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31 January, 10:57

An empty parallel plate capacitor is connected between the terminals of a 10.1-V battery and charges 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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  1. 31 January, 11:10
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    V' = 20.2 V

    Explanation:

    The formula for the capacitance of a parallel plate capacitor, with air as a medium between the plates, is given by the following formula:

    C = A∈₀/d

    where,

    V = Capacitance of capacitor

    A = Area of plate

    ∈₀ = permittivity pf free space

    d = distance between plates

    Keeping all the other variables constant, the relationship between capacitance and the distance between plates is given as:

    C α 1/d

    Therefore, when the distance between plates is doubled, it implies that the capacitance of the capacitor becomes half. That is:

    C' = C/2

    Now the formula for capacitance of a capacitor, is also given as:

    Q = CV

    C = Q/V

    where,

    Q = charge stored in capacitor

    V = Voltage across capacitor

    Keeping the charge constant the relationship between capacitance and voltage is:

    C α 1/V

    Therefore, when the capacitance is halved, the voltage must be doubled, keeping the charge constant:

    V' = 2V = (2) (10.1 V)

    V' = 20.2 V
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