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CBSE Class 12 Physics 2017 Outside Delhi Set 1 Paper

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Question : 22 of 26
Marks: +1, -0
Two identical parallel plate capacitors AA and BB are connected to a battery of VV volts with the switch boSbo 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 KK. 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 =12CV2(=12Q2C)= \frac{1}{2} C V^2 \left( = \frac{1}{2} \frac{Q^2}{C} \right)
Net capacitance with switch SS closed
=C+C=2C= C + C = 2 C
Energy stored =12×2C×V2=CV2= \frac{1}{2} \times 2 C \times V^2 = C V^2
After the switch S is opened, capacitance each capacitor of =KC= K C
∴\therefore Energy stored in capacitor A=12KCV2A = \frac{1}{2} K C V^2
For capacitor B,
Energy stored =12Q2KC=12C2V2KC= \frac{1}{2} \frac{Q^2}{K C} = \frac{1}{2} \frac{C^2 V^2}{K C} =12CV2K= \frac{1}{2} \frac{C V^2}{K}
∴\therefore Total Energy stored
=12KCV2+12CV2K= \frac{1}{2} K C V^2 + \frac{1}{2} \frac{C V^2}{K} =12CV2(K+1K)= \frac{1}{2} C V^2 \left( K + \frac{1}{K} \right)
=12CV2(K2+1K)= \frac{1}{2} C V^2 \left( \frac{K^2+1}{K} \right)
∴\therefore Required ratio =2CV2⋅KCV2(K2+1)=2K(K2+1)= \frac{2 C V^2 \cdot K}{C V^2 (K^2+1)} = \frac{2 K}{ (K^2+1)}
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