# Charge distribution of a spherically symmetric electric potential

Here we have a spherically symmetric electric potential: $$V(r)=\frac{e^{-\lambda r}}{r},$$ and now we want to know the charge distribution which triggers the potential. Of course we can use the Poisson’s equation $$\nabla^2 V(r) = -\frac{\rho}{\varepsilon_0}$$ to get the charge distribution. However, when I applied the Laplacian Operator to $$V(r)$$, I got $$\frac{\lambda^2 e^{-\lambda r}}{r}$$, which is positive. This resulted a negative charge distribution. But I think the charge distribution should be positive, because the electric potential distribution is positive, and the potential is getting greater when $$r$$ is getting smaller. Where am I wrong? Can you help me find out it?

If you had a positive point charge at the origin, the potential would be proportional to $$1/r$$, but your potential is decreasing faster for increasing $$r$$. To achieve this, spherically symmetric layers of negative charges are needed to shield part of the point charge, which means your mathematical result seems to make sense. The positive point charge, however, is not included in this result, because the derivative of your potential $$V(r)$$ does not exist at the origin.
• Note also that to find the value of the point charge $q_o$ at the origin, you can find the the electric flux over a sphere of size $\delta$, and then take the limit as $\delta \to 0$:$$q_o = - \epsilon_0 \lim_{\delta \to 0} \iint_{|\vec{r}| = \delta} (\nabla V) \cdot d\vec{a}.$$This will work out to be something positive and non-zero, I'm pretty sure. Commented Apr 14, 2022 at 14:25