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31 March, 22:50

A 43 V battery charges a 4750 pF air gap capacitor. After disconnecting. porcelain (k-7) is placed between the plates. What is the resulting charge of the plates? O 154 HC O 144 AC O 1.73 HC O 1.23 AC s

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  1. 1 April, 00:48
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    1.2255 * 10⁻⁶ μC

    Explanation:

    Given

    Capacitance, C = 4750 pF = 4750*10⁻¹² F

    Potential difference, V = 43 V

    Relative permittivity of the dielectric, k = 7

    now we know

    charge Q = CV

    ⇒Q = 4750*10⁻¹² * 43 = 2.0425 * 10⁻⁷ C

    Now, the charge (Q') due to the placing of the dielectric in between the gap

    Q' = C'V

    where,

    C' = kC = 7 * 4750*10⁻¹² = 33250 * 10⁻¹² F

    substituting the C' in Q'

    we get

    Q' = 33250 * 10⁻¹² * 43 = 1.42975 * 10⁻⁶ C

    Thus,

    Resulting charge will be = Q' - Q

    ⇒Resulting charge = 1.42975 * 10⁻⁶ C - 2.0425 * 10⁻⁷ C = 1.2255 * 10⁻⁶ C

    Resulting charge = 1.2255 * 10⁻⁶ μC
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