Electric potential energy of charges localized to the surface of an object For a 3D object with a uniform density of surface charge, is there a particular shape which minimizes the total electrical potential energy? For instance, one could consider an object of fixed volume that could be deformed to assume shapes with different surface areas, with charge allowed to redistribute freely. Each geometry would present a different set of distances between points on the surface, and the potential energy could be represented by a continuous integral over all possible surface area elements. However, it may not be necessary to carry out such optimization; perhaps there is a more intuitive result based on symmetries.
 A: 
For instance, one could consider an object of fixed volume that could be deformed to assume shapes with different surface areas, with charge redistributing to maintain uniform charge density regardless of shape.

If you make a continuous deformation, the charges will readjust their distribution on the surface in order to minimize electrostatic energy. It might be that the distribution may not remain uniform. For example, for a spherical metallic surface, the charges will be uniformly distributed but not on a shape with pointed edges. For example, at the pointed edge (vertex) of a conical shaped metallic surface, will have higher charge density which follows from Poisson's equation $\nabla^2\phi=-\rho/\epsilon_0$. 
A: The answer to many of these minimization/extremization questions for solids and surfaces is either that the extremum does not exist (an infinitely stretched or folded object is the answer), or that the shape you are looking for is the sphere (or its surface). This is because since the statements of these problems obey rotational symmetry, one expects the shapes that obey this symmetry to play a special role in the spectrum. The only shape that is symmetric to all rotations is the sphere, so this is always the first guess for the potential minimum. 
However, in certain special cases where the minimized problem is nonlinear, it might happen that there is a whole subspace of potential-minimizing shapes that breaks the rotational symmetry. This would be called spontaneous symmetry breaking (the subspace would essentially be rotations of the same symmetry-breaking shape, note that the sphere is a spontaneous breaking of the translational symmetry of the problem and its arbitrarily positioned copies are such a family). On the other hand, various examples show us that the sphere (the "original symmetric groundstate") becomes a local maximum of such symmetry-breaking problems. So a good heuristic strategy is to take a look at the sphere, find a small parametrized deviation from it and the total potential corresponding to the parametric family, and if the sphere is at the potential minimum of this family, you are probably in the "usual", "non-broken" case. I have done this numerically in your case and the sphere does seem to be the minimum. However, if the sphere is a maximum, you have to bring the big guns and solve the full variational problem. 

The full variational problem would look something like this 
$$\Phi[\Sigma,\lambda] = \int_{\Sigma^2} \frac{\sigma}{|\vec{r} - \vec{r}'|} d \Sigma(\vec{r}) d \Sigma(\vec{r}') - \lambda \Psi[\Sigma]$$
where $\sigma$ is the constant charge distribution times some constants, and $\lambda$ is the Lagrange multiplier associated to the constraint $\Psi[\Sigma] = 0$. In the case of the total volume being constant and equal to $V_0$, this would be $\Psi[\Sigma] = V_{\rm encl}[\Sigma] - V_{0}$. The variational problem is then given by
$$\frac{\delta \Phi}{\delta \Sigma} = 0,\;\frac{\delta \Phi}{\delta \lambda} = 0$$
This is all awfully abstract. In concrete terms you would assume local coordinates on the surface $\xi,\eta$, and end up solving non-linear partial differential equations for the three two-variable function $\vec{r}(\xi,\eta)$. I will not go into this but I wanted to show you at least in principle.
