Past exam of the electromagnetism course of the University of Oxford / 2023 / 1 / d

Translate the concentric spheres so that their centre is $(0,0,-a)$, take $b=a+d$, and keep the point charge fixed at $(0,0,h)$. Its distance from the centre is $r_0=a+h$. As $a\to \infty$, the upper portions of the inner and outer spheres approach the planes $z=0$ and $z=d$. The preceding total induced electric charge becomesThe outer sphere has total induced electric charge $Q_b=-q-Q_a$: a Gaussian surface within its conducting material has zero electric flux, and the grounded exterior is field-free by uniqueness of Poisson equation. Therefore $Q_b\to -qh/d$.

One can justify the total-charge limit without assuming that local convergence of electric fields preserves their surface integrals. Use induced charge as harmonic measure: if a harmonic function $w$ is one on the selected grounded boundary and zero on the other, Green's second identity with the electric potential givesIndeed, ${\nabla }^2\varphi =-q{\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)/{ϵ}_0$, $\varphi =0$ on the boundaries, and their unit normal outward from the gap satisfies $\sigma ={ϵ}_0{\partial }_n\varphi$. For the spheres the weighting harmonic function is $w_a(r)=a(b-r)/(r(b-a))$, exactly the factor already obtained. For the slab the two weighting harmonic functions are $w_0=1-z/d$ and $w_d=z/d$. Applying the same identity to the decaying grounded-slab electric potential, whose boundary term at lateral infinity vanishes, establishes directlyTheir sum is $-q$, and their relative sizes measure how close the point charge lies to each grounded plane.