= Solution
The charge and its proposed <image charge> give the exterior <electric potential>
$$
\Phi(r,z)=\frac q{4\pi\varepsilon_0}\left(\frac1{\sqrt{r^2+(z-d)^2}}-\frac1{\sqrt{r^2+(z+d)^2}}\right).
$$
At $z=0$ the terms cancel, so $\Phi=0$ everywhere on the plane and its tangential <derivatives> vanish. Thus \b[the image construction satisfies the earthed-conductor boundary conditions]. It has the correct point-charge singularity, is harmonic elsewhere in $z>0$, and decays at infinity; electrostatic uniqueness identifies it with the exterior solution.
Back to article page
