Potential Energies of Two Charged Cylinders

[Matt Chu]

Homework Statement

Problem 1.24 (this is unimportant; it's just a different way of calculating the potential energy of a solid cylinder) gives one way of calculating the energy per unit length stored in a solid cylinder with radius a and uniform volume charge density $\rho$. Calculate the energy here by using $U = \frac{\epsilon_0}{2} \int_{entire \ surface} E^2 dv$ to find the total energy per unit length stored in the electric field. Don’t forget to include the field inside the cylinder.

You will find that the energy is infinite, so instead calculate the energy relative to the configuration where all the charge is initially distributed uniformly over a hollow cylinder with large radius $R$. (The field outside radius $R$ is the same in both configurations, so it can be ignored when calculating the relative energy.) In terms of the total charge $\lambda$ per unit length in the final cylinder, show that the energy per unit length can be written as $\frac{\lambda^2}{4\pi\epsilon_0}\left(1/4+ln(R/a)\right)$

(It's important to note that the potential energy involved in this problem is NOT the potential energy of a particle in the field created by the charged cylinders, but the potential energy of the charged cylinders themselves.)

Homework Equations

$U = \frac{\epsilon_0}{2} \int_{entire \ surface} E^2 dv$

The Attempt at a Solution

The first part of the problem, involving solving for the energy of a solid cylinder, is pretty simple. For this, I got that the potential energy density per length of the inside of the cylinder being:

$U_{in}/h = \frac{\pi \rho^2 R^4}{16 \epsilon_0}$

The external potential energy of the cylinder goes to infinity, as the problem states, as you get:

$U_{out}/h = \left. \frac{\pi \rho^2 R^4}{4 \epsilon_0} \ln(r) \right|_R^\infty$

The second part of the problem, though, is confusing to me. It would seem that the potential energy of the hollow cylinder is zero, because the electric field inside is zero. In addition, the problem says that you can "ignore" the field outside $R$, but I'm not sure how exactly that can be possible if the total potential energy is the sum of the potential energies of the internal and external areas.

I thought that maybe I could calculate the potential energy inside the hollow cylinder by doing it piece by piece, i.e. with differential areas rather than with the equation given, but logically it should still be zero as well.

 

[kuruman]

It would seem that the potential energy of the hollow cylinder is zero, because the electric field inside is zero.

It would seem that way. However, it takes some energy to assemble all those charges on the surface of a cylinder of radius R in a region of space where there was nothing initially.