Two small identical balls P and Q, each of mass $\sqrt{3}$ /10 gram, carry identical charges and are suspended by threads of equal lengths $l=30cm$. At equilibrium, they position themselves as shown in figure. What is the charge on each ball. Given = 9 x 10⁹ Nm² C^-2 and take g = 10 ms^-2. 

 Let us consider forces on a ball, say, Q. Three forces act on it: (i) tension T in the thread, (ii) force ‘mg’ due to gravity and (iii) force F due to Coulomb repulsion along + ve x-direction. For equilibrium, the sum of the x and y components of these forces must be zero, 

i.e.             T cos 60°$-$F = 0 

and            T sin 60°$-$mg = 0 · 

These equations give F = mg cot 60° =$\frac{\sqrt{3}}{10}\times {10}^{-3}\times 10\times \frac{1}{\sqrt{3}}={10}^{-3}N$ . Now 

   $F=\frac{1}{4{\mathrm{\pi \epsilon }}_0}·\frac{q^2}{r^2}$

Putting F = 10^{- 3} N, r = 0.3 m and $\frac{1}{4{\mathrm{\pi \epsilon }}_0}$= 9 x 10⁹, we get Q= 10^{- 7} coulomb.