Charge `q_(2)` of mass m revolves around a stationary charge `q_(1)` in a circular orbit of radius r. The orbital periodic time of `q_(2)` would be

Options

A. `[(4pi^(2)mr^(3))/(kq_(1)q_(2))]^(1//2)`
B. `[(kq_(1)q_(2))/(4pi^(2)mr^(3))]^(1//2)`
C. `[(4pi^(2)mr^(4))/(kq_(1)q_(2))]^(1//2)`
D. `[(4pi^(2)mr^(2))/(kq_(1)q_(2))]^(1//2)`

Answer

A

Explanation

Force between the both charges,
`F=(1)/(4pi epsilon_(0)).(q_(1)q_(2))/(r^(2))`
and centripetal force, `F=mr omega^(2)`
`therefore" "(1)/(4piepsilon_(0))(q_(1)q_(2))/(r^(2))=mromega^(2)=(4pi^(2)mr)/(T^(2))`
`rArr" "T^(2)=((4pi epsilon_(0))r^(2)(4pi^(2)mr))/(q_(1)q_(2))`
`T=[(4pi^(2)mr^(3))/(kq_(1)q_(2))]^(1//2)`