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Potential of a metal sphere with changing radius

  1. Apr 29, 2008 #1
    1. The problem statement, all variables and given/known data

    An inflatable metal balloon assumed to be spherical with radius R is charged to a potential of 1000 V. After all the wires and batteries are disconnected, the balloon is inflated to a new radius 2R. Does the potential of the balloon change as it is inflated? If so, by what factor? If not, why not?


    2. Relevant equations

    V(pt. charge) = kQ/R


    3. The attempt at a solution

    I think the answer should be that yes, V does change by a factor of 1/2 since R increases by 2 and V is proportional to 1/R. However, I also want to think the potential is infinite at a point on the sphere. I think I understand that we can treat the sphere as a point charge, but what I don't understand is what happens when a charged particle is on the sphere. Why doesn't potential go to infinity? It seems that since the distance between some bit charge dQ of the sphere and the test charge is 0 this would blow up to infinity. I'm probably over-thinking the question but I seem to have dug myself into a hole of thorough confusion. Can someone help explain this to me? Any help is appreciated!
     
  2. jcsd
  3. Apr 29, 2008 #2

    Hootenanny

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    Are we talking about the potential relative to a point inside the sphere or outside the sphere?
     
  4. Apr 29, 2008 #3
    I'm assuming when they say 1000V that's at point a point on R relative to infinity. So outside.
     
  5. Apr 29, 2008 #4

    Hootenanny

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    Then you are correct, for a distance r>R you can treat the spherical shell as a point charge located at the centre of the sphere. For a point inside the spherical shell (r<R) the potential can be found to be constant by Gauss' law. At the boundary, the potential is taken to be the constant value inside the sphere in order to ensure that the potential function is piecewise smooth. See the image below,

    [​IMG]
    Image taken from HyperPhysics
     
  6. Apr 29, 2008 #5
    Cool, thanks for your help, I really appreciate it! Those pictures were very useful! I'm still confused though. Why the function has to be piecewise smooth?
     
  7. Apr 29, 2008 #6

    Hootenanny

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    Note that the electric field can be expressed as the gradient of the potential,

    [tex]\underline{E} = \nabla V[/tex]

    Therefore, the potential must be continuously differentiable (at least once) in order to be physically meaningful, i.e. in order to associate an electric field with the potential, we must be able to differentiate it at least once. Therefore, the potential must be [piecewise] smooth.
     
  8. Apr 29, 2008 #7
    Alright, I think that makes sense. Thanks again for all your help!
     
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