# Resistance between coaxial cylinders

1. Mar 3, 2006

### Reshma

Two coaxial cylinders, inner radius a and outer radius b are separated by a material of conductivity given by $\sigma (r) = k/r$ for some constant 'k'. Find the resistance between the cylinders.

Here the conductivity is a function of position and the charge density is not zero in the resistive medium, and the electric field $\vec E$ does not go as 1/r. So how is this problem solved?

2. Mar 3, 2006

### Meir Achuz

Assuming steady state, div j=0 so j=j_0/r and E=j/\sigma=constant.

3. Mar 4, 2006

### Reshma

The charge density is not zero here, so $\vec \nabla \cdot \vec E = {1\over \sigma}\vec \nabla \cdot \vec J = 0$ will not hold good.

4. Mar 4, 2006

### Meir Achuz

Div E is NOT zero. Div j is zero because rho is constant in time.
$\vec \nabla \cdot \vec E = \vec \nabla \cdot [\vec J/\sigma] ,$
but Div j does =0.
Do you want to argue or to do the problem?

Last edited: Mar 4, 2006
5. Mar 5, 2006

### Reshma

Ok, here goes...
I need to evaluate the current and the voltage in order to find the resistance.
$$\vec \nabla \cdot \vec E = \vec \nabla \cdot [\vec J/\sigma]$$

Integrating over a volume dV.
$$\int \vec \nabla \cdot \vec E dV= \int \vec \nabla \cdot [\vec J/\sigma] dV$$

By Divergence theorem,

$$\int \vec E \cdot d\vec a = \int (\vec J /\sigma)\cdot d\vec a$$

$${Q_{enc}\over \epsilon_0} = \int (\vec J /\sigma)\cdot d\vec a$$

I'm getting stuck here. I don't know how to proceed.

Anyway, I found the current.
But by Ohm's law, $\vec J = \sigma \vec E$
Current I is given by;
$$I = \int \vec J \cdot d\vec a = \int {k\over r}\vec E \cdot d\vec a$$

But, I need $\vec E$ to compute the voltage. How do I do that?

Last edited: Mar 5, 2006
6. Mar 5, 2006

7. Mar 6, 2006

### Reshma

$$\vec \nabla \cdot \vec J = 0$$
$$J = {J_0\over r}$$
$$E = {J\over \sigma} = {J_0\over r} {r\over k} = {J_0\over k}$$

$$I_{enc} = \int \vec J \cdot d\vec a = \int_a^b {J_0\over r} 2\pi r dr = J_0 2\pi (b-a)$$

$$V_a - V_b = -\int_b^a \vec E \cdot d\vec r = {J_0\over k}(b-a)$$

Resistance R will be:
$$R = \frac{V_a - V_b}{I} = \left({J_0\over k}(b-a)\right)\left({1\over (J_0 2\pi (b-a)}\right)$$
$$R = {1\over 2\pi k}$$