We know that if charges were to be placed within a conductor, they would start to rearrange themselves until they reach an electrostatic equilibrium where all charges are 'still' and no $E$-field is present within or on the surface of the conductor.

Would they still reach an electrostatic equilibrium if Coulomb's law was not an inverse square law?

My intuition tells me that this equilibrium is independent of the inverse square law, if we had a law that falls like $r^{3}$, same charges would still repel each other, and they would keep doing so until all the mutual repulsive forces cancel each other, i.e., until all charges reach a stationary position. I have the impression that this equilibrium is solely due to the mutual repulsion of same charges, regardless of how strong they repel each other.


1 Answer 1


Yes, indeed they would reach a similar equilibrium i.e. all particles will fly to the surface of the conductor.

In fact, this is motivated only by the fact that all the particles follow Newton's law which will push a system to states of lower potential energy. Because Coulomb's law results in lower potential energy when particles are farther apart, the configuration with lowest potential energy must be the one in which all of them go to the edges i.e. are the farthest apart.

It is unclear what the effect on the electric field would be if Coulomb's law were changed as in many formalisms, Coulomb's law is the result of combining Maxwell's equations and the Lorentz force, the former of which defines the E field. You would have to clarify specifically which of these two you change to discuss how the E field would change.

  • $\begingroup$ I think the problem is perfectly well defined if we just discuss alteration of the Coulomb force law, as the OP suggested, as we are just talking about position dependent forces on charge carriers here and about equilibrium. Any kind of electrodynamical considerations can be left out. I think it's a beautiful problem. I sat down for a while but couldn't show anything rigorously. I'm 50/50 between that the equilibrium is altered or unaltered. $\endgroup$
    – Georg E.
    Commented Sep 4, 2019 at 14:36

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