CBSE Class 12 Physics 2017 Outside Delhi Set 1 Paper

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Question : 22
Total: 26
Two identical parallel plate capacitors A and B are connected to a battery of V volts with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.
Solution:  
Formula for energy stored
Energy stored before
Energy stored after
Ratio
Energy stored =‌
1
2
C
V2
(=‌
1
2
‌
Q2
C
)

Net capacitance with switch S closed
‌‌‌‌‌=C+C=2C
Energy stored =‌
1
2
×2C
×V2
=CV2

After the switch S is opened, capacitance each capacitor of =KC
∴ Energy stored in capacitor A=‌
1
2
K
C
V2

For capacitor B,
Energy stored =‌
1
2
‌
Q2
KC
=‌
1
2
‌
C2V2
KC
=‌
1
2
‌
CV2
K

∴ Total Energy stored
‌=‌
1
2
K
C
V2
+‌
1
2
‌
CV2
K
=‌
1
2
C
V2
(K+‌
1
K
)

‌=‌
1
2
C
V2
(‌
K2+1
K
)

∴ Required ratio =‌
2CV2â‹…K
CV2(K2+1)
=‌
2K
(K2+1)
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