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Question

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

A
r
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B
2r
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C
r2
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D
r4
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Solution

## The correct option is D r4As 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...(1) ​ At the point of closest approach, the K.E of the particle is momentarily zero. Substituting the values in (1), we get 0+12mv2=KQqr+0 ⇒12mv2=KQqr...(2) For case 2: 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...(3) ​ At the point of closest approach, the K.E of the particle is momentarily zero. Substituting the values in (3), we get 0+12m(2v)2=KQqr′+0 ⇒12m(2v)2=KQqr′...(4) Using eqaution (2) and (4) we get, r′=r4 Hence, option (d) is the correct answer.

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