Consider the 3-torus which arises from taking $R^3$ and identifying two points $x\equiv x+nL$ whenever $n$ is a vector with integer components. I'm curious about finding the electrostatic potential resulting from some arrangement of charges in this space. It seems the correct answer should be the infinite sum:
$$\phi(x)=\sum_{n}\int_{[0,L]^3}\frac{\rho(y)}{4\pi||x-y-nL||}d^3y$$
Where we add up the contribution of charges within the box $[0,L]^3$ and also the image charges from infinite copies of the box displaced integer multiplies of the box length away.
Naively, this expression appears to diverge: the terms in the infinite sum diminish roughly as fast as the harmonic series as you add up the contributions from the image charges further and further away. My thinking is that if the total charge in the box is zero, as the image charges get further and further away their contributions to the potential end up cancelling more and more perfectly, so that the expression converges after all. Is this correct?
Can this be extended to the case where the total charge is non-zero? By the same reasoning, it seems the following expression should converge: $$\phi(x)=\sum_{n}\int_{[0,L]^3}\frac{\rho(y)-\rho_{avg}}{4\pi||x-y-nL||}d^3y$$ Where: $$\rho_{avg}=\frac{1}{L^3}\int_{[0,L]^3}\rho(y)dy$$ And the $\rho_{avg}$ should just shift the otherwise divergent expression by an infinite constant, so that it produces the same electric field.