.
02:00
And then we're just going to solve this equation.
02:02
So these two things equal each other.
02:03
We can see the four pi cancels, and we're going to get that e equals c over three epsilon naught r.
02:15
So that's inside.
02:16
How about outside? outside we'll take our gaussian surface to be this shell.
02:24
Again we'll call the radius r, but now we're considering r bigger than a.
02:28
So e out, which is r bigger than a...