Past exam of the physics 2 paper of the University of Oxford / 2022 / 4

Choose zero electrostatic potential at infinity. By Coulomb's law and the superposition principle, the two point charges giveHere ${ϵ}_0$ is the vacuum permittivity. For a nondegenerate zero-equipotential curve, the electric charges must have opposite signs. Let $d_1,d_2$ be the two distances. The equation $\Phi =0$ becomes $q_1{d}_2=-q_2{d}_1$, and henceFor a circle $r=\rho >0$ centred at the origin, its $\cos \theta$ coefficient must vanish: $q_1^2{r}_2=q_2^2{r}_1$. If $r_1,r_2>0$ and $r_1\ne r_2$, this givesSubstitute $q_2^2/q_1^2=r_2/r_1$ in the constant term:ThereforeThis is an Apollonius circle: the distance ratio $d_2/d_1$ is constant. The distinct-position assumption matters. For example, $r_1=r_2=1$, $q_2=-q_1$ gives zero electrostatic potential everywhere away from the coincident point charges, so there is no distinguished zero circle. If only one position is at the origin and the other is not, no nontrivial positive-radius circle centred at the origin is possible.