= Fall time of a charge toward a grounded plane
{title2=$t_{\rm hit}=\pi\sqrt{2\pi\varepsilon_0 m d^3}/|q|$}
The image of a charge $q$ at height $z$ is $-q$ at height $-z$. Their separation is $2z$, giving force $-q^2/(16\pi\varepsilon_0z^2)$. Starting at rest at height $d$, energy conservation gives $\dot z^2=q^2(1/z-1/d)/(8\pi\varepsilon_0m)$. Integrating the reciprocal speed to the plane yields the stated finite time in the electrostatic Newtonian model. The induced-conductor potential energy is $-q^2/(16\pi\varepsilon_0z)$; using the full image-pair energy would double the force incorrectly.
Back to article page
