Question

Two identical small equally charged conducting balls are suspended from long threads secured at one point. The charges and masses of the balls are such that they are in equilibrium when the distance between them is a (the length of the thread L >> a). One of the balls is then discharged. Again for the certain value of distance b (b << $l$) between the balls, the equilibrium is restored, the value of $\left(a^3/b^3\right)=?$

Answer 1

For equilibrium, $T\sin \theta =F=\frac{q^2}{4\pi {ϵ}_0{a}^2}....\left(1\right)$
and $T\cos \theta =mg.....\left(2\right)$
$\left(1\right)/\left(2\right),\tan \theta =\frac{q^2}{4\pi {ϵ}_0{a}^{2}mg}$
For small $\theta ,\tan \theta \sim \frac{a/2}{L}$
$\therefore \frac{a/2}{L}=\frac{q^2}{4\pi {ϵ}_0{a}^{2}mg}$
or $a^3=\frac{q^{2}L}{4\pi {ϵ}_{0}mg}....\left(3\right)$
When one of the balls is discharged, the balls come closer and touch each other and again separate due to repulsion. Due to charge induction,the charge on each ball after touching each other is $\frac{q}{2}$.
Replacing $q$ with $q/2$ in $\left(3\right)$, we get
$b^3=\frac{(q/2{)}^{2}L}{4\pi {ϵ}_{0}mg}....\left(4\right)$
$\left(3\right)/\left(4\right),\frac{a^3}{b^3}=4$