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## Homework Statement

"An infnitely long hollow cylinder of radius ##a## has surface charge density ##σ_a##.

It is surrounded by a coaxial hollow cylinder of radius ##b## with charge density ##σ_b##. The charge densities are such that the total confguration is electrically neutral. Using whatever method you choose, calculate the work per unit length needed to assemble the system.

## Homework Equations

$$\oint \vec{E} \cdot d\vec{a} = \frac{Q_{enc}}{\epsilon _0}$$

$$W = \frac{\epsilon _0}{2} \int E^2 d \tau$$

## The Attempt at a Solution

First of all, I assume that the charge density is uniform, which I think is a good assumption but I'm not sure.

Then I find ##\vec{E}##:

When ##r<a, \vec{E} = 0## (by symmetry)

When ##r>a, \vec{E} = 0## (given)

When ##a<r<b##: Use Gauss' Law $$\oint \vec{E} \cdot d\vec{a} = \frac{Q_{enc}}{\epsilon _0}$$

As my surface, I use a cylinder of radius r and length L:

$$\oint \vec{E} \cdot d\vec{a} = E2 \pi rL$$

and

$$Q_{enc} = \sigma_a 2 \pi a L$$

so

$$E2 \pi rL = \frac{\sigma_a 2 \pi a L}{\epsilon_0}$$

so

$$\vec{E} = \frac{\sigma_a a}{r\epsilon_0} \hat{r}$$

Now here's where I start to get into trouble. To find the energy I would use the equation for work above, with ##d\tau =rdrd\theta dz##, but how do I turn this into an energy per unit length? Integrating the ##dz## over all space gives an infinity obviously.

What I tried to do was just divide the whole thing by L and simply ignore the z differential, which I have no idea if I can do:

$$\frac{W}{L} = \frac{\epsilon_0}{2L} \int_{a}^{b} \int_{0}^{2\pi} (\frac{\sigma_a a}{r \epsilon_0})^2 rdrd\theta$$

Which leads to: $$\frac{W}{L} = \frac{\pi \sigma_a^2 a^2}{L \epsilon_0} ln(\frac{b}{a})$$

I'm not sure if this makes sense. Is it allowed for me to just ignore the z differential like that? I don't really see a way around it going to infinity.