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Homework Help: Potential inside (and outside) a charged spherical shell

  1. Sep 28, 2008 #1
    [solved] Potential inside (and outside) a charged spherical shell

    1. The problem statement, all variables and given/known data
    Use the integral (i) to determine the potential V(x) both inside and outside a uniformly charge spherical surface, with total charge Q and radius R.


    2. Relevant equations
    (i) V([tex]\vec{x}[/tex]) = [tex]\frac{1}{4\pi\epsilon_{0}}\int\frac{\rho(\vec{x'})d^{3}x'}{|\vec{x}-\vec{x'}|}[/tex]
    [tex]\rho(\vec{x'}) = \frac{Q}{4\pi R^{2}}[/tex]


    3. The attempt at a solution
    if the x vector is along the z axis (can exploit spherical symmetry), then |x-x'| = [tex]\sqrt{x^{2} + R^{2} - 2xRcos\theta}[/tex]. (using cosine law)
    Then
    V([tex]\vec{x}[/tex]) = [tex]\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2}\int_{0}^{\pi}\frac{sin\theta d\theta}{\sqrt{x^{2} + R^{2} - 2xRcos\theta}}[/tex]

    V(x) = [tex]\frac{Q}{4\pi\epsilon_{0}x}[/tex]
    or
    V(r) = [tex]\frac{Q}{4\pi\epsilon_{0}r}[/tex]
    Which is correct for outside the spherical shell, but I can't figure out how |x - x'| would be any different (or what else should be different) for inside the spherical shell to get
    V(r) = [tex]\frac{Q}{4\pi\epsilon_{0}R}[/tex]

    I realize there are better/easier ways of doing this, but the question says to use that integral specifically...
     
  2. jcsd
  3. Sep 28, 2008 #2

    gabbagabbahey

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    Shouldn't you have:

    [tex]V(x)=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2} \int_0^{x} \int_{0}^{\pi }\frac{sin \theta ' dx' d \theta '}{\sqrt{(x')^{2} + R^{2} - 2(x')Rcos \theta '}}[/tex]

    I understand that the integration over [itex]\phi '[/itex] gave you [itex]2 \pi[/itex] , but what happened to the integration over [itex]x'[/itex]?
     
  4. Sep 28, 2008 #3
    Well [tex]d^{3}x'[/tex] turns into [tex]r^{2}sin\theta dr d\theta d\phi[/tex] and since r is constant at R (spherical shell) then an R^2 comes out of the integral and cancels the R^2 in the denominator from the charge density rho = Q / (4 pi R^2).

    I also figured out the problem, after integration:
    [tex]V(x) = \frac{Q}{4\pi \epsilon_{0}} \frac{1}{2xR} [\sqrt{(R + x)^{2}} - \sqrt{(R - x)^{2}}][/tex]
    and I forgot to consider the different cases for [tex]\sqrt{(R - x)^{2}}[/tex] when x > R (outside spherical shell) and x<R (inside).
     
    Last edited: Sep 28, 2008
  5. Sep 28, 2008 #4

    gabbagabbahey

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    Since your dealing with a surface charge, and not a volume charge, you should have:

    [tex] V(\vec{x})=\int \frac{\sigma (\vec{r'}) d^2 x'}{|\vec{x}-\vec{x'}|}[/tex]

    where [itex]\sigma (\vec{r'})= Q/4 \pi R^2 [/itex] is the surface charge density.
     
  6. Sep 28, 2008 #5
    Yes, you're right. The form I used was ambiguous, sorry about that!
    I guess I just jumped right to using [tex]R^{2}sin\theta' d\theta' d\phi[/tex] as the surface area element.
    I have solved the problem though, thanks for the assistance!
     
  7. Sep 28, 2008 #6

    gabbagabbahey

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    No problem, but I wasn't much assistance:smile:
     
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