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

A parallel-plate capacitor is connected to a battery as shown in the figure. A conducting plate is inserted mid-way between the two plates. Take the plate area as A. The distance between each plate in the new configuration is d.

If the middle and upper plates are short circuited, the extra charge flown through the battery is

A
ε0AVd
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Zero
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
ε0AV2d
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2ε0AVd
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C ε0AV2d
We have,
Ci=ε0A2d

Qi=CiVi
Qi=ε0A2d×V

Now upper plates are short circuited,

So, only one plate remains in picture and the potential difference across it is V.
Cf=ε0Ad
Qf=CfVf
Qf=ε0Ad×V

So, the extra charge flown is
QfQi
ε0AdVε0AV2d
=ε0AV2d

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