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Question

# A point charge −q of mass m is released with negligible speed from a distance √3R on the axis of a fixed uniformly charged ring of charge Q and radius R. Find out the velocity of point charge when it reaches the centre of ring.

A

KQq2mR

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B

KQqmR

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C

2KQqmR

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D
KQq3mR
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Solution

## The correct option is B √KQqmR As potential due to uniformly charged ring along its axis (at x distance) is : V=kQ√R2+x2 [∵k=14πε0] Let A be the point where −q charge is placed along the axis. So, potential at point A due to ring: VA=kQ√R2+3R2 ⇒VA=kQ2R Potential at point B due to ring: VB=kQR So, potential energy of charge −q at point A, P.EA=−kQq2R...(1) So, potential energy of charge −q at point B: P.EB=−kQqR...(2) Now by energy conservation: P.EA+K.EA=P.EB+K.EB...(3) Since, the particle is released with negligible speed at point A, so its kinetic energy at this point will be zero i.e., K.EA=0...(4) Let the speed at the centre ( at point B) be v. K.EB=12mv2...(5) Using (1),(2),(3),(4) and (5), we get −kQq2R+0=−kQqR+12mv2 ⇒12mv2=kQq2R ∴v=√kQqmR Hence, option (b) is the correct answer.

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