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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//4pi in_(0) mr)`

B

`sqrt(Qqm//4pi in_(0) r)`

C

`0`

D

cannot determine

Text Solution

Verified by Experts

The correct Answer is:
B
Applying conservation of energy, we get
`(kQq)/(r ) = (KQq)/(2r) + (1)/(2)mv^(2)`
`(1)/(2)mv^(2) = (KQq)/(2r) rArr V = sqrt((KQq)/(mr))`

Force required to keep `Q` fixed = Coulombic repulsion between the charges
`=` Force on charge `q`
`q. mv = sqrt((mkQq)/(r )) = sqrt((Qqm)/(4pi in_(0)r))`
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Knowledge Check

  • A charged particle of mass m and charge q is released from rest in an electric field of constant magnitude E . The kinetic energy of the particle after time t is

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