00:01
Okay, in this problem we have a non -conducting spherical shell, and it has an inner radius a, an outer radius b, and the charge is found in between there.
00:18
So for r less than a, there's no charge, and for r greater than b, there's also no charge.
00:24
It has a charge density of row equals a constant c over r squared.
00:34
And then like i said row equals zero between or for r less than a and for r greater than b okay so we first want to find the total charge enclosed and yeah okay so one total charge so what is the total charge enclosed by a sphere with a radius a less than r less than b.
01:07
So we're drawing a, what we're doing basically is we're getting ready to use gauss's law.
01:11
So we have this, we're making this gaussian sphere radius r and we want to figure out how much charge is enclosed.
01:20
So the formula for enclosed charge is the integral of row dv.
01:30
So the charge density times the volume will give you the total charge.
01:36
But the charge changes as r changes, so we do have to integrate.
01:42
So this turns into a integral and spherical coordinates of row times dv.
01:51
So dv in spherical coordinates is r squared sine theta, d -fai, where r goes from.
02:04
So here we want to think about what areas are charged.
02:08
So we don't want to go from zero to r because because there's no charge from 0 to a.
02:16
So we want to go from a to r.
02:19
And maybe a better way to write this would be put primes on these r's inside the integral because, yeah.
02:32
Okay, so we're integrating from a to r.
02:35
In theta, we go from 0 to pi, and phi we go from 0 to 2 pi in physics.
02:41
Okay.
02:45
So, that looks good.
02:49
This integrates out to be 4 pi times c times r minus a.
03:00
Okay, so that is the charge enclosed by a sphere of radius r.
03:11
Okay.
03:13
And this integral is not too bad because you're just, the r -primes cancel, the r -primes squareds cancel, i mean, and you're just integrating sine theta, theta, d, d -fi.
03:26
So, fairly easy integral in spherical coordinates.
03:31
Okay, that's the total charge enclosed by the sphere.
03:34
Now we're going to find the magnitude of the electric field on the surface of that sphere in terms of a, b, c, r, and constants.
03:43
Okay, so what's the electric field here? you apply gauss's law.
03:47
The integral of, surface integral of e .da equals q enclosed over epsilon not.
03:59
Well, everywhere on that sphere, the electric field will be radially outward, so it's parallel to da...