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Q.

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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a

r

b

r/2

c

r/4

d

2r

answer is D.

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Detailed Solution

KE=PE;12mv2=14πε0ze2dv1v22=d2d1⇒v2v2=d2r⇒d2=r4

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