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

Let $r_0=|{\mathbf{r}}_0|$ and let ${\Omega }_0=({\theta }_0,{\phi }_0)$ describe its direction in spherical coordinates; write $\Omega =(\theta ,\phi )$. The given Dirichlet Green function between concentric spheres has normalization ${\nabla }^{2}G=-4\pi {\delta }^{(3)}$, so the electric potential is $\varphi =qG/(4\pi {ϵ}_0)$. On the inner sphere the unit normal from metal into the gap is $+{\mathbf{e}}_r$, hence its induced surface charge density isNear this boundary $r_{<}=r$ and $r_{>}=r_0$. The needed radial derivative isIt cancels the factor $2\ell +1$ in the Dirichlet Green function. Therefore the answer in spherical harmonics isThis spherical harmonic expansion converges on the boundary: after the angular sum, the two radial factors decay geometrically with ratios $a/r_0<1$ and $a{r}_0/b^2<1$, while the angular factors grow at most polynomially. This justifies the boundary derivative and the subsequent surface integral.

Write the constant spherical harmonic $Y_{00}=C$. Its orthonormality gives $4\pi |C{|}^2=1$, independently of its phase convention. Orthogonality to $Y_{00}$ givesOnly the $\ell =m=0$ term contributes to the surface integral of the surface charge density. Multiplying by the surface-area factor $a^2$ givesThus the spherical harmonic calculation reproduces the total induced electric charge: