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29 November, 23:47

A 3.65-pF capacitor is connected in series with an 8.55-pF capacitor and a 400-V potential difference is applied across the pair. What is the charge on the first (smaller) capacitor?

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  1. 30 November, 00:16
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    2.56*10⁻¹⁰ C

    Explanation:

    Note: When capacitor's are connected in series, They act like resistance connected in parallel.

    Therefore,

    1/Ct = 1/C1 + 1/C2 ... Equation 1

    Where Ct = Total Capacitance, C1 = Capacitance of the the first capacitor, C2 = Capacitance of the second capacitor.

    Given: C1 = 3.65 pF, C2 = 8.55 pF.

    Substitute into equation 1

    1/Ct = 1/3.65 + 1/8.55

    1/Ct = 0.274+0.117

    1/Ct = 0.391

    Ct = 2.56 pF.

    Recall,

    Q = CtV ... Equation 2

    Where,

    Q = charge, V = potential difference.

    Note: Since the capacitor are connected in series, the same charge flows through the first and the second capacitor.

    Given: V = 400 V, Ct = 2.56 pF = 2.56*10⁻¹² F

    Substitute into equation 2

    Q = 2.56*10⁻¹² (100)

    Q = 2.56*10⁻¹⁰ C.
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