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JEE Main Physics Class 12 Electrostatic Potential and Capacitance Part 1 Questions
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© examsnet.com
Question : 42
Total: 100
A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20V. If a dielectric material of dielectric constant
k
=
5
3
is inserted between the plates, the magnitude of the induced charge will be:
[2018]
1.2 n C
0.3 n C
2.4 n C
0.9 n C
Validate
Solution:
Charge on Capacitor,
Q
i
=
C
V
After inserting dielectric of dielectric constant
=
K
Q
f
=
(
k
C
)
V
Induced charges on dielectric
Q
ind
=
Q
f
−
Q
i
=
K
C
V
−
C
V
(
K
−
1
)
C
V
=
(
5
3
−
1
)
×
90
p
F
×
2
V
=
1.2
n
c
© examsnet.com
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