00:01
In this problem, you are told you have a spherical shell, and you're told you have certain surface charged density, area densities, meter squared, not volume densities, peter squared.
00:16
And the first thing they ask is, what is the field at four centimeters? and your idea might be, okay, here's my gaussian surface inside here.
00:33
And, well, there's no charging closed, so the field must be zero.
00:37
Well, that's not the case.
00:40
Let's talk about this.
00:46
If you're talking, this is metal, this is a conductor, electrostatic equilibrium, excess charge must be on the outer surface.
00:58
Those charges distribute themselves so that the electric field in the interior of that conductor is zero at all points.
01:06
That means i could take a knife and a spoon and chop out any or all of the interior and nothing changes.
01:16
This charge cannot be there.
01:21
If we're talking about electrostatic equilibrium, this is induced charge from a charge in the interior.
01:34
That's why it arises on that inner surface.
01:38
It's an induced charge.
01:39
If we're talking about just a plain old conductor by itself, electrostatic equilibrium, this cannot be there.
01:49
This cannot exist.
01:53
As an induced charge from q, it can.
01:59
This q will cause an induced charge on the inner surface, also on the outer surface, even though there is excess charge also.
02:10
There is excess charge also in terms of, this but that we'll see later it's not worry about that just yet so let so there's got to be a charge queue there and i'll show you even by galsals as long that if it wasn't there we'd have results that we know are wrong so let me draw galsian surface in the metal now call this i'll call this surface one e.
02:54
D .a charging closed or epsilon 0.
03:00
And let's say for the sake of argument, all we had was the charge on the inner surface.
03:06
Sigma a4 pi a squared over epsilon 0.
03:13
If i stop there, this is non -zero.
03:18
That would tell me i have an electric field inside the metal, and that can't be.
03:23
E is zero inside the metal in an electrostatic equilibrium.
03:29
E is zero in here.
03:42
Trostatic cladilibrium.
03:45
So that's zero.
03:46
Well, if that's zero, then that would make sigma a zero.
03:50
But that's not zero.
03:52
So there's got to be something more.
03:56
And that's the cue i was talking about.
04:00
There's that cue.
04:01
So now we can find the q what's inside.
04:06
Otherwise, it makes no sense.
04:08
Either you have a non -zero field in the metal, or you have a kind, you say you have a zero service charge, density, which we know is not the case.
04:20
So you have to have this.
04:22
So 0 is equal to sigma a 4 pi a squared plus q.
04:31
So that gives me q is equal to minus sigma a 4 pi a squared, minus 250 times 10 and minus 9 kouloms per meter cubed or times 4p3 .2 .5 .5 .0 .000 pi, don't forget to convert to centimeters to meters.
04:56
And this works out to be 7 .85 times 10 and minus 9 coulomes, which is 7 .85 nanoculms.
05:08
So that's the charge in the interior.
05:13
Now, field it 4 centimeters.
05:18
We'll draw our gaussian surface.
05:25
I'll call this surface 2.
05:30
So, e .da, charge enclosed, epsilon zero, which is just q over epsilon zero.
05:48
Now, da vector is da r -had.
05:53
R -hat's a radially pointing vector, outward -pointing vector, unit vector.
05:59
Now, what i do, e.
06:00
Dot, r -hat, that gives me er, the r component of the electric field, the radio component.
06:10
So this becomes er, da, q over epsilon 0, but er is the same from the symmetry.
06:20
E .r is the same at all points on this, on that gaussian surface.
06:26
So i can bring that out.
06:27
So this becomes er integral da, q over epsilon 0, but da is just 4 pi r squared.
06:37
Er, 4 pi r squared.
06:40
Is q over epsilon 0.
06:43
So it gives me er is equal to q over 4 pi epsilon 0 r squared.
06:50
So it's a point charge field.
06:58
Now, this is q we know is positive.
07:02
So this whole quantity is positive.
07:04
Greater than 0, which is greater than 0 implies er, or let me not do it that way.
07:14
E is outward pointing.
07:28
So we know that much, and we still have to calculate the field.
07:33
We'll calculate the field magnitude.
07:36
E is equal to er 8 .99 times 10 to 9, newton, meter squared, coulum squared, 7 .85 times 10 to the 9, over 0 .04 meters squared, and this is 4 .41 times 10 to the 4 newtons per cooler.
08:07
So that's the electric field magnitude, which is the same as the radio component because its radio component is positive.
08:18
So that is 4 centimeters...