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Question
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.
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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
`"V"_1 = "V"/"K"`
Initial energy in both capacitors = `"CV"^2/2 + "CV"^2/2 = "CV"^2`
Final energy of capacitor A = `"KCV"^2/2`
Final energy of capacitor B = `"KCV"^2/"2K"^2 = "CV"^2/"2K"`
Total final energy of both capacitors = `"KCV"^2/2 + "CV"^2/"2K" = (("K"^2 + 1)/"2K") "CV"^2`
Ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric = `"CV"^2/((("K"^2 + 1)/"2K") "CV"^2) = "2K"/("K"^2 + 1)`
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