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Electrostatics Boundary Conditions (finding potential)

  1. Sep 21, 2014 #1
    1. The problem statement, all variables and given/known data

    I have a hollow, grounded, conducting sphere of radius R, inside of which is a point charge q lying distance a from the center, such that a<R. The problem claims, "There are no other charges besides q and what is needed on the sphere to satisfy the boundary condition".

    I have to calculate V and I'm having trouble coming up with the boundary conditions.

    2. Relevant equations

    \bigtriangledown ^2 V = \frac{- \rho}{ \epsilon }

    Uniqueness Thm: Knowing ρ and V at the surface means an answer I get satisfying Poisson's eq. and these conditions means it's the only one

    Properties of grounded conductors: V=0 on surface (others?)

    3. The attempt at a solution

    It's grounded, so I know V=0 along the entire surface of the sphere. I also thought I knew the charge density ρ since I have q within the volume V the sphere occupies. However, I'm thrown off by the statement "... and the charge on the sphere to satisfy the BCs". At first I assumed no charge on the surface but the line implies there should be some, so I think there must be some caveat of the setup I'm missing. If there is indeed charge on the sphere, does this affect ρ, since the sphere is infinitely thin I'm not sure if it would count as being 'within' the volume?

    Griffiths does a similar problem, except with the charge outside of the sphere; I attempted to adapt his solution to mine by adding an image charge outside of the sphere to at least calculate V inside of the shell, but I can't see how to prove that it equals zero everywhere on the surface as he did (he cited some ancient ratio equation or something...)

    Any tips? We just started doing things like this involving BCs with Laplace/Poisson's equation but this is my first exposure. I can't think of how the field/potential should behave either inside or outside because I can't decide how the charge, if there is any, affects my situation.
  2. jcsd
  3. Sep 21, 2014 #2


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    Welcome to PF!
    There will be some induced surface charge on the conductor because the conductor wants to remain grounded. When a point charge is present, the conductor has to place some charge on its surface so that the whole thing remains grounded.
    The volume is whatever you choose it to be. Don't worry about the infinitesimal surface of charge on the sphere, just know that ##V=0## there.
    That sounds like a good idea, but you don't have to prove it (in fact, I think Griffiths leaves it as a separate problem).
    Last edited: Sep 21, 2014
  4. Sep 21, 2014 #3

    For my outside solution, I tried using Laplace's eqn. coupled with: V=0 everywhere on the shell and rho = 0 everywhere outside of the shell. I got a solution of form ##V = -C/r + D## (because there is no angular dependence) and I'm assuming that the -C/r term comes from the point charge, so then D = kq/R. However I am not sure if I can completely assume this due to the charge induced on the sphere--for all I know it could equal -q in total and so V=0 everywhere outside of the shell. Any ideas how to look at it?

    edit: I think actually MVT implies that the potential is actually 0 everywhere outside of the shell, since I know it's the same for any angle and it has to average to the value at the center (which is also 0 by MVT applied to the shell)
  5. Sep 21, 2014 #4


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    ##V = -C/r + D## is correct, and you have ##V(R)=0##, but you are missing a boundary condition. Remember, you have to specify the values of ##V## on the entire boundary of the volume you are considering.

    EDIT: Don't assume ##C/r## is due to the point charge -- it's not. The point charge is "shielded" from the outside by the conductor.
    Last edited: Sep 21, 2014
  6. Sep 21, 2014 #5
    Is infinity considered the other bound of the volume outside of the sphere? I'm not sure if it should be zero or finite at infinity--both are satisfied anyways.
  7. Sep 21, 2014 #6


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    It's convention to set ##V(\infty)## to be 0.
  8. Sep 21, 2014 #7
    I thought so, so it all worked out--the potential is 0 everywhere outside of the sphere, and I was able to prove the image charge use by plugging in all points on the radius to show the terms cancel, so I believe I've solved it completely. Thank you very much for the help!
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