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Can you add potentials if charge redistributes?

  1. Feb 11, 2014 #1
    Let say we have charged conductor M and we know its potential energy function Vm(r) when M is isolated from any charges. We also have charged conductor N with potential energy function Vn(r) when it is isolated.

    Now we put objects M and N close together, the charges on their surfaces redistribute. I am interested in potential energy at every point in space, can I still add potential energy functions (Vm + Vn) to find that?

    I'd like to think that it is possible to add functions like that, but I can't find a way of proving it mathematically yet. Any ideas of how to show this?
     
  2. jcsd
  3. Feb 11, 2014 #2
    No, you can't do that. When the charges redistribute, the associated potential functions change. The new potentials can be added but they will be different from the original potentials.
     
    Last edited: Feb 11, 2014
  4. Feb 12, 2014 #3
    Could you show that mathematically or find some kind of contradiction, because I am having difficulties finding easy-calculable one.
     
  5. Feb 12, 2014 #4

    Vanadium 50

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    If you move the charge distribution, the potentials created by those charges will change as well.
     
  6. Feb 12, 2014 #5

    AlephZero

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    The issue is not about mathematics, it's about physics. Of course if you assume the charge does not redistribute, you can use superposition. But if the charges are distributed over the surface of finite-sized conductors, that is probably a bad assumption.
     
  7. Feb 12, 2014 #6

    Dale

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    The contradiction is that if you simply add the isolated potentials you will wind up with a non-zero E field inside the conductor. This will lead to a current and therefore a redistribution of charges.
     
  8. Feb 12, 2014 #7

    UltrafastPED

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    The physics and the math for working with potentials and conductors can be found in Griffiths' "Introduction to Electrodynamics", chapters 2 and 3.

    The explanations for each of the comments above will be found there as well.
     
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