A charged particle q is shot towards another charged particle Q which is fixed, with a speed v. It approaches Q upto a closest distance r and then returns. If q is shot with speed 2v, the closest distance of approach would be

Options

A. `(r)/(4)`
B. `(r)/(2)`
C. 2r
D. r

Answer

A

Explanation

`(1)/(2) mv^(2) = (kQq)/(r_(1))" "...(i)`
`(1)/(2)m.4v^(2) = (kQq)/(r_(2))" "...(ii)`
From Eqs. (i) and (ii), we can say
`(r_(1))/(r_(2)) = 4`
`r^(2) = (r)/(4)`