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When charges are released on sphere, what is the shape made by charges?

Two charges are on opposite points of one diameter of the sphere.

Three charges make a shape of an equilateral triangle.

Four charges gives tetrahedron.

What shall five and more give?

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  • $\begingroup$ Related MO.SE question: mathoverflow.net/q/187063/13917 $\endgroup$
    – Qmechanic
    Oct 8 '12 at 15:20
  • $\begingroup$ @Qmechanic I'll just add that the main important result from that question was $N^2/2$ is the $1/r$ summation for uniformly distributed, infinite, points. I'm having fun with some of the links posted here, they do indeed limit to the N^2/2, but there are some interesting lower order terms, not to mention some deviations that can't be approximated with calculus approaches, which is fascinating. $\endgroup$ Oct 8 '12 at 15:58
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This problem with $N$ point charges on a sphere is a famous problem in electrostatics known as the Thomson problem. For large $N$, it is in general an open problem still under active research.

References:

  1. Wikipedia.org

  2. Mathworld.wolfram.com

  3. Mathpages.com

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This has been a problem since Thomson proposed the arrangement of electrons and positive charges (nucleus was not known at that time) in rigid electron shells of atom which is what called Plum-pudding model of atom. He suggested that electrons are arranged in a symmetrical pattern with respect to the center of sphere which is applicable only to smaller elements in periodic table (Old-timer wandered a lot after discovering the $e/m$ ratio).

While googling, I found this applet which generates some arbitrary patterns (up to 5000). I think there are many algorithms which can be used to solve these kind of patterns up to some finite value.

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  • $\begingroup$ It's not my downvote, but I can see some issues. If we're looking at a finite # of electrons, the electric field inside the sphere isn't zero perfectly. I don't know why you say the charges would stick to the surface of a non-conductor. I'm not sure what it would mean for a non-conductor to have a charge anyway... $\endgroup$ Oct 8 '12 at 16:02

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