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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
By energy conservation :
Initially :`0 + 1/2 mv^2 = (kQq)/r`
Finally, `1/2, (2 v)^2 = (kQq)/r`
So, `(4 kQq)r = (kQq)/r'`
or `f' = r/4`.
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