Q. Two balls of same mass and carrying equal charge are hung from a fixed support of length $l$. At electrostatic equilibrium, assuming that angles made by each thread is small, the separation, $x$ between the balls is proportional to :
Solution:
In equilibrium, $F_{e} = T \,sin\, \theta$
$mg = T\,cos\,\theta$
$tan\,\theta = \frac{F_{e}}{mg} = \frac{q^{2}}{4\pi\,\in_{0}\,x^{2}\times mg}$
also $tan\, \theta\,\approx\, sin = \frac{x/2}{\ell}$
Hence, $ \frac{x}{2\ell } = \frac{q^{2}}{4\pi \,\in _{0}\,x^{2}\times mg}$
$\Rightarrow x^{3} = \frac{2q^{2}\ell}{4\pi \,\in_{0}\,\times mg}$
$\therefore x = \left(\frac{q^{2}\ell }{2\pi \,\in _{0}\,\times mg}\right)^{1/3}$
Therefore $x \,\propto \,\ell^{1/3}$
