Past exam of the physics B course of the University of Cambridge / 2011 / Paper 2 / Section B / B6 / ii / Solution

For the vacuum half-space keep the actual charge $Q$ at $(a,0,0)$ and put its image at the reflected point $(-a,0,0)$. ThenThe reflected singularity lies outside this region. The real singularity has the correct charge strength, and the interface coefficients obey the two matching equations in the parent Solution at every $y,z$. With decay at infinity, uniqueness of an electrostatic dielectric boundary-value problem establishes the construction throughout both half-spaces.