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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 was given a speed 2v, the closest distance of approach would be:

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Solution

The correct option is **C** r/4

As initially q is at infinite distance from Q, thus P.Ei=0

As initially q is at infinite distance from Q, thus P.Ei=0

For case 1: the closest distance given is r and velocity of q is v initially.

Applying conservation of energy, P.Ei+K.Ei=P.Ef+K.Ef

0+12mv2=KQqr+0

⟹12mv2=KQqr ................(1)

For case 1: Let the closest distance be r' and velocity of q is 2v initially.

Applying conservation of energy, P.Ei+K.Ei=P.Ef+K.Ef

0+12m(2v)2=KQqr′+0

⟹12m(2v)2=KQqr′ ................(2)

Solving (1) and (2), r′=r4

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