= Solution
The <method of images> solves certain <electrostatics> problems by replacing a conductor by fictitious <electric charges> outside the region where the solution is required. The replacement must satisfy the actual boundary values of the <electric potential> as well as the prescribed sources. The <uniqueness of Poisson equation> then establishes that the constructed <electric potential> equals the physical one in that region. The image locations are mathematical devices; they are not real charges inside the metal.
For a <point charge> $q$ at $(0,0,a)$ above an infinite grounded conducting plane $z=0$, put an image charge $-q$ at $(0,0,-a)$. In a <cylindrical coordinate system> with coordinates $(s,z)$ the potential in the physical region is
$$
\phi(s,z)=\frac{q}{4\pi\epsilon_0}\left[\frac1{\sqrt{s^2+(z-a)^2}}-\frac1{\sqrt{s^2+(z+a)^2}}\right],\qquad z>0.
$$
This vanishes at $z=0$, has the required singularity at the real charge, and tends to zero at infinity. With <electric field> $\mathbf E=-\nabla\phi$, the induced <surface charge density> follows from the normal field immediately above the conductor:
$$
\sigma(s)=\epsilon_0E_z(s,0^+)=-\frac{qa}{2\pi(s^2+a^2)^{3/2}},\qquad
\int_0^\infty\sigma(s)\,2\pi s\,ds=-q.
$$
The <force> on $q$ is the field of the image evaluated at the real charge:
$$
\boxed{\mathbf F=-\frac{q^2}{16\pi\epsilon_0a^2}\hat{\mathbf z}.}
$$
The <electrostatic energy> relative to infinite separation is $U=\tfrac12q\phi_{\rm image}=-q^2/(16\pi\epsilon_0a)$. The factor $1/2$ accounts for the fact that the induced charge changes as the real charge is assembled; treating the image as an independently fixed physical charge would give the wrong <energy>.
A second standard geometry is a grounded conducting sphere of radius $b$ with a charge $q$ a distance $d>b$ from its centre. Its exterior field is generated by an image $q'=-qb/d$ at distance $b^2/d$ from the centre on the line towards the real charge. On the sphere the distances to the real and image charges are in the ratio $d/b$, making their potentials cancel. For an isolated neutral sphere, an additional central image $+qb/d$ makes the total induced charge zero and leaves the surface an <equipotential>. The construction is powerful for suitable symmetric boundaries, but arbitrary boundaries do not in general admit a finite image system.
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