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A charged particle of charge 'Q' is held...

A charged particle of charge 'Q' is held fixed and another charged particle of mass 'm' and charge 'q' (of the same sign) is released from a distance 'r'. The impulse of the force exerted by the external agent on the fixed charge by the time distance between 'Q' and 'q' becomes `2 r` is

A

` sqrt ((Qq)/ (4 pi in_0 mr))`

B

`sqrt ((Qqm)/(4pi in_0r ))`

C

` sqrt( (Qqm) /(pi in_0 r))`

D

` sqrt ((Qqm)/(2 pi in _0 r))`

Text Solution

Verified by Experts

The correct Answer is:
B
From conservation of energy
Loss of PE = Gain of `KE`
` 1/(4 pi in_0) (Qq)/r - 1/(4 pi in_0) (Qq)/(2r) = P^2/(2m)`
` 1/2 xx 1/(4 pi in_0) (Qq)/r = P^2/(2m)`
` P^@ = (mQq)/(4 pi in_0r) rArr p + sqrt (mQq)/(4 piin_0 r)` ltbr charge in linear momentum is equal to impulse received by free chare. Same impulse is on fixed charge.
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