A particle of mass `m` and carrying charge `-q_(1)` is moving around a charge `+q_(2)` along a circular path of radius `r`. Period of revolution of the charge `-q_(1)` about `+q_(2)` is

Options

A. `4sqrt((pi^(3)epsilon_(0)mr^(3))/(q_(1)q_(2)))`
B. `sqrt((pi^(3)epsilon_(0)mr^(3))/(q_(1)q_(2)))`
C. `2sqrt((pi^(3)epsilon_(0)mr^(3))/(q_(1)q_(2)))`
D. `3sqrt((pi^(3)epsilon_(0)mr^(3))/(q_(1)q_(2)))`

Answer

A

Explanation

Suppose that the charge `-q_(1)` moves around the charge `q_(2)` along a circular path of radius `r` with spedd `v`.
The necessary centripetal force is provided by the electronic force of attraction between the two charges i.e.,
`(1)/(4piepsilon_(0)).(q_(1)xxq_(2))/(r^(2))= (mv^(2))/(r)`
Or `v= (1/(4piepsilon_(0)).(q_(1)xxq_(2))/(mr))^(1//2)`
If `T` is period of revolution of the charge `-q_(1)` about `q_(2)` then
`T=(2pir)/(v)`
Subsituting for `v` we get `T = sqrt((16pi^(3)epsilon_(0)mr^(3))/(q_(1)q_(2)))`