Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 2 / 15

Inside the sphere, the electric field has the form $\mathbf{E}=\nabla [-k{r}^2/(2a^3)]$, including at the origin. Thus it is a conservative vector field. Outside, use the gradient of a radial function to obtainThis is also an explicit potential of a conservative vector field, single-valued throughout the exterior region. Equivalently, the curl vanishes in both regions: for a radial coefficient $h(r)$, $\nabla \times (h\mathbf{r})=\nabla h\times \mathbf{r}+h\nabla \times \mathbf{r}=0$ because $\nabla h$ is radial. The interior formula avoids any apparent $1/r$ issue at the origin. The vector-field potentials here have the opposite sign from the electric potential convention used in (b).

For a radial electric potential, $\mathbf{E}=-V^{\prime }(r)\mathbf{r}/r$. Comparing radial coefficients givesIntegrating these one-variable derivatives gives the general local expressionsThe two constants are initially independent because these integrations were performed in separate radial regions.

The condition at infinity gives $C_{\mathrm{out}}=0$. Matching the electric potential at $r=a$ givesConsequentlyThis function is finite and differentiable at the origin. At the joining radius, both $V$ and $V^{\prime }$ match, giving $V(a)=-k/a$ and $V^{\prime }(a)=k/a^2$. Thus the electric field is continuous there as well as being conservative on each side. This is the potential of a uniform spherical source, with source coefficient $3k/a^3$.