# Electric field in two concentric conducting cylinders

## Homework Statement

Two concentric cylindrical conducting shells of length L are separated by a vacuum. The inner shell has surface charge density +σ and radius ra. The outer shell has radius rb. Using Gauss’ Law, as a function of radius r find: The direction and magnitude of electric field inside and outside the shells. Be sure to clearly state the Gaussian surfaces that you are using!
Find an expression for the voltage between the shells.

## Homework Equations

Electric flux = ∫ EdA = qencl / ε0
σ = q/A

## The Attempt at a Solution

I'm using a cylindrical Gaussian surface. I understand that when r < ra the field E = 0. Here's what I've gotten so far for when r is between ra and rb:
E = qencl / ε0⋅Agauss = qencl / ε0⋅2πrl
⇒ E = σ2πral / ε0⋅2πrl
⇒ E = σra / rε0

That doesn't seem very... right. I don't think there should be a dependence on ra for starters. Would the expression for the flux through a Gaussian surface with r > rb be the same also, as there isn't any non-induced charge on it? Also, for the last part of the question I don't understand why there would be a voltage between the shells when the charge on the outer shell is just induced by the inner cylinder, ie the net charge on the outer cylinder is 0. Isn't it? Sorry if I'm not making much sense, I'm almost through my exams and my brain is extra melty. TSny
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Gold Member
⇒ E = σra / rε0

That doesn't seem very... right. I don't think there should be a dependence on ra for starters.
I don't see anything wrong. Does your result give the correct expression for E at the surface of the inner cylinder?
See item 3 here http://www.math.union.edu/~barbanej/IMP13S02/PhysicsNotes/Gauss's Law III.pdf

Would the expression for the flux through a Gaussian surface with r > rb be the same also, as there isn't any non-induced charge on it?
Yes. (The outer cylinder has no net charge.)

Also, for the last part of the question I don't understand why there would be a voltage between the shells.
If there is an electric field between the cylinders, then there will be a potential difference between the cylinders. How do you get ΔV from E?

I don't see anything wrong. Does your result give the correct expression for E at the surface of the inner cylinder?
See item 3 here http://www.math.union.edu/~barbanej/IMP13S02/PhysicsNotes/Gauss's[/PLAIN] [Broken] Law III.pdf

Yes. (The outer cylinder has no net charge.)

If there is an electric field between the cylinders, then there will be a potential difference between the cylinders. How do you get ΔV from E?

Ok, for the first part: when r = ra that leaves E = σ/ε0 which is correct - thank you for pointing out the sanity check :) Thanks for the link too - very helpful!

To get V from E, you multiply E by r (to get electric potential energy) and then divide by q (to get electric potential) - is that right? So since the outer cylinder is at zero potential, the potential difference is just the electric potential of the inner cylinder, which would be:
Er/q = V = σr / ε0q
= σr / ε0σ(2πrl )
= 1 / 2πlε0
Does that seem reasonable? The dimensional analysis seems to check out!

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TSny
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Gold Member
The electric field between the cylinders is not uniform, it varies with r. So, you will need to use some calculus.