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25 March, 21:34

A parallel plate capacitor is created by placing two large square conducting plates of length and width 0.1m facing each other, separated by a 1 cm gap with vaccum between the plates. We connect the capacitor to power supply, charge it to a potential difference VO 5 kV, and disconnect the power supply. We then insert a sheet of insulating plastic material between the plates, completely filling the space between them. We find that the potential difference decreases to 1 kV while the charge on each capacitor plate remains constant. Find the original capacitance CO. 885pf 17.7pf 8.85pf 1.77pf

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  1. 25 March, 21:58
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    8.854 pF

    Explanation:

    side of plate = 0.1 m,

    d = 1 cm = 0.01 m,

    V = 5 kV = 5000 V

    V' = 1 kV = 1000 V

    Let K be the dielectric constant.

    So, V' = V / K

    K = V / V' = 5000 / 1000 = 5

    C = ε0 A / d = 8.854 x 10^-12 x 0.1 x 0.1 / 0.01 = 8.854 x 10^-12 F

    C = 8.854 pF
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