Two similar pith balls of mass m are hung from a common point with the help of two long silk threads . The balls carry similar charges q . Assuming that `theta` is very small , then find the expression of x .

Explanation

According to Laml.s theorem , we get `(F)/(AC) = (w)/(OC)= (T)/(AO) implies F = (w)/(OC) xx AC`
`F = (mg AC)/(OC) = (mg l sin theta)/(l cos theta) = mg tan theta " "` … (i)
Now , repulsive force `F = (1)/(4 pi epsi_(0)) (q^(2))/(x^(2)) " "`... (ii)
From Eqs. (i) and (ii) , we get
`implies (1)/(4pi epsi_(0)) (q^2)/(x^2) = mg tan theta`
`implies x^(2) = (1)/(4 pi epsi_(0)) xx (q^(2))/(mg tan theta) " "`.... (iii)
Now , for small angle , tan `theta= theta` and `theta = ((x/2))/(l) = (x)/(2l)`
So, Eq. (iii) will be reduced as `x^(2) = (1)/(4 pi epsi_(0)) (q^(2) 2l)/(mgx)`
`x^(3) = (q^(2) l)/(2pi epsi_(0) mg) implies x = ((q^(2) l)/(2pi epsi_(0) mg))^(1//3)`