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The space ${\mathbb{R}}^3$ is simply connected, and the electric field has zero curl. Hence it is a conservative vector field. More explicitly, Stokes theorem gives zero circulation around any closed loop, since such a loop bounds a spanning surface in ${\mathbb{R}}^3$. Therefore the line integral between two fixed endpoints is independent of the chosen path.

Fix a reference point ${\mathbf{r}}_0$ and define the electric potentialTaking a short straight terminal segment in each coordinate direction shows ${\partial }_j\phi =-E_j$. ConsequentlyThe electric potential is determined up to an additive constant; the later condition at infinity fixes that constant for the finite charge distribution in part (b).

Set $r=R$ in the enclosed electric charge formula:Outside the entire charge distribution, Gauss's law gives $E_r(r)=Q/(4\pi {ϵ}_0{r}^2)$. Integrating $E_r=-{\phi }^{\prime }$ with $\phi (\infty )=0$ givesThus the exterior electric potential and electric field are exactly those of a point charge $Q$ placed at the center. This is the electrostatic shell theorem: outside a spherically symmetric charge distribution, only its total electric charge affects the electric field, not how that electric charge is distributed radially. If positive and negative electric charges cancel to give $Q=0$, both exterior quantities vanish.

A charge distribution invariant under rotations about one center. For its spherically symmetric potential, Gauss's law gives the radial electric field $\mathbf{E}=Q(r){\mathbf{e}}_r/(4\pi {ϵ}_0{r}^2)$, where $Q(r)$ is the enclosed electric charge.