In the given figure two tiny conducting balls of identical mass `m` and identical charge `q` hang from non-conducting threads of equal length `L`. Assume that `theta` is so small that than `theta~~ sin theta`, then for equilibrium `x` is equal to

Options

A. `((q^2L)/(2piepsilon_0mg))^(1//3)`
B. `((qL^2)/(2piepsilon_0mg))^(1//3)`
C. `((q^2L^2)/(2piepsilon_0mg))^(1//3)`
D. `((q^2L)/(4piepsilon_0mg))^(1//3)`

Answer

A

Explanation

In equilibrium `F_e = T sin theta`
`mg = T cos theta`
`tan theta=F_e/(mg)`
`implies sin theta=F_e/(mg)`
`implies(x/2)/L=(Kq^2)/(x^2mg)implies L=(x^3mg)/(2Kq^2)`
`implies x^3 =(2Lq^2)/(4piepsilon_0mg)implies x = [(Lq^2)/(2piepsilon_0 mg)]^(1/3)`