= Solution
For the vacuum half-space keep the actual charge $Q$ at $(a,0,0)$ and put its image at the reflected point $(-a,0,0)$. Then
$$
\boxed{\phi_+(x,y,z)=\frac{Q}{4\pi\epsilon_0}\left[\frac1{r_+}-\frac{\epsilon_r-1}{\epsilon_r+1}\frac1{r_-}\right]\quad(x>0).}
$$
The 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.
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