= Solution
Let $r_0=|\mathbf r_0|$ and let $\Omega_0=(\theta_0,\varphi_0)$ describe its direction in <spherical coordinates>; write $\Omega=(\theta,\varphi)$. The given <Dirichlet Green function between concentric spheres> has normalization $\nabla^2G=-4\pi\delta^{(3)}$, so the <electric potential> is $\phi=qG/(4\pi\epsilon_0)$. On the inner sphere the <unit normal> from metal into the gap is $+\mathbf e_r$, hence its induced <surface charge density> is
$$
\sigma(a,\Omega)=-\frac q{4\pi}\left.\partial_rG(r,\Omega;\mathbf r_0)\right|_{a^+}.
$$
Near this boundary $r_<=r$ and $r_>=r_0$. The needed radial <derivative> is
$$
\left.\frac{d}{dr}\left(r^\ell-\frac{a^{2\ell+1}}{r^{\ell+1}}\right)\right|_{r=a}
=\ell a^{\ell-1}+(\ell+1)a^{\ell-1}=(2\ell+1)a^{\ell-1}.
$$
It cancels the factor $2\ell+1$ in the <Dirichlet Green function>. Therefore the answer in <spherical harmonics> is
$$
\boxed{\sigma(a,\Omega)=-q\sum_{\ell=0}^{\infty}\sum_{m=-\ell}^{\ell}
\frac{a^{\ell-1}\overline{Y_{\ell m}(\Omega)}Y_{\ell m}(\Omega_0)}{1-(a/b)^{2\ell+1}}
\left(\frac1{r_0^{\ell+1}}-\frac{r_0^\ell}{b^{2\ell+1}}\right).}
$$
This <spherical harmonic> expansion converges on the boundary: after the angular sum, the two radial factors decay geometrically with ratios $a/r_0<1$ and $ar_0/b^2<1$, while the angular factors grow at most polynomially. This justifies the boundary <derivative> and the subsequent <surface integral>.
Back to article page
