Cylindrical Surface Charge Density

In summary, the surface charge density inside the hollow cylinder is calculated to be -20.103 nC/m^2.
  • #1
dinnsdale
6
0

Homework Statement



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.

PhysImage.jpg


(part 5 of 6)
What is the surface charge density inside the hollow cylinder?
Answer in units of nC/m^2.

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

Homework Equations



(sigma) = Q/A

A=2(pi)rL

Flux = EA = Qenc/E0

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!
 
Physics news on Phys.org
  • #2
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
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
Qin is not 6nC :smile: 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
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
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
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
If it's a mastering physics assignment look carefully at what units and decimals it wants.
 
  • #9
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
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
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. =)
 

1. What is cylindrical surface charge density?

Cylindrical surface charge density is a measure of the amount of electric charge per unit area on the surface of a cylindrical object. It is commonly denoted by the symbol σ and has units of coulombs per square meter (C/m²).

2. How is cylindrical surface charge density calculated?

Cylindrical surface charge density can be calculated by dividing the total charge on the surface of a cylinder by the surface area. The equation for this is σ = Q/A, where Q is the charge and A is the surface area.

3. What is the difference between surface charge density and volume charge density?

Surface charge density is a measure of charge per unit area on the surface of an object, while volume charge density is a measure of charge per unit volume within an object. In other words, surface charge density looks at the charge on the surface, while volume charge density looks at the charge within the object.

4. How does cylindrical surface charge density affect electric fields?

Cylindrical surface charge density plays a role in determining the strength of the electric field around a charged cylinder. The electric field is directly proportional to the surface charge density, so a higher surface charge density will result in a stronger electric field.

5. Can cylindrical surface charge density be negative?

Yes, cylindrical surface charge density can be negative. This occurs when the charge on the surface of a cylinder is negative, indicating an excess of electrons. However, in most cases, cylindrical surface charge density is positive as most objects have a net positive charge on their surface.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
449
  • Introductory Physics Homework Help
Replies
26
Views
476
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
232
  • Introductory Physics Homework Help
Replies
11
Views
309
  • Introductory Physics Homework Help
Replies
15
Views
3K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
6K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Back
Top