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A charge particle 'q' is shot towards an...

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

A

r

B

2r

C

r/2

D

r/4

Text Solution

Verified by Experts

The correct Answer is:
D

From the given data, using energy conservation
`1/2 mv^(2)=(KQq)/r`
when particle is shot with a speed eV, let distance of closest approach =x
`1/2 m.4v^(2)=(KQq)/xrArr x=r/4`
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