# Cylindrical Surface Charge Density

1. Jul 14, 2010

### dinnsdale

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

The figure shows a portion of an infinitely long, concentric cable in cross section. The inner conductor carries a charge of 6 nC/m and the outer conductor is uncharged.

(part 5 of 6)
What is the surface charge density inside the hollow cylinder?

Knowns:

Central solid conducting cylinder of radius 0.013 m and charge 6 nC/m.

Inner radius of cylindrical shell = 0.0475 m

Outer radius of cylindrical shell = 0.067 m

Charge of shell = 0 nC/m

Qenc = Q1 + Qin

2. Relevant equations

(sigma) = Q/A

A=2(pi)rL

Flux = EA = Qenc/E0

3. The attempt at a solution

I constructed a Gaussian Cylinder between the outer and inner parts of the shell. I know that E inside the shell is zero (since it is a conductor), so Qenc/E0 = 0, which means Qenc = 0.

However, Qenc = Q1 + Qin, where Qin is the charge on the inside surface of the shell.

The charge density (sigma) relies on Qin, and (sigma) = Qin/A

However, A = 2(pi)rL, and L is not given; rather, L is infinite! I know r = 0.0475 m. Help!

2. Jul 14, 2010

### bp_psy

If you know the charge per unit length of the inner conductor, you can use any L you like to solve for the area.The simplest L you can chose is obvious.

3. Jul 14, 2010

### dinnsdale

I thought so, too, so I tried L = 1 m.

(sigma) = -6 nC / 2(pi)rL m^2 = -3 / (pi)x0.0475x1 m^2 = -20.1038 nC/m^2

However, that's apparently incorrect.

Unless I'm missing something...

4. Jul 15, 2010

### hikaru1221

Qin is not 6nC The problem didn't give you any value of the charge; it only gave you the surface charge density of the inner shell 6 nC/m.

5. Jul 15, 2010

### dinnsdale

All right... I did notice that and was a bit confused as to how to use that information. I suppose that means there's a way to get the charge Q given (lambda) (where (lambda) = Q/L ). To me it seems that Q = 6 nC, if I'm to take an arbitrary L of 1 m, so Qin = -6 nC. But when I do that calculation (as above, -6 nC/2(pi)(Rin)(L) ), I get the wrong answer.

I appreciate all the help so far, by the way! I just feel at a loss in this class; we're moving very quickly.

6. Jul 15, 2010

### hikaru1221

Okay, I see. I got the same answer as you too. I think there is nothing wrong here, so you'd better check the given answer.

7. Jul 15, 2010

### dinnsdale

I went to my school's tutoring center, and they said everything looked correct, as well.

They helped me see more clearly that length doesn't matter. Equating (sigma)(area) = (lambda)(length) removes the length and leaves (sigma) = (lambda)/2(pi)r^2

-20.103 is incorrect, though; the computer rejects it as wrong. I can't check the actual answer until the due date for the entire assignment has passed. Any more insights would be helpful (am I misinterpreting "inside"?); otherwise, once I know the answer I'll post it here with the explanation given.

8. Jul 15, 2010

### bp_psy

If it's a mastering physics assignment look carefully at what units and decimals it wants.

9. Jul 15, 2010

### dinnsdale

I'm sure I have the units correct. It wants nC/m2; I'm given 6 nC/m on the inner cylinder.

This makes me wonder if, perhaps, I've been calculating sigma on the inner cylinder, rather than the inner surface of the outer cylinder. Still, for the latter I'd need the charge of the inner cylinder, and though I believe I have sigma (and am given lambda) of the inner cylinder, I have no way to convert that into the charge. Is there a way I'm unaware of, or overlooking?

10. Jul 15, 2010

### lightgrav

I would bet, since the diameter of the inner conductor is given (and it is not small)
that its "6 nC/m" is supposed to be "6 nC/m^2" .

By the way, since lambda * L = sigma * 2pi r L , the L cancels.

11. Jul 15, 2010

### dinnsdale

I went to the tutoring center at my school. Twice. Finally was able to see the professor in the late afternoon. He checked the problem; it's got the wrong solution attached. Thank you everyone, you helped a lot; I was doing it right, -20.104 is the answer, but the site was wrong. It's a relief, to be sure, but talk about frustrating and confusing!

Anyway. Thanks again. =)