# Find charge field within coaxial cables

1. Apr 15, 2013

### physninj

1. The problem statement, all variables and given/known data
two very long coaxial cylinders. Inner has radius a and is solid with charge per unit length of λ. Volume is also uniform but not defined by a parameter. Outer is hollow with inner radius b and outer radius c. Outer cylinder is a conductor with charge per unit length of -2λ. Find E(r) for all r in terms of given parameters. you can also view attached picture.

2. Relevant equations
∫E.dA=qenclosed0

3. The attempt at a solution
I'm hung up right away on what to do with the inner radius from 0->a Ive got the left side of the equation I think, but I don't know how to get qenclosed in terms of r, if it needs to be. I expect to integrate on that part. Anywho here's where I'm at.

E2∏rh=(1/ε0)∫dq

I've got λ=dq/dh where h represents the length of the cable being considered. All the examples seem to use charge density rather than charge per unit length, using this gives me:

E2∏rh=(1/ε0)∫λdh which is not in terms of r. I suppose I could still integrate it and cancel h on both sides, let me know if that's wrong.

E2∏rh=(λh/ε0)→E2∏r=(λ/ε0)

E(r)=λ/(2∏r*ε0)

Is this wrong for any reason? It doesn't feel right to not use the radius on the right side.

#### Attached Files:

• ###### cylinders.JPG
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2. Apr 15, 2013

### domenico

Let's consider the cable's length finite, L. The total charge of the inner cylinder would be Q = λL. Applying Gauss' theorem ∫E*dA = λL/ε°, since E is uniform and perpendicular to dA: E∫dA = λL/ε° <=> E(2πrL) = λL/ε° <=> E = λ/2πε°r.
The units on both sides are also ok, so your formula is right.

Last edited: Apr 15, 2013