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

A parallel-plate capacitor has plates of area A and separation d and is charged to a potential difference V. The charging battery is then disconnected and the plates are pulled apart until their separation is 2d. What is the work required to separate the plates?

A
20AV2d
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0AV2d
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
30AV22d
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0AV22d
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 0AV22d
Let C1 and C2 be the capacitance of capacitor at separation d and 2d respectively.
C1=Aϵ0d and C2=Aϵ02d
As there is no transfer of charge in this case thus, charge will remain conserved.
q1=q2=q
We know that q = CV so,
q=Aϵ0d×V---------(i)
We know that work done is given by change in potential energy
W.D=ΔU=UfUi
W.D=q22C2q22C1
using equ (i) and putting the values of C1 and C2 we get
W.D = ϵ0AV22d

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon