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19 August, 05:48

A 6.0-μF air capacitor is connected across a 100-V battery. After the battery fully charges the capacitor, the capacitor is immersed in transformer oil (dielectric constant = 4.5). How much additional charge flows from the battery, which remained connected during the process?

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  1. 19 August, 05:59
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    21 x 10^-4 C

    Explanation:

    C = 6 micro Farad = 6 x 10^-6 F

    V = 100 V

    Charge, q = C V = 6 x 10^-6 x 100 = 6 x 10^-4 C

    As the dielectric is inserted, the capacitance is increased.

    C' = K C = 4.5 x 6 x 10^-6 = 27 x 10^-6 F

    As the battery remains connected, so the voltage across the plates remains same and charge becomes q'.

    V = q' / C'

    q' = C' V = 27 x 10^-6 x 100 = 27 x 10^-4 C

    Amount of additional charge = q' - q = 27 x 10^-4 - 6 x 10^-4 = 21 x 10^-4 C
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