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

An air capacitor of capacitance 10 μF is connected to a battery of 10 V. Now the space between its plates is completely filled with an insulator (K=2). The magnitude of induced charge on the dielectric is

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

The correct option is A 100 μC
Given:
C=10 μF=10×106 F
V=10 V; k=2

Initial charge on capacitor will be

Q=CV=10 μF×10 V=100 μC
Now, as battery is connected and dielectric of k=2 is introduced in between the plates, the capacitance will increased to kC, thus, the final charge on the plates will be
Qf=kCV=2×10×10=200 μC

Induced charge is given by the formula,

Qin=Qf(11K)

Qin=200(112)=100 μC

Hence, option (b) is correct.
Key concept- Induced charge on the surface of dielectric inside a capacitor.

flag
Suggest Corrections
thumbs-up
5
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon