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A charged particle is shot with speed V...

A charged particle is shot with speed V towards another ifxed charged particle Q .It approaches Q up to a closest distnce r and then returns.If q were given a speed 2V, the closest distance 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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