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A charged particle q is shot from a larg...

A charged particle q is shot from a large distance towards another charged particle Q which is fixed, with speed v. It approaches Q up to as closed distance r and then returns. If q were given a speed 2v, the distasnce of approach would be

A

r

B

2r

C

`r/2`

D

`r/4`

Text Solution

Verified by Experts

The correct Answer is:
D
At closest distance, all of KE becomes PE of system.
`therefore 1/2 mv^2= (kQq)/r`……..(i)
When speed is 2V,
`1/2 m(2v)^2=(kQq)/r'`…..(ii)
Dividing equation (i) by equation (ii) we have
`1/4=r'/r implies r' r/4`
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