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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.

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Solution

The correct option is **B**

As potential due to uniformly charged ring along its axis (at x distance) is :

√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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