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E-field and charge density outside two coaxial cylinders

  1. Feb 18, 2015 #1
    I am working on a problem of electrostatics, and I am having troubles in trying to figure out one part of it.

    1. The problem statement, all variables and given/known data

    It consists of an inner solid cylinder of radius ##a## with a voltage ##V_A##, and an outer coaxial cylindrical shell of inner radius ##b## and outer radius ##c## charged with a voltage ##V_B##. Find ##V(\mathbf{r})##, ##\mathbf{E}(\mathbf{r})## everywhere, and the surface change density at the surfaces of the conductors.

    2. Relevant equations
    Maxwell's equations

    3. The attempt at a solution
    I have calculated the potential field ##V## as well as the ##\mathbf{E}## field for ##r < a##, ##a < r < b##, and ##b < r < c##. I also calculated the surface charge density for the inner cylinder and the inner layer of the cylindrical shell. But can't figure out anything for ##c \leq r##.

    I tried using Gauss's law, taking the origin of potential at infinity (##V(\infty) = 0##), finding the ##\mathbf{E}(\mathbf{r})## field for ##c<r## and calculating ##\displaystyle V_B = \int_{\infty}^{c} \mathrm{d}V = -\int_{\infty}^{c} \mathbf{E}(\mathbf{r} )\mathrm{d}\mathbf{r}## in order to find the value of surface charge density at ##r = c##. Also, I tried using that outside the shell we can assume the space is dielectric, so as ##-\nabla ^{2} V = 0##. None of this yields reasonable values, or at least I am doing something wrong in the procedure.

    Procedure used:

    • Using a Gauss cylinder of radius ##r>c##, ##Q_{in} = Q = 2\pi c L \sigma_{c}##, where ##L## is the length of the cylinder.
    • Through Gauss's law, ##\Phi = E(r) Area = E \cdot 2 \pi r L = Q /\varepsilon_0 = \dfrac{2\pi c L \sigma_{c}}{\varepsilon_0}## ##\Rightarrow## ##\mathbf{E}(r) = \dfrac{\sigma_{c}}{\varepsilon_0}\dfrac{c}{r}\hat{r}##.
    • Finding ##\sigma_c##: ##\displaystyle \left. V_B = \int_{\infty}^{c} \mathrm{d}V = -\int_{\infty}^{c} \mathbf{E} \mathrm{d}\mathbf{r} = - \dfrac{\sigma_c}{\varepsilon_0} c \ln r \, \right|_{\, \infty}^{\, c} = \infty##, which doesn't make sense.
     
    Last edited: Feb 18, 2015
  2. jcsd
  3. Feb 19, 2015 #2

    BvU

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    Gold Member

    There is something with the potential from an infinitely long line charge that makes it impossible to choose V = 0 at infinity. You bumped into that. But you can still use Gauss law for a a cylindrical surface around the outer cylinder as you did. Just don't go as far as infinity.
     
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