Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 3 / 17C (Electromagnetism)

Let the smaller charge be $q$ at $\mathbf{a}$ and the other charge be $-Q$ at $\mathbf{b}$, where $0<q<Q$; an overall sign reversal changes no zero-potential locus. Put $\lambda =q/Q\in (0,1)$. The electrostatic potential vanishes exactly when $|\mathbf{x}-\mathbf{a}|=\lambda |\mathbf{x}-\mathbf{b}|$. Completing the square gives the Apollonius sphereThe distances from its center to the charge positions are $|\mathbf{c}-\mathbf{a}|=\lambda R<R$ and $|\mathbf{c}-\mathbf{b}|=R/\lambda >R$. Thus it encloses exactly the charge of smaller magnitude. Unequal magnitudes are essential: the equal-magnitude case instead gives the perpendicular bisector plane.

Write $R=|\mathbf{p}|>a$. In SI units, the exterior electrostatic potential solveswith the point-charge singularity $q/(4\pi {\epsilon }_0|\mathbf{x}-\mathbf{p}|)$. The image charge for a grounded conducting sphere choiceslie inside the sphere. On $|\mathbf{x}|=a$, direct squaring gives $|\mathbf{x}-{\mathbf{p}}^{\prime }|=(a/R)|\mathbf{x}-\mathbf{p}|$. Thereforesatisfies the source equation, grounded boundary value and decay at infinity. The Uniqueness of the Dirichlet problem, applied to the harmonic difference with its singularity removed, identifies this with the physical exterior potential. Thus its gradient gives the required exterior electric field.

The charge feels the induced field, equivalently the field of the image charge, excluding its own singular field. Their separation is $(R^2-a^2)/R$, so Coulomb's law givesIts magnitude is $q^{2}aR/[4\pi {\epsilon }_0(R^2-a^2{)}^2]$ and its direction is toward the sphere's center, independent of the sign of $q$. No factor one-half belongs in this force; that factor occurs in induced electrostatic energy instead.