 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.