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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/4
As 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=KqQrr=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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