I've been trying to find the scalar potential that would correspond to the charge density of a ground state hydrogen atom. The result is known, and the inverse of my problem can be found e.g. in Jackson's electrodynamics problem 1.5 or many questions here on this site.
The problem asks you to find the charge density that corresponds to the following potential: $$ \Phi(r) = \frac{q \exp{(-\alpha r)}}{4 \pi \epsilon _0 r}\left(1+\frac{\alpha r}{2}\right) .$$
As far as I can tell, according to Poisson's equation you basically just need to get the Laplacian of this potential. This is not hard to do, and the result is what you would expect:
$$ \rho(r) = \frac{-q\alpha^3}{8\pi}\exp{(-\alpha r)}. $$
- My question is the inverse: given $\rho(r)$, how would you find $\Phi(r)$?
The most obvious approach that I had is to use the integral you get from Coulomb's law:
$$ \Phi(r) = \int \frac{\rho(r')}{|r-r'|}d^3r' $$
However, I have not been able to solve the integral by hand and Mathematica can not tell me the result either. My guess is that this potential never goes to zero, so the direct integration is not feasible? If so, then how else would you go about solving this problem?
(My next task would be to solve the same thing for a Gaussian density, for which I can again find the result on Wikipedia. Is that problem easier or harder than this one?)