wiz-icon
MyQuestionIcon
MyQuestionIcon
4
You visited us 4 times! Enjoying our articles? Unlock Full Access!
Question

A capacitor having C=10μF; Potential across plates, V = 10V; Distance between plates 10mm; And area of plates 10cm2, has air between the plates. What is the force on one of the plates.


A
4.65×105N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
5.65×105N
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
6.65×105N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 5.65×105N
Let’s first find the charge on the plate,
Q = CV
Q=10×106×10
Q=104C
Now we know that the force will simply be F=EQ
And since E is approximately constant between the plates and is given by
E=σ2ε0
E=Q2Aε0
F=Q22Aε0
F=(104)22×10×104×8.854×1012
F=5.65×105N

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Parallel Plate Capacitor
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon