An empty parallel plate capacitor is connected between the terminals of a 9.0-V battery and charged up. The capacitor is then disconnected f

Question

An empty parallel plate capacitor is connected between the terminals of a 9.0-V battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor

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Thành Công 3 years 2021-08-14T23:12:15+00:00 1 Answers 123 views 0

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  1. thienthanh
    0
    2021-08-14T23:13:23+00:00
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    Answer:

    The new voltage between the plates of the capacitor is 18 V

    Explanation:

    The charge on parallel plate capacitor is calculated as;

    q = CV

    Where;

    V is the battery voltage

    C is the capacitance of the capacitor, calculated as;

    C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}

    q = \frac{\epsilon _0A V}{d}

    where;

    ε₀ is permittivity of free space

    A is the area of the capacitor

    d is the space between the parallel plate capacitors

    If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;

    q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d} = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V

    Therefore, the new voltage between the plates of the capacitor is 18 V

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