# Electric potential and field due to a continuous charge distribution

(1) The electric potential due to a continuous charge distribution is:

$$\psi=\int_V \dfrac{\rho}{r}\ dV$$

To calculate this integral $$\rho$$ must be continuous over $$V$$. But $$\rho$$ is discontinuous at the boundary of $$V$$.

Why does it not prevent us from carrying out the integral?

(2) I have read that discontinuity in $$\rho$$ prevents us from computing the field (via potential) at the point of discontinuity. Why is it so?

If you want to calculate the potential and field at a point where the charge density $$\rho$$ is discontinuous, there is no problem if the discontinuity is a simple one (meaning there is a change from one smooth function to another smooth function across a similarly smooth surface). As an example of this, consider the field inside and outside a uniformly charged solid sphere. The electric field is well defined and continuous at the boundary, as can be verified by Gauss's Law, and the potential is continuous and differentiable at the boundary, meaning that we can indeed calculate the electric field as the gradient of the potential.