Question #9efd0

Aug 24, 2017

Let each charged sphere have a charge $q$, have a mass $m$ and make an angle $\theta$ with the vertical.

In the initial position of equilibrium electrical force of repulsion $F$ and weight of each sphere is connected through expression

$\frac{F}{m g} = \tan \theta$

Using Coulomb's Law and other values we write this as

$\frac{{k}_{e} {q}^{2} / {d}^{2}}{m g} = \frac{\frac{d}{2}}{{l}^{2} - {\left(\frac{d}{2}\right)}^{2}} ^ \left(\frac{1}{2}\right)$
$\implies q = {\left(\frac{m g}{2 {k}_{e}} {d}^{3} / {\left({l}^{2} - {\left(\frac{d}{2}\right)}^{2}\right)}^{\frac{1}{2}}\right)}^{\frac{1}{2}}$

As the charge starts leaking let $x$ be instantaneous distance between the spheres. The above expression reduces to
$q = \sqrt{\frac{m g}{2 {k}_{e}}} {x}^{\frac{3}{2}} / {\left({l}^{2} - {\left(\frac{x}{2}\right)}^{2}\right)}^{\frac{1}{4}}$

Differentiating both sides with respect to time $t$ we get
$\frac{\mathrm{dq}}{\mathrm{dt}} = \sqrt{\frac{m g}{2 {k}_{e}}} \frac{d}{\mathrm{dt}} \left[{x}^{\frac{3}{2}} / {\left({l}^{2} - {\left(\frac{x}{2}\right)}^{2}\right)}^{\frac{1}{4}}\right]$
$\implies$
$\frac{\mathrm{dq}}{\mathrm{dt}} = \sqrt{\frac{m g}{2 {k}_{e}}} \left[\frac{\frac{3}{2} {x}^{\frac{1}{2}}}{{l}^{2} - {\left(\frac{x}{2}\right)}^{2}} ^ \left(\frac{1}{4}\right) + \frac{\left({x}^{\frac{3}{2}}\right) \left(- \frac{1}{4}\right) \left(- \frac{x}{2}\right)}{{l}^{2} - {\left(\frac{x}{2}\right)}^{2}} ^ \left(- \frac{5}{4}\right)\right] \frac{\mathrm{dx}}{\mathrm{dt}}$

$\frac{\mathrm{dq}}{\mathrm{dt}}$ is rate of leakage of charge and velocity $\frac{\mathrm{dx}}{\mathrm{dt}} = v$. Above expression becomes
$\frac{1}{v} \frac{\mathrm{dq}}{\mathrm{dt}} = \sqrt{\frac{m g}{2 {k}_{e}}} \left[\frac{3 {x}^{\frac{1}{2}}}{2 {\left({l}^{2} - {\left(\frac{x}{2}\right)}^{2}\right)}^{\frac{1}{4}}} + \frac{{x}^{\frac{5}{2}}}{8 {\left({l}^{2} - {\left(\frac{x}{2}\right)}^{2}\right)}^{- \frac{5}{4}}}\right]$

is the required expression.