Question

Two identical pith balls, each of mass $m$ and carrying identical charges $q$ each, are suspended from a common point by two strings of equal length $l$. The angle between the string is $2\theta$, and initially changes at a rage $\left|\frac{d\theta }{dt}\right|={\omega }_0$, due to leakage of charge at a very slow rate. Find the rate of leakage of charge (i.e.$\left|\frac{dq}{dt}\right|$).

Answer 1

$T\cos \theta =mg$ (equation $1$)

$T\sin \theta =\frac{k{q}^2}{l^2{\sin }^2\theta }$ (equation $2$)

Dividing equation $2$ by equation $1$

$\tan \theta =\frac{k{q}^2}{mg{l}^2{\sin }^2\theta }$

$\tan \theta \cdot {\sin }^2\theta =\frac{k}{mg{l}^2}{q}^2$

$({\sin }^2\theta \cdot {\sec }^2\theta +2\sin \theta \cos \theta \cdot \tan \theta )\frac{d\theta }{dt}=\frac{k}{mg{l}^2}2q\frac{dq}{dt}$

$\left|\frac{dq}{dt}\right|=\frac{mg{l}^{2}w}{k\left(2q\right)}({\tan }^2\theta +2{\sin }^2\theta )$