00:01
In this example, we're going to be determining the electric field at all points in space for the following configuration.
00:08
Okay, what we have is a conducting spherical shell with a point charge q at its center.
00:15
So here's q at the center and then that's surrounded by a conducting spherical shell.
00:27
Okay, it's got inner radius a, outer radius b.
00:37
Okay, this is supposed to be a circle, but you can see that it's not quite a circle, but we're going to pretend it's perfectly spherical.
00:44
Okay, this shell has a charge on it as well and this charge is going to be equal to negative three q.
00:54
Okay, this charge in the center has charge q.
01:00
So i want to find the electric field in three regions.
01:03
I want to find it for r less than a, k, r in between a and b, so less than b but greater than a.
01:19
And then i want to find it for r is greater than a.
01:25
Okay, so first we're going to be looking at the region inside the hollow spherical shell, the outside of that charge q in the center.
01:36
So this will be my first region.
01:39
Then our second region is from a to b, so we're inside the shell.
01:43
And then my third region, this is region two, my third region is everything outside of the shell.
01:50
Okay, so let's start calculating these fields.
01:54
We're going to be using gauss's law for all three locations.
02:01
Okay, gauss's law says that the e dot da equals q enclosed over epsilon naught.
02:12
Okay, so for my first region when we're inside this hollow spherical shell, my q enclosed is just q.
02:24
My surface integral, that's just e is perpendicular to the surface everywhere because we have that spherical symmetry so we can use gauss's law.
02:34
So e dot da is just e times the surface of the sphere for pi r squared.
02:42
Okay, so e for r less than a equals q over four pi epsilon naught r squared.
02:59
Okay, now we want to look at region two.
03:02
That's between the two walls of the shell.
03:08
Okay, the shell is a conducting shell.
03:10
What do we know about conductors? we know that there's no electric field within them.
03:15
Okay, all of the charge resides on the surface.
03:18
Okay, the two surfaces.
03:20
So we're going to have, we have a positive charge in the center.
03:23
So we're going to have negative charge distributed on the inner surface of the shell.
03:27
We'll have positive distributed on the outside.
03:31
Okay, so for region two, e equals zero.
03:37
Then for region three, now we're outside of the shell.
03:42
Okay, so again we can use gauss's law here.
03:46
Okay, now we have some enclosed charge...