Past exam of the physics B course of the University of Cambridge / 2010 / Paper 2A / Section B / B9 / Solution

The method of images replaces conductors by fictitious charges chosen to reproduce their boundary potentials in the physical domain. The uniqueness theorem for the electrostatic potential then identifies the resulting physical exterior field; the fictitious interior field is not the field inside the conductor.

For a source at $d\hat{\mathbf{z}}$, let $r^{\prime }=a^2/d$ and $q^{\prime }=-qa/d$. At a point on $r=a$, the distances $R$ and $R^{\prime }$ to source and image obey $R^{\prime }=(a/d)R$. Hence $q/R+q^{\prime }/R^{\prime }=0$ everywhere on the sphere. This proves the image charge for a grounded conducting sphere construction.

With real charges at $\pm d\hat{\mathbf{z}}$, superpose two images of charge $-qa/d$ at $\pm a^2\hat{\mathbf{z}}/d$. For the right-hand charge, take positive force away from the centre. Its electrostatic force isPutting $x=d/a$, the equal-charge balance around a grounded sphere requiresFor large $x$, the right side is approximately $2/x^3$, so $x\simeq 8$. This is an approximate equilibrium, not an exact cancellation at $x=8$.

To hold the sphere at electric potential $V$, add a central image $Q_0=4\pi {ϵ}_{0}aV$. The grounded pair already contributes zero potential on its surface. ThereforeAt $d=8a$ this is $F_0(8a)+qV/(64a)$, or approximately $qV/(64a)$ if the small grounded residual is neglected as in the preceding approximation. The charge on the sphere is $Q_0-2qa/d$; fixing its potential is different from independently fixing its net charge.