Two small equally charged spheres, each of mass $m$, are suspended from the same point by light silk threads of length $l$. The separation between the spheres is $x\ll l$. Calculate the rate $\frac{dq}{dt}$ with which the charge leaks off each sphere if their velocity of approach varies as $v=\frac{\alpha }{\sqrt{x}}$ where $\alpha$ is a positive constant.

In this problem, let us first calculate the value of $x$ in terms of other known parameters. Let us consider the particles to be in equilibrium, then

$T\cos \theta =mg$ …..(1)

$T\sin \theta =\frac{q^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{x}}^2}$ ……(2)

$\Rightarrow \tan \theta =\frac{q^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{mgx}}^2}$

Since $x\ll l$

$\tan \theta$ is very small and hence $\tan \theta =\theta =\frac{x}{2l}$

$\Rightarrow \frac{x}{2l}=\frac{q^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{mgx}}^2}$

$\Rightarrow x^3=\frac{q^{2}l}{2{\mathrm{\pi \epsilon }}_0\mathrm{mg}}$ …..(3)

$\Rightarrow x={\left(\frac{q^{2}l}{2{\mathrm{\pi \epsilon }}_0\mathrm{mg}}\right)}^{\frac{1}{3}}$

Take derivative of (3) w.r.t time, we get

$3x^2\frac{dx}{dt}=\left(\frac{l}{2{\mathrm{\pi \epsilon }}_0\mathrm{mg}}\right)2q\left(\frac{dq}{dt}\right)$

But according to the problem, $v=\frac{dx}{dt}=\frac{\alpha }{\sqrt{x}}$

$\Rightarrow 3x^2\frac{\alpha }{\sqrt{x}}=\left(\frac{x^3}{q^2}\right)2q\left(\frac{dq}{dt}\right)$

$\Rightarrow \frac{dq}{dt}=\frac{3\alpha }{2}\left(\frac{q}{x^{3/2}}\right)$

But $\frac{x^3}{q^2}=\frac{l}{2{\mathrm{\pi \epsilon }}_0\mathrm{mg}}$

$\Rightarrow \frac{dq}{dt}=\frac{3\alpha }{2}\sqrt{\frac{2{\mathrm{\pi \epsilon }}_0\mathrm{mg}}{l}}$