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Earnshaw's Theorem: Why Can't I Contain a Charged Particle Electrostatically?

  1. Jul 3, 2010 #1
    I'll be honest, I've never really understood the "why" of Earnshaw's Theorem. Why can't a charged particle be held in equalibrium with only electrostatic forces? If I have six charges arranged in at the corners of a hexagon and put a charge at the center, why can't it just stand there since the electric field is zero at that point? What is the reasoning/justification for this theorem? Does it apply to gravitostatic forces as well?
     
  2. jcsd
  3. Jul 3, 2010 #2

    Doc Al

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    The particle would not be in stable equilibrium.
    The simplest way to get stable equilibrium is to arrange it so that the field always points inward toward the particle. Then, any slight deviation from its position will push it back. But that would imply a nonzero divergence of the field in free space, contrary to Gauss's law.
    Yes. It applies to any inverse square force law.
     
  4. Jul 3, 2010 #3
    As I see it you will need extra forces to hold the six charges at the corners of your hexagon and this goes beyond the scope of Earnshaw because as I understand it the theorem applies to the whole system of charges and to electrostatic forces only.
     
  5. Jul 3, 2010 #4

    Doc Al

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    It certainly applies to the complete set of charges, but it also applies to an individual charge held in equilibrium by electrostatic fields only.
     
  6. Jul 3, 2010 #5
    Isn't this an application of the hairy ball theorem?
     
  7. Jul 3, 2010 #6
    Thanks for the clarification Doc Al.
     
  8. Jul 3, 2010 #7
    Perhaps I'm a little slow but, I'm still not getting it... stable or unstable equilibrium point, its still an equilibrium point...which means that if I put a charge in that exact position it should just stay there. Suppose I simplify this a bit. Imagine I have a triangular pool ball rack with three positively charged pool balls in the corners. If I take another positively charged pool ball and put it in the exact center, I would think that the system could just sit there like that. The total electric force on the middle ball should be zero since the vector sum of the electric fields from the corner balls is zero, but Earnshaw's Theorem says my thinking is wrong, why? Since its true for gravity too, why can't I just stick a satillite at a Lagrange point for the Earth/Moon/Sun system and have it just stay there? Does it have something to do with the potential?
     
  9. Jul 3, 2010 #8
    That, precisely is the error.

    You cannot balance a pencil on it's tip.
     
  10. Jul 3, 2010 #9
    In post #2 everything is written very shortly Doc Al.
    The first thing that only stable equilibrium state is considered.
    The nonstable equilibrium state there of course exists.
    May be you could find a situation when a stable equilibrium state exist?
    That is when you slightly push a charge from this stable equilibrium state there appears
    a force returning to this stable point. Of course this stable point is without charges.
     
    Last edited: Jul 3, 2010
  11. Jul 3, 2010 #10

    Doc Al

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    You misunderstand Earnshaw's theorem, which prohibits stable electrostatic equilibrium. Just plain old equilibrium--zero net force--is trivially possible.
     
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