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Two identified 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.
623081_88547843f3444aa096cfa4e12d3d9bb8.17

Solution
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Initially charge on either capacitor is QA=QB=CV .
After dielectric is introduced, new capacitance of either capacitor =KC
After opening switch potential across capacitor A is V volts.
Let potential across capacitor B be V1
Therefore QB=CV=C1V1=KCV1
V1=VK
Initial energy in both capacitors =CV22+CV22=CV2
Final energy of capacitor A =KCV22
Final energy of capacitor B =KCV22K2=CV22K
Total final energy of both capacitors = KCV22+CV22K=(K2+12K)CV2
Ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric= CV2(K2+12K)CV2=2KK2+1

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