wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

KQqmR

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

2KQqmR

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
KQq3mR
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B

KQqmR


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

V=kQR2+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=kQR2+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.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon