 Electric field of a uniformly charged ball

Distribute a positive electric charge $Q$ uniformly through a closed ball of radius $R$ in vacuum, with no external field. Its volume charge density is $3Q/(4\pi R^3)$. For $0<r<R$, the enclosed fraction is $(r/R{)}^3$. Applying Gauss's law on a concentric Gaussian surface givesThe electric field grows linearly inside and decays as an inverse square outside. With $E_R=Q/(4\pi {\epsilon }_0{R}^2)$, its values at $r/R=1/2,1,2$ are $E_R/2,E_R,E_R/4$. The field vanishes at the centre by spherical symmetry and is continuous at the surface because this ideal volume distribution has no singular surface layer. Its interior behavior differs from a conductor in electrostatic equilibrium, where mobile charge redistributes.