Two similar balls are suspended from a point by two silk thread, each of length l. Each ball of mass m contains q amount of charge. If the angle between the two threads is very small, show that the distance between the centres of the two balls at equilibrium is `x=((2q^(2)l)/(mg))^(1//3)`

Explanation

Let A and B be the equilibrium positions of the two ball. `OA=OB=l` and `AB=x`. Suppose th angle of inclination of the two threads with the vertical `=theta`

At equilibrium, there forces act on each ball : 1. weight of the ball mg, 2. tension in the string T and 3. mutual repulsive force betwene the charged ball, F.
Here `G sin theta=F=(q^(2))/(x^(2)),T cos theta =mg`
`:.tan theta=(q^(2))/(mgx^(2))`
or `sin theta=(q^(2))/(mgx^(2)) [ :' theta` is small `tan theta~~sin theta]`
or `(x//2)/l,(q^(2))/(mgx^(2))` or `x=((2q^(2)l)/(mg))^(1//3)`