Two identical balls each of mass m are hung from a point by two silk threads of length l. Each of them has charge q. If the angle between the two threads are negativebly small, show that equillirium distance between the centres of the balls will be ,x`=((2q^(2)l)/(mg))^(1//3)`

Explanation

Let at equilibrium, the two balls will be at A and B [Fig. 2.36] OA = OB = l and AB = x.
Also, let the angle made by each thread with the vertical be `theta`.
Three forces act on each ball at equilibrium (i) weight of the ball mg, (ii) tension on threadT and (iii) repulsive force F acting between the balls.

Here, `Tsintheta=F=(q(2))/(x^(2)),Tcostheta=mg`
`thereforetantheta=(q^(2))/(mgx^(2))` or, `sintheta=(q^(2))/(mgx^(2))`
`[because` is small, `tantheta=sintheta]`
or, `x/(2l)=(q^(2))/(mgx^(2))`
`thereforex=((2q^(2)l)/(mg))^(1//3)`