- 2

- 0

The published solution of the problem states that by symmetry each will end up with the same kinetic energy when the charges have moved to infinity. So the procedure is to calculate the potential energy of the configuration, divide that by 3, set that equal to one-half m v squared, and solve for v, where v is the speed of one charge.

I am having trouble seeing the notion of “symmetry” here since the charges are not equidistant from each other. I also pose the argument that what if a fourth charge were to be included, one that would be orders of magnitude away from the original 3 and would consequently experience far less force due to the other particles. Would the individual charges each still have 1/4 the potential energy of the assembly of four? The only rationale I can muster is that the symmetry is not local symmetry but symmetry at infinity, and I have difficulty picturing that. Hope you can help me sort this out.