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Question

A charged 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 2v, the closest distances of approach would be

A
r
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B
2 r
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C
r2
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D
r4
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Solution

The correct option is D r4
Charge q will momentarily come to rest at a distance r from charge Q when all it's kinetic energy converted to potential energy i.e. 12mv2=14πϵ0.qQr
Therefore the distance of closest approach is given by
r=qQ4πϵ0.2mv2r1v2

Hence if v is doubled, r becomes one fourth.

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