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Two Identical Parallel Plate Capacitors a and B Are Connected to a Battery of V Volts with the Switch S Closed.

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प्रश्न

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.

योग
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उत्तर

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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2016-2017 (March) All India Set 1

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