Question

# A charged particle q is fired 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 is given speed 2v, the closest distance of approach would be: [AIEEE-2004]

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 D r/4As the electrostatic force is a conservative force, we can conserve total energy. (K.E)i+(P.E)i=(K.E)f+(P.E)f Let us consider the position from which 'q' is fired as our reference point, because what only matters is the change in P.E. with speed v: 0+12mv2=KqQr+0 12mv2=KqQr⇒r=2KqQmv2 ...(1) with speed 2v: 12m(2v)2+0=0+KqQr1 where r1 is the new closest distance of approach ⇒2mv2=KqQr1 r1=KqQ2mv2 from (1),r1=r4

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