Electrostatics - charges in stable equilibrium

  1. Mar 8, 2006 #1
    We have a charge distribution in which all the charges are in equilibrium due to electrostatic forces . Can we prove that none of these charges will be in stable equilibrium ?
     
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  3. Mar 8, 2006 #2

    Galileo

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    Yes, since the potential due to the other charges satisfies [itex]\nabla^2 V=0[/itex]. V is harmonic with has the property that is has no local maxima or minima.
     
  4. Mar 8, 2006 #3

    Meir Achuz

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    It is called Thomson's theorem.
     
  5. Mar 8, 2006 #4

    Galileo

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    Do you mean Earnshaw's theorem?
     
  6. Mar 8, 2006 #5

    Meir Achuz

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    Whoops, Galileo is right again. I meant Earnshaw.
    Thomson had several theorems (besides his home run), but not that one.
    Sorry, and thank you.
     
  7. Mar 8, 2006 #6
    Yes, since the potential due to the other charges satisfies [itex]\nabla^2 V=0[/itex]is being generated. Reload this page in a moment.. V is harmonic with has the property that is has no local maxima or minima.

    thats what i wanted to ask . how can we prove that [itex]\nabla^2 V=0[/itex]
     
  8. Mar 9, 2006 #7

    Galileo

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    It's just Maxwell's (first) equation: [itex]\vec \nabla \cdot \vec E =\rho/\epsilon_0[/itex], or [itex]\nabla^2 V=-\rho/\epsilon_0[/itex].
    In the region where there is no charge density you have [itex]\nabla^2 V=0[/itex].
     
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