Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2007 / Paper 2 / 2 / Grounded hemisphere

Reflect the source across the equatorial plane and subtract the two ball kernels. This gives the Dirichlet Green function inside a hemisphere $G_{\mathrm{hemi}}(\mathbf{x};{\mathbf{x}}_0)=G_B(\mathbf{x};{\mathbf{x}}_0)-G_B(\mathbf{x};{\mathbf{x}}_0^{-})$. It is zero on the spherical boundary because each ball kernel is zero there, and on the plane because reflection makes their values equal.

For a point charge $e$, Poisson's equation in SI units is $\Delta V=-e\delta (\mathbf{x}-{\mathbf{x}}_0)/{\epsilon }_0$, so $V=-e{G}_{\mathrm{hemi}}/{\epsilon }_0$. Writing ${\mathbf{x}}_0^{\ast -}$ for the reflected inverse point givesThe construction uses the real charge, its negative reflected image, a negative inversion image of magnitude $ea/r_0$, and its positive reflected image. Reflection and inversion commute. All three images lie outside the open upper half-ball, and the prescribed zero boundary values determine the electrostatic potential uniquely.