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Question

A charged particle q is shot towards another charged particle (from infinity) Q which is fixed, with a speed v. It approaches Q up to the 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
2r
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C
r2
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D
r4
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Solution

The correct option is D r4
When charge q is shot with velocity V:
Let r be the distance of closest approach. A/c to the law of conservation of K.E. of charge q = Electrostatic energy of charges q and Q
or, 12mv2=14πϵ0Qqr(i)
When charge q is shot with velocity 2v:
Let r' be the distance of closest approach. In this case;
12m(2v)2=14πϵ0Qqror,4×(12mv2)=14πϵ0Qqr(ii)
From (i) and (ii), we have
4×14πϵ0Qqr=14πϵ0Qqr
or, r=r4

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