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Electrostatics ofa shell question

  1. Feb 24, 2010 #1
    1. The problem statement, all variables and given/known data
    A spherical shell with potential:
    [tex]V(r=R)=V_0cos\theta[/tex]
    (a) Please solve the potential inside and outside the shell.
    (b) if there is a point charge at the center and the potential of the shell keeps the same, solve the potential inside and outside.

    2. Relevant equations
    Laplace equation:
    [tex]\nabla^2V=0[/tex]

    3. The attempt at a solution
    Part a) I can do. Using the separation of variables in spherical coordinates I found that:
    [tex]\frac{V_0 r}{R} cos\theta ; r\leq R[/tex]
    and
    [tex]\frac{V_0 R^2}{r^2} cos\theta ; r\seq R[/tex]

    Part b) I'm having some trouble.
    I think we should apply the uniqueness theorem since the potential is specified on the boundary (the sphere) but not sure how. Does that mean that the potential is the same as before inside the sphere? I was told that this is not the case by upperclassman who can't tell me why. They only know the answer is:
    [tex]V(r,\theta)=\frac{V_0}{R} cos \theta +\frac{q}{4 \pi \epsilon_0}(\frac{1}{r}-\frac{1}{R})[/tex]
    I think that this is indeed the right answer since it satisfy the boundary condition, but I want to know why. It looks like that the first term accounts for the charge distribution on the sphere, but if that is the case shouldn't the second term be the point charge in ther center namely:
    [tex]\frac{q}{4 \pi \epsilon_0}\frac{1}{r}[/tex]
    instead?
    THe outside is even more tricky. Can I even apply the uniqueness theorem? The boundary where the potential is defined doesn't enclose the area outside the sphere. What can I do?
     
  2. jcsd
  3. Feb 24, 2010 #2

    gabbagabbahey

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    Gold Member


    No, the potential will be the same outside the sphere; it still satisfies Laplace's equation, and has the same values at its boundaries, [itex]r=R,\infty[/itex] , as before.

    However, inside the sphere there is now a point charge and so the potential in that region doesn't satisfy Laplace's equation, but rather Poisson's equation corresponding to a point charge at the origin:

    [tex]\nabla^2 V_{in}=-q\frac{\delta^3(\textbf{r})}{\epsilon_0}[/tex]


    You can use the superposition principle here and say that the potential is a sum of two terms; one, [itex]V_1[/itex], that satisfies Laplace's equation, is finite at the origin, and takes on the value [itex]V_0\cos\theta[/itex] at [itex]r=R[/itex]; and a second, [itex]V_2[/itex], which satisfies Poisson's equation corresponding to a point charge at the origin, and takes on the value of zero at [itex]r=R[/itex] (so that the total potential takes on the correct value there). That way, you have

    [tex]\nabla^2 V_{in}=\nabla^2 V_1+\nabla^2 V_2=-q\frac{\delta^3(\textbf{r})}{\epsilon_0}\;\;\;\text{and}\;\;\; V(r=R)=V_0\cos\theta[/tex]

    as required.

    The uniqueness theorem tells you that [itex]V_1[/itex] is exactly the same as the inside potential from part (a), while [itex]V_2[/itex] must be the potential due to a point charge at the origin, with reference point set at [itex]r=R[/itex] rather than infinity (so that [itex]V_2(r=R)=0[/itex] as required).
     
  4. Feb 24, 2010 #3
    Thanks I think I got it now. Just to make sure, so outside the sphere is the same as before because of the uniqueness theorem? I thought it only applies to the potential inside a volume bounded the a boundary where the boundary conditions are defined (inside the sphere). Here, are we treating infinity as the boundary too? If so then can I argue that no matter what I put inside the sphere won't affect the potential outside as long as the BC remain unchanged? But if I put some charge outside the sphere the external potential will be changed. Am I correct?
     
  5. Feb 24, 2010 #4

    gabbagabbahey

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    Yes, that's correct. :approve: Of course, if you allowed the potential at [itex]r=R[/itex] to change as you added charge inside the shell (as it normally would for most shell materials), then one of the boundary conditions would change and hence so would the potential outside.
     
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