CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A mass m=20 g has a charge q=3.0 mC. It moves with a velocity of 20 m/s and enters a region of electric field of 80 N/C in the same direction as the velocity of the mass. The velocity of the mass after 3 sec in this region is–

A
80 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
56 m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
44 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
40 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 56 m/s
Given:-
m=20 g=20×103 kg
q=3 mC=3×103 C
E=80 N/C
u=20 m/s

In an electrostatic equilibrium,

Electric force =ma

So, qE=ma

or, a=qEm=12 m/s2

From first equation of the motion we get,

v=u+at

=20+12×3

=56 m/s

flag
Suggest Corrections
thumbs-up
4
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Visualising Electric Fields - Electric Field Lines
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon