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Homework Statement
Suppose the conductivity of the material separating two coaxial cylinders, of radius a and b (a < b) and held at a potential difference ##V##, would not be uniform. Specifically, ##\sigma(s) = k/s##, for some constant ##k##. Find the resistance between the cylinders. [Hint: Because ##\sigma## is a function of position, ##\nabla\cdot\vec{E}≠0##, the charge density is not zero in the resistive medium, and ##\vec{E}## does not go like ##1/s##. But we do know that for steady currents ##I## is the same across each cylindrical surface. Take it from there.]
[Image of the problem is attached below.]
Homework Equations
Ohm's law: ##\vec{J} = \sigma\vec{E}## where ##\sigma## is the conductivity.
Gauss' law: ##\nabla\cdot\vec{E} = \frac{\rho}{\epsilon_0}## where ##\rho## is the volume charge density.
Definition of the electric potential: ##V(a)  v(b) = \int_b^a \vec{E}\cdot\vec{dl}##
A handy product rule I used: ##\nabla\cdot(f\vec{A}) = f(\nabla\cdot\vec{A}) + \vec{A}\cdot(\nabla f)## where ##f## is a scalar function and ##\vec{A}## is a vector function.
For steady currents: ##\nabla\cdot\vec{J} = 0##
The Attempt at a Solution
I began by calculating the charge density in the conductive volume: [tex]\nabla\cdot\vec{E} = \nabla\cdot\vec{\frac{1}{\sigma}\vec{J}} = \frac{1}{\sigma}(\nabla\cdot\vec{J}) + \vec{J}\cdot(\nabla \frac{1}{\sigma}) = \vec{J}\cdot(\nabla \frac{s}{k}) = \frac{J}{k} = \frac{\rho}{\epsilon_0}[/tex] From this, ##\rho = \frac{J\epsilon_0}{k}##. Now, by symmetry, both ##\vec{J}## and ##\vec{E}## point only in the ##\hat{s}## direction  i.e. perpendicular to the axis of the cylinders. Thus, I thought about using Gauss' law with ##\lambda## being the line density in the inner cylinder: [tex]E2\pi sL = \frac{1}{\epsilon_0}\left(\lambda L + \frac{J\epsilon_0L}{k}(s^2a^2)\right)[/tex] After substituting in the above equation ##E = \frac{1}{\sigma} = \frac{s}{k}## and solving for ##J## I get: [tex]\vec{J} = \frac{k\lambda}{\pi\epsilon_0 (s^2+a^2)}\hat{s}[/tex] By definition: ##V = \int_b^a \vec{E}\cdot\vec{dl} = \int_a^b \vec{\frac{s}{k}\vec{J}}\cdot\vec{dl}##. From this, we can get [tex]\lambda = \frac{2\pi\epsilon_0 V}{ln[(a^2+b^2)/(2a^2)]}[/tex] Substituting, [tex]\vec{J} = \frac{2kV}{(s^2+a^2)ln[(a^2+b^2)/(2a^2)]}[/tex] I could go further and find ##I## from its definition via ##J## but I already see a problem: we assumed ##\nabla\cdot\vec{J} = 0## but it is clearly not the case for the above formula... Something clearly went wrong and I can't figure out what, the argument seems sound to me (obviously...).
Sorry if this is a bit long but I got utterly confused here... Any suggestion/help/comment will be greatly appreciated on this one!
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