Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 1 / 16A / iii / Solution

As the real charge descends to height $z(t)$, its instantaneous image charge is at $-z(t)$. The force is thereforeThe image is not kept at its initial position. Multiplying by $\dot{z}$ and using $z(0)=d$, $\dot{z}(0)=0$ givesFor the falling charge above a grounded conducting plane, the arrival time isPut $z=d{\sin }^2\theta$; the integral becomes $2d^{3/2}{\int }_0^{\pi /2}{\sin }^2\theta d\theta =\pi d^{3/2}/2$. HenceThis is the finite arrival time in the specified ideal force model, for $q\ne 0$. The field-force derivation also avoids incorrectly using the full real-image pair interaction energy: the conductor's effective energy is $-q^2/(16\pi {\epsilon }_{0}z)$, with the factor one-half appropriate to induced charge.