electrostatics - How would charge be distributed in charged conductors if the Coulomb law was not ${1}/{r^2}$?

Would the excess charge on a conductor move to surface until the electric field inside become zero if the Coulomb law was for example $\frac{1}{r^3}$? If yes, would the distribution $\sigma(x,y)$ be different from when it is $\frac{1}{r^2}$?

Answer

Suggestion to the question (v3): Generalize the question to a $1/r^s$ potential law in $n$ spatial dimensions! Then according to Henry Cohn's mathoverflow answer here, the charges rush to the boundary iff $s\leq n-2$. So in OP's example $(s=2,n=3)$, the charges don't rush to the boundary, in contrast to the real world $(s=1,n=3)$.