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Energy per unit length of a cylindrical shell of charge

  1. Oct 16, 2016 #1
    1. The problem statement, all variables and given/known data
    "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.

    2. Relevant equations
    $$\oint \vec{E} \cdot d\vec{a} = \frac{Q_{enc}}{\epsilon _0}$$
    $$W = \frac{\epsilon _0}{2} \int E^2 d \tau$$

    3. 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.
     
  2. jcsd
  3. Oct 16, 2016 #2

    haruspex

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    I don't know if there is an easier way, but I supposed that at some time we have a charge +Q on some length of the inner cylinder and a charge -Q on the outer cylinder. Now consider moving ΔQ from the outer to the inner. We know the potential at each radius due to each cylinder, so we can compute the work done.
     
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