 Solution

Gauss law states that for outward-oriented boundary $S=\partial V$,For a continuously differentiable field and continuous density, the divergence theorem turns the left side into ${\int }_V\nabla \cdot \mathbf{E}$. Thus the differential law implies the integral law. Conversely, if the integral law holds for every suitable volume, the integral of $\nabla \cdot \mathbf{E}-\rho /{ϵ}_0$ vanishes on every small ball. A nonzero continuous value would keep one strict sign in some small ball, a contradiction. Hence $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. Weaker regularity gives the corresponding almost-everywhere or distributional formulation.

For the radial field, the flux on a radius-$r$ sphere is $4\pi r^{2}E(r)$, while its enclosed charge is $4\pi {\int }_0^r{s}^2\rho (s)ds$. ThereforeThis uses the stated density as the full charge distribution, with no additional point charge at the origin. Since it vanishes beyond $a$, the total charge isand for $r>a$ the field is $\boxed{\mathbf{E}=Q{\mathbf{e}}_r/(4\pi {ϵ}_0{r}^2)}$.