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27 October, 13:38

A parallel-plate capacitor of capacitance 20 µF is fully charged by a battery of 12 V. The battery is then disconnected. A dielectric slab of K = 4 is slipped between the two plates of the capacitor:

(a) Find the change in potential energy of the capacitor.

(b) Does the potential energy increase or decrease? Explain

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  1. 27 October, 14:22
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    capacitance = 20 x 10⁻⁶ F.

    potential V = 12 V

    charge = CV

    = 20 x 10⁻⁶ x 12

    Q = 240 x 10⁻⁶ C

    energy = Q² / 2C

    = (240 x 10⁻⁶) ² / 2 x 20 x 10⁻⁶

    = 1440 x 10⁻⁶ J

    b)

    In this case charge will remain the same but capacity will be increased 4 times

    new capacity C = 4 x 20 x 10⁻⁶

    = 80 x 10⁻⁶

    energy = Q² / 2C

    = (240 x 10⁻⁶) ² / 2 x 4 x 20 x 10⁻⁶

    = 360 x 10⁻⁶ J.

    potential energy will decrease from 1440 x 10⁻⁶ J to 360 x 10⁻⁶ J
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