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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.

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

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