# Coaxial metal cylinders

## Homework Statement

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Two long, coaxial metal cylinders are separated by a material of conductivity sigma and dielectric constant epsilon. The radius of the inner cylinder is a, the radius of outer cylinder is b, and the length of both is L.

Suppose that the inner conductor is held at a potential Vo with respect to the outer one. What is the current density J(r) between the two?

## Homework Equations

J(r) = I/A = sigma*E
C=Q/V
R=s/(A*sigma)

## The Attempt at a Solution

First attempt:
I found resistance of the dielectric to be ln(b/a)/(2pi*L*sigma). As a function of radius r, R(r) = ln(r/a)/(2pi*L*sigma).

Current = V/R = Vo / (ln(b/a)/(2piLsigma)) = Vo*2pi*L*sigma / ln(r/a) as a function of r.
Then divide by Area=2pi*L*r to get current density J(r) = Vo*sigma / (r*ln(r/a)).

Second attempt:
Vo = Q/C. I found capacitance of the configuration to be 2pi*epsilon_o*L*epsilon / ln(b/a) using the expression for resistance. Then Q = Vo*2pi*L*epsilon*epsilon_o / ln(b/a).
EA = Q/epsilon_o, so E = Vo*epsilon / (r*ln(b/a)). Using J(r) = E*sigma, I get J(r) = sigma*Vo*epsilon / rln(b/a).

The difference in the two is the presence of epsilon (dielectric constant) in the numerator, and b rather than r in the ln term of the denominator. Are either of these approaches correct?

rude man
Homework Helper
Gold Member
Can bound charges support current flow?
Better stick with method 1.

haruspex
Homework Helper
Gold Member
2020 Award
Current = V/R = Vo / (ln(b/a)/(2piLsigma))
Looks reasonable (but I've not checked it in detail).
= Vo*2pi*L*sigma / ln(r/a) as a function of r.
That I don't understand. The same total current has to flow through any given radius, no?

Ok, maybe instead I'll do J(r) = current (constant) / area (function of r). In this case I just get J(r) = V*sigma/(r*ln(b/a)). Does this sound right?

rude man
Homework Helper
Gold Member
Ok, maybe instead I'll do J(r) = current (constant) / area (function of r). In this case I just get J(r) = V*sigma/(r*ln(b/a)). Does this sound right?
Sounds right, looks right ... I'm buying!

haruspex