Question

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

A. `((q^(2)L)/(2piepsilon_(0) mg))^(1/3)`
B. `((qL^(2))/(2piepsilon_(0) mg))^(1/3)`
C. `((q^(2)L^(2))/(4piepsilon_(0) mg))^(1/3)`
D. `((q^(2)L)/(4piepsilon_(0) mg))^(1/3)`

Answer 1

Correct Answer - a

In equilibrium `F_(e)= T sin theta`….(i)
`mg= T cos theta`….(ii)
`tan theta=(F_(e))/(mg)= (q^(2))/(4piepsilon_(0)x^(2)xxmg) also tan theta~~ sin theta= (x//2)/(L)`
Hence `(x)/(2L)= (q^(2))/(4piepsilon_(0)x^(2)xxmg)`
`implies x^(3)= (2q^(2)L)/(4piepsilon_(0)mg)implies x=((q^(2)L)/(2piepsilon_(0)mg))^(1//3)`