00:01
This problem asks us some questions about a coaxial cable.
00:04
And i think a picture actually is going to help a lot.
00:07
So let me show you a picture that i made of this coaxial cable.
00:11
And what you'll see here is that i have an inner conducting rod or wire.
00:18
And this whole thing, by the way, is infinitely long.
00:20
I mean, really really long.
00:22
Doesn't need to be infinite long.
00:23
But this wire, i'm going to call that a.
00:26
And the radius of that wire is equal to two centimeters.
00:33
And then surrounding a i have, first of all, this empty space, which is this white space.
00:40
And then there is this conducting shell, which is a blue, is made in blue here.
00:45
And the shell is, i want to call that b.
00:48
Okay.
00:49
And the radius of b is 10 centimeters.
00:53
Now, when i say radius, i mean the outer radius of the interrace.
00:57
Radius.
00:57
This is a very thin shell.
01:00
So this blue shell has almost, you know, very little thickness.
01:04
It's not infinitely thin, but it's really thin.
01:07
So we can take the inner radius and the outer radius has been the same thing.
01:12
They're both equal to 10 centimeters.
01:13
And then what i'm told that on the exterior of this coaxial conductor, i have a surface charged density on that outer surface there.
01:28
And the surface charge density there is equal to positive 0 .5 nanocoules per meter squared.
01:37
So that's the surface charge density on that outer conductor.
01:42
Now, we're also told that this outer shell, shell b is uncharged.
01:49
So b is uncharged.
01:51
Now that means that there's an interior surface charge density here, i'll call that the inner one.
01:58
Okay, in the inner surface charge density is equal to negative 0 .5 nanocoultz per meter squared.
02:08
And that's because there's no net charge on that blue conductor.
02:13
Well, why does it have a surface charge density at all? well, that's because of what's going on with the wire a.
02:21
Okay.
02:22
And what we're asked to do then is we're asked to find what is the surface charge density on wire a.
02:30
Now, we're going to solve this problem using gouselaw, and i've drawn in here a gaussian surface, so that is this black rectangle here, that goes through the center of the outer shell, that blue shell, and it goes through the center of the inner wire, which is the red wire.
02:51
And both the inner wire and the outer shell, they're both conducting materials, so remember that.
02:57
And what that means that they're conducting materials is that the electric field in their interior is zero.
03:05
Now, we're also, remember that this is an infinitely long wire, so that means that the electric field through this side and that side, both of those are zero.
03:15
And because there's no electric field within either conductor b or conductor a, that means that the total flux through that gaussian surface is equal to zero.
03:32
Now, what that means is that the charge enclosed, because that's equal to the enclosed charge and that gaussian surface divided by the permittivity constant.
03:44
And that means that the enclosed charge is zero.
03:48
Now, you might think, well, that, no duh, that means that sigma a is equal to sigma b, that's not quite true.
03:57
What that really means that the linear charge density, the charge per unit length on wire a is equal to the charge per unit length on wire b.
04:10
Because that's what's really the total charge enclosed, let's say, is equal to it's actually equal to negative that.
04:21
That's equal to the charge for unit length or a times the length of this thing.
04:28
I'll call that little l.
04:31
So that's the length here times l plus lambda b times l.
04:39
Okay, and those sum equals zero.
04:43
So that means that lambda a is equal to negative of lambda b.
04:47
So now we're going to look at the relationship between sigma's and lanzas.
04:54
Now, the charge per unit area in general is equal to the total charge on something divided by the area.
05:05
So that can be the charge on a certain length of wire, right? so that's equal to then charge divided by 2 pi r, that's the perimeter of a circle going around loopwise.
05:25
Let me see about this.
05:27
Let me see if i can draw a circle in here.
05:29
I probably can't.
05:30
But let's say if i drew a circle here, the perimeter of that circle is 2 pi r.
05:36
And then if i multiply that by a little length l, i'm going to change, i'll call that h, a little length h.
05:47
So that's what the charge for unit area, because the area of that little thing is equal to 2 pi r times h.
05:55
Or i can make an l.
05:59
So here, what that is then is that's equal to 1 over 2 pi r times the charge per unit length.
06:09
Okay.
06:09
So that's equal to lambda divided by 2.
06:16
Therefore, lambda is equal to 2 pi r times sigma.
06:27
Okay.
06:28
Now we are told, we know that lambda a is equal.
06:34
To negative lambda b and let me remind you too that what i mean by lambda b is the lambda on the inner surface of this here okay so that's equal to negative and this is i guess it's the inner it's the inner lambda b which is equal to positive lambda the outer okay so so then i can i can use this and say okay well lambda a is equal to the 2 pi r a times sigma a, and that is equal to 2 pi r b times sigma b, and here i mean the outer surface charge density.
07:24
And therefore, the surface charge density, sigma a, is equal to the two pies cancel.
07:33
So i have rb over ra times sigma b.
07:44
And rb over ra is just equal to 5.
07:48
So that's equal to 5 times 2 times 0 .5, which is equal to 2 .5 nanocouons per meter squared.
08:01
So that's the surface charge density on surface a.
08:07
Let me clean this up a little bit here.
08:10
It's a little more legible.
08:13
That's going to be 2 .5 nanocouons per meter squared.
08:19
So that's the surface charge density on the inner surface.
08:24
Okay, now the next question asks for an expression for the electric field inside the outer cylinder and distance r from the center.
08:38
So central axis, i guess you'd call it...