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Q.

A parallel plate capacitor with width 4 cm, length 8 cm and separation between the plates of 4mm is connected to a battery of 20 V. A dielectric slab of dielectric constant 5 having length 1cm, width 4 cm and thickness 4 mm is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be………0 J. (Where 0 is the permittivity of free space)

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answer is 240.

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Detailed Solution

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Ceff=ε0(7×4)4/10+5ε0(1×4)4/10×102Ceff=1.2ε0 Energy =12CeffV2=12(1.2)ε0(20)(20)=240ε0

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