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.
√KQqmR
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.