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Question:

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.

Solution:

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=V/K
Initial energy in both capacitors = CV²/2 + CV²/2 = CV²
Final energy of capacitor A = KCV²/2
Final energy of capacitor B = (CV²/K) /2 = CV²/2K
Total final energy of both capacitors = KCV²/2 + CV²/2K = (K² + 1)/2K * CV²
Ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric = CV² / ((K² + 1)/2K * CV²) = 2K/(K² + 1)