Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 1 / 16A (Electromagnetism)

In electrostatic equilibrium, the electric field inside a perfect conductor is zero. Apply the Faraday's law $\oint \mathbf{E}\cdot d\mathbf{l}=0$ to a thin rectangular loop crossing the surface, with its long sides tangential and its short sides shrinking to zero. It yields continuity of the tangential electric field. Since its interior value is zero, $E_x(0+)=E_y(0+)=0$. Earthing fixes the electric potential at the plane to zero. For completeness, a thin Gauss law pillbox gives $E_z(0+)=\sigma /{\epsilon }_0$; the normal component can be nonzero because of induced surface charge. These are the electrostatic boundary conditions at a conductor.

The charge and its proposed image charge give the exterior electric potentialAt $z=0$ the terms cancel, so $\Phi =0$ everywhere on the plane and its tangential derivatives vanish. Thus 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.

At instantaneous height $z$, the image charge is a distance $2z$ below the real charge, giving force $m\ddot{z}=-K/z^2$ with $K=q^2/(16\pi {\epsilon }_0)$. Multiplying by $\dot{z}$ and using rest at $z=d$ givesThe induced-conductor potential energy is $-K/z$, not the full energy of a pair of independently movable charges. On the falling branch, integrate the reciprocal speed:The last integral is $\pi /2$ by $s={\sin }^2\theta$. Hence the fall time of a charge toward a grounded plane in the electrostatic Newtonian model isFor $q=0$ there is no force and the particle remains at rest.