 Solution

For the potential of an inverse-radius spherical charge density, assume there is no additional point charge at the origin and choose $\varphi (\infty )=0$. The enclosed charge is $Q(r)=4\pi A{\int }_0^r{r}^{\prime }d{r}^{\prime }=2\pi A{r}^2$ for $r<a$, and $Q(a)=2\pi A{a}^2$ outside. Gauss's law then givesIntegrating the field inward from infinity,The potential and its first radial derivative are continuous at $a$, since there is no surface charge there. The integrable density singularity has a finite potential limit $\varphi (0^{+})=Aa/{ϵ}_0$.