# Gauss's Law 2 concentric cylinders in electrostatic equilibrium

1. Jan 5, 2009

### shale

1. The problem statement, all variables and given/known data

Consider the cross sections of two, very long, concentric, metallic, hollow cylinders placed in a vacuum. The small cylinder has inner radius A and outer radius B while the larger cylinder has inner radius 2A and outer radius 2B. Initially the small and big hollow cylinders have charge densities of +2λ and -2λ respectively.

1) AT electrostatic equilibrium what is the total charge of the inner surface of the big cylinder if the cylinders both hav e length L?

2) what is the electric field in tha cavity between the two cylinders? se r as the distance measured from the center of the spheres.

2. Relevant equations

relevant equations would be the flux which is EA=Qenco

3. The attempt at a solution

for number 1

since this is at electrostatic equilibrium, the net charge AT the CYLINDER is ZERO. therefore Qin=0... but inner cylinder has +2λL, and so the inner surface of the larger cylinder would be -2λL... and for the outside surface.... (will it be +2λL for the net charge on the conductor to be zero.?? or there will be no charge because the said "net" charge in that conductor should be -2λL??) i'm confused....

and is it correct for me to multiply the charge densities to L??

for number 2

I set up an equation like this...

EA= E2πrL = ([2λ(πr2L)] - [-2λ(πR2L)]) /εo

we cancel and it becomes like this

E = [λπ(r2+R2)] /εor

where r is the outer radius of the smaller cylinder and R is inner radius of larger one.. i think i'm wasting my effort because all this is wrong :-ss please help.

Last edited: Jan 5, 2009
2. Jan 5, 2009

### Redbelly98

Staff Emeritus
Welcome to PF.

Yes.

No, because the net charge on the outer conductor is NOT zero.

Yes. There is zero charge on the outer surface of the outer cylinder, for exactly the reason you said here.

Yes, if you want to get the total charge in a length L of a cylinder.

Not quite, but you're not far off.

Think about the surface you are defining, and the fact that you need to use the charge enclosed by that surface here.

• Does the surface contain the inner cylinder?
• Does the surface contain the outer cylinder?

Based on the answers to those 2 questions, what is the charge enclosed within the surface?

3. Jan 6, 2009

### shale

do mean the gaussian surface? ---- 1) Yes, and 2) No..

so total charge enclosed is +2λL

should i still use EA=Qenco...? if so then...

EA= E2πrL = 2λL/εo

(I think I was wrong in my first attempt because i multiplied it to the volume and not the area, λ is always linear density right? because in this question it's not clearly stated)

E = λ/πrεo

is this one correct now? or is there a different approach on how this is solved? :-ss

4. Jan 6, 2009

### Redbelly98

Staff Emeritus
Yes.
I think so, but I didn't really think about that and just assumed that it was the linear density. You're right, it should be clearly stated.

Looks good!

You're welcome.