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

The plates of a parallel plate capacitor are charged upto 100V. now, after removing the battery, a 2mm thick plate is inserted between the plates. Then, to maintain the same potential difference, the distance between the capacitor plates is increased by 1.6mm. The dielectric constant of the plate is :

A
5
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1.25
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2.5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 5
As battery is disconnected, so charge will remain the same.
It is given that final potential is the same.
So final capacitance should be C1=C2
ε0Ad=ε0A(1.6+d)t(11/K)K=5

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