# Potential of hydrogen atom solution of the Laplacian: Missing boundary condition to fix integration constant $c_1$

I have following problem,

I want to calculate the classical potential $$\phi(r)$$ of the hydrogen atom in its ground state.
The charge density is known: $$\rho(r)=\frac{-e_{0}}{\pi a^3}e^{-\frac{2r}{a}}$$ with $$a$$ being the Bohr radius, $$e_{0}$$ the elementary charge and $$r$$ being the distance from the hydrogen atom.

My approach: I use the Laplace equation in spherical coordinates leading to: $$\Delta\phi=-4\pi\rho(r)$$ $$\frac{\partial^2\phi}{\partial r^2} +\frac{2}{r}\frac{\partial\phi}{\partial r }=\frac{4e_{0}}{a^3} e^{-\frac{2r}{a}}$$

by using reduction of order the general solution is:

$$\phi(r)=\frac{e_{0}}{r}e^{-\frac{2r}{a}}+\frac{e_0}{a}e^{-\frac{2r}{a}}+\frac{c_1}{r}+c_2.$$ If I set $$\phi(\infty)=0$$ we obtain that $$c_2=0$$.
Whats bothering me is that I don't know what I should do with $$c_1$$.
My guess is that there may be a boundary condition I'm not aware of.
Any help would be appreciated.

Hint: By spherical symmetry the electric field $$E_r=-\frac{d\phi}{dr}$$ should vanish at $$r=0$$, which leads to $$c_1=-e_0$$.