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Express electrostatic energy in terms of both charges

  1. Sep 2, 2011 #1
    1. The problem statement, all variables and given/known data
    For two concentric conducting spheres (radius a and b, b>a) that form a capacitor with charge q on the inner sphere and -q on the outer sphere, express the electrostatic energy in terms of q and -q and the potential difference between them.


    2. Relevant equations
    Gauss's Law, the equation for electrostatic potential, the equation for the energy stored in a static electric field.


    3. The attempt at a solution
    I have the field
    [tex]
    \vec{E}=\frac{q}{4\pi\epsilon_0 r^2}\hat{r}
    [/tex]
    between the conductors, but when I calculate the energy, should I only integrate between the spheres?
    [tex]
    W=\frac{\epsilon_0}{2}\int E^2 d\tau=\frac{q^2}{8\pi\epsilon_0}\int_{a}^{b}\frac{1}{r^2}dr=\frac{q^2}{8\pi\epsilon_0}\left( \frac{1}{a}-\frac{1}{b} \right)
    [/tex]

    Then to express it in terms of the original charges and the potential difference,
    [tex]
    \Delta\phi=\frac{q}{4\pi\epsilon_0}\left(\frac{1}{a}-\frac{1}{b}\right)
    [/tex]
    then
    [tex]
    W=\frac{q}{2}\Delta\phi
    [/tex]
    but how would I write this in terms of the charges? Does it want me to split it up like
    [tex]
    q=\frac{1}{2}(q-(-q))
    [/tex]
    and put this in the above equation?

    Thanks in advance,
     
  2. jcsd
  3. Sep 3, 2011 #2

    I like Serena

    User Avatar
    Homework Helper

    Isn't the electric field zero at every other point?

    You have calculated a 1-dimensional integral here.
    Shouldn't you integrate over 3-dimensional space?


    The phrasing of your problem implies that q and -q are equal and opposite.
    It's a bit weird that the problem asks for you to use q and -q, since "just" q should suffice.
     
  4. Sep 3, 2011 #3
    Yeah that seems reasonable, from Gauss' law.

    I integrated the phi and theta parts in the background, sorry I wasn't more explicit, the constants out front should already reflect those integrals being done.

    yeah, I don't know how to interpret the question about the q and -q, that's why I brought it here lol.
     
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