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Coaxial Cable/charge density

  • Thread starter critter
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  • #1
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1. A long coaxial cable consists of an inner cylindrical conductor with radius "a" and an outer coaxial cylinder with inner radius "b" and outer radius "c." The outer cylinder is mounded on insulating supports and has no net charge. The inner cylinder has a positive charge per unit length λ. Calculate the electric field (A) at any point between the cylinders a distance "r" from the axis and (B) at any point outside the outer cylinder. (C) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.



2. Parts A and B seem really simple, but maybe I'm looking at this wrong. I don't understand how the outer cylinder has any charge density since it was already stated that the outer cylinder has no net charge.



3. I think A and B might have the same answer: (2kλ)/r. This equation was given in the chapter summary in my textbook, but I somehow think they might be wanting a more extensive proof. I have no idea how to approach C since the information given in the problem seems contradictory.
 

Answers and Replies

  • #2
Doc Al
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I don't understand how the outer cylinder has any charge density since it was already stated that the outer cylinder has no net charge.
Just because the net charge is zero doesn't mean the surface charge densities are zero.
 
  • #3
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How would I solve for the density with what I'm given? Would they both be equal to λ?
 
  • #4
Doc Al
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How would I solve for the density with what I'm given?
Use Gauss's law.
Would they both be equal to λ?
In magnitude.
 
  • #5
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I think I'm starting to understand. There is some charge q enclosed between the two cylinders. The inner cylinder produces the electric field E=(2kλ)/r, and the total flux is equal to 0, so the outer cylinder must produce a field E=(-2kλ)/r. In that way, the magnitudes of the densities would be the same. Am I thinking about this correctly?
 
  • #6
Doc Al
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I think I'm starting to understand. There is some charge q enclosed between the two cylinders. The inner cylinder produces the electric field E=(2kλ)/r, and the total flux is equal to 0, so the outer cylinder must produce a field E=(-2kλ)/r. In that way, the magnitudes of the densities would be the same. Am I thinking about this correctly?
I'm not exactly sure what you are referring to when you say outer cylinder, since it has two surfaces. (I think you have it, but are expressing it unclearly.)

The inner cylinder (radius a) produces a field E=(2kλ)/r for r > a.

Since the field within the outer conducting cylinder (b < r < c) must be zero, Gauss's law tells you that the total charge enclosed must be zero. Thus the inner surface of the outer shell (at r = b) must have a charge equal and opposite to λ.
 

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