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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:
41627_ae16cd2be1f54e709110d4bd70ba40bd.png

A
r
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B
2r
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C
r/2
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D
r/4
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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
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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