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

A particle of mass 0.01 kg and charge 106 C is placed at rest in a uniform electric field 2×106 N/C and then released. The kinetic energy attained by the particle after moving a distance of 2 m is
[Assume gravity free space]

A
3 J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2 J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1 J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4 J
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 4 J
Given:
m=0.01 kg; q=106 C
E=2×106 N/C; d=2 m

So, the force F experienced by charge in the electric field

F=qE=106×2×106=2 N

On applying work - energy theorem,

Work done by all forces = change in kinetic energy

Welectric=K.EfK.Ei

Welectric=12mv2f12mv2i

charge released from rest, so vi=0

F.d=12mv2f0=K.Ef

K.Ef=2×2

K.Ef=4 J

Hence, option (d) is the correct answer.

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