metal die charged, where do the charges go? I'm puzzled by this question: I have a metal die (a regular cube, completely filled with homogeneous iron for instance); I charge it with some µCoulombs. Where do the charges go?
Can the faces of the die have some charge?
Or do they go to the eight corners?
 A: They accumulate in the corners. 
Free excess charges (electrons) in a conductor (metals) will move freely towards the surface until the internal electric field is zero. If this was not the case, the non-balanced field will further move them until this is the case.
But the geometry of the dice makes the faces very unstable place for charges to accumulate, since field components parallel to the faces are not balanced with stability.
So they will end up accumulating in edges where non balanced field along them will push them towards the corners equally distributed.
I made this simple graph, it should look similar to this...


Expanding on why charges don't populate the faces...
Imagine 6 equal charges centered on the faces of the die, because of symmetry conditions they will be static (assuming the surface has a much larger potential and electrons cannot escape). The problem with these electrons in this positions is that is a very unstable configuration: if you move one of them by a tiny bit, all will be unbalanced and fall to the corners. However, in the corners, having three surface potentials holding them, there is no other place in the cube to fall back to, and small displacements of the electrons will not unbalance the configuration as a whole.
A: I created a discrete model of 56 equal charges distributed on the faces of a cube of edge length 3 as shown below

the parameter I can change is d. I computed a value proportional to the electrostatic potential energy of the distribution of charges as a function of d. The result is as follows

So the charges do repel each other slighty toward the corners.
