A charge 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 were given a speed of '2v' the closest distances of approach would be

Options

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

Answer

d

Explanation

At closest distance r its whole Ke is converted into PE
`therefore (1)/(2)mv^(2)=(1)/(4piepsilon_(0))(Q.q)/(R )rArr r=(1)/(4piepsilon_(0))(Q.q)/(mv^(2))`
In next case, `r'=(1)/(4piepsilon_(0))(Qq)/(m(2v)^(2))
`rArr r'=(1)/(4)(1)/(4piepsilon_(0)).(Qq)/(mv^(2))rArr r'=r//4`