 Solution

Fix the electric potential to vanish at infinity and assume a localized, nonsingular charge density of finite electrostatic energy. Assemble the charge by scaling ${\rho }_s=s\rho$ from $s=0$ to $1$. Linearity gives ${\varphi }_s=s\varphi$, so the work isThe factor $1/2$ prevents double counting of pair interactions. Using Gauss law, $\rho =-{ϵ}_0\Delta \varphi$, and integration by parts,The boundary integral at infinity vanishes for a localized distribution since $\varphi =O(r^{-1})$ and ${\partial }_r\varphi =O(r^{-2})$. Singular point-charge self-energies require a separate regularization and are not finite instances of this identity.

For the uniform solid sphere, spherical symmetry and Gauss law giveIntegrating inward from infinity and matching the continuous electric potential at $R$ givesThe value of the electric field at the centre is the zero vector.

Writing $Q=4\pi \rho R^3/3$, the electrostatic energy of a uniformly charged solid sphere follows either from the charge-potential integral or from the field integral including the exterior:For the nuclear scaling, $Q=Ze$ while constant volume per proton gives $R\propto Z^{1/3}$. Therefore the electric contribution scales as $Q^2/R\propto Z^{5/3}$ in this uniform-density model.