Question

Two similar balls of mass $m$ are hung from silk threads of length $\ell$ and carry similar charges $+q$ and $+q$.
(a) Prove that when the system is in equilibrium, separation between the balls is $x={\left(\frac{q^2\ell }{2\pi {\epsilon }_{0}mg}\right)}^{1/3}$, assuming the angle made by the threads with the vertical line passing through the point of suspension is small.
(b) Specify the rate $\left(dq/dt\right)$ with which the charge from each sphere leaks off if their velocity of approach varies as $v=\frac{a}{\sqrt{x}}$, where $a$ is a constant.

Answer 1

(a) Force of electrostatic repulsion:
$F=\frac{q^2}{4\pi {\epsilon }_0{x}^2}$
For equilibrium of each sphere:
$Tcos\theta =mg$
$T\sin \theta =F=\frac{q^2}{4\pi {\epsilon }_0{x}^2}$
$⟹\frac{F}{mg}=\tan \theta$
$⟹=\frac{q^2}{4\pi {\epsilon }_0{x}^2\left(mg\right)}=\tan \theta \approx \sin \theta =\frac{x}{2\ell }$ $[\because \theta \to 0]$
$⟹x={\left(\frac{q^2\ell }{2\pi {\epsilon }_{0}mg}\right)}^{1/3}$ ...(i)
(b) From equation (i)
$q^2=\frac{2\pi {\epsilon }_{0}mg{x}^3}{\ell }$
Differentiating both sides with respect to time:
$2q\frac{dq}{dt}=\frac{2\pi {\epsilon }_{0}mg}{\ell }.3x^2\frac{dx}{dt}$
Now, $\frac{dx}{dt}=$velocity of approach $=v=\frac{a}{\sqrt{x}}$
${\left(\frac{2\pi {\epsilon }_{0}mg}{\ell }\right)}^{1/2}.x^{3/2}\frac{dq}{dt}=\frac{3\pi {\epsilon }_{0}mg}{\ell }{x}^2\left(\frac{a}{\sqrt{x}}\right)$ Using equation (i)
$⟹\frac{dq}{dt}=\frac{3}{2}a\sqrt{\frac{2\pi {\epsilon }_{0}mg}{\ell }}$