Given an n-dimensional electrically neutral, solid metal ball (a point for n=0; a rod, n=1; a disc, n=2; a solid ball, n=3; ...), place N=(n+1)! identical ions on the ball. As one of my favorite physics professors used to say, forget the mathematics, intuitively I expect the ions to equilibrate to the vertices of an n-D permutohedron inscribed in the n-ball (e.g., a hexagon inscribed in the circumference of a disc with 6 ions and a truncated octahedron for n=3 ). Similarly with N= [2(n+1)]! /[(n+1)! (n+2)!] (the Catalan numbers), I expect to see an associahedron (a Stasheff polytope, e.g., a pentagon for the disc with 5 ions).

Is my intuition correct? Has anyone seen this worked out mathematically, as an extremum problem? An electrostatics simulation for n=3 would be interesting also. (See OEIS A019538 for refined f-vectors of permutohedra / permutahedra and A133437 for associahedra.)

After reviewing the very interesting, proposed answers, I think I need to clarify and qualify my question: I initially was thinking of a real physical situation, a conductive metal ball in 3 dimensions or disc in 2 with Coulomb's inverse-square law applying. Then, if I remember my physics correctly, free charges must always reside on the suface with the potential vanishing in the interior of the ball and with the gradient of the potential at the surface being normal to the surface in the equilibrium state. The number of ions for the 3-D ball is constrained as stated above to either 4!=24 for the 3-D permutahedron or 8!/(4! 5!) = 14 for the 3-D associahedron. Given either number, how will the ions distribute themselves on the surface? Whether one should or can sensibly modify the repulsion for higher dimensions n and whether analogous surface (n-1) physics and equilibrium topology would occur are the next logical considerations; however, the number of "ions" would still be restricted to the number of vertices of either the n-D permutohedron or associahedron.