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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.

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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)`
संबंधित प्रश्न
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| A capacitor is a system of two conductors separated by an insulator. The two conductors have equal and opposite charges with a potential difference between them. The capacitance of a capacitor depends on the geometrical configuration (shape, size and separation) of the system and also on the nature of the insulator separating the two conductors. They are used to store charges. Like resistors, capacitors can be arranged in series or parallel or a combination of both to obtain the desired value of capacitance. |
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OR
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