 Solution

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.