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13 April, 12:13

A parallel-plate capacitor is charged and then disconnected from the battery. By what factor does the stored energy change when the plate separation is then doubled?

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  1. 13 April, 12:33
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    U/U₀ = 2

    (factor of 2 i. e U = 2U₀)

    Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected

    Explanation:

    Energy stored in a capacitor can be expressed as;

    U = 0.5CV^2 = Q^2/2C

    And

    C = ε₀ A/d

    Where

    C = capacitance

    V = potential difference

    Q = charge

    A = Area of plates

    d = distance between plates

    So

    U = Q^2/2C = dQ^2/2ε₀ A

    The initial energy of the capacitor at d = d₀ is

    U₀ = Q^2/2C = d₀Q^2/2ε₀ A ... 1

    When the plate separation is increased after the capacitor has been disconnected, the charge Q of the capacitor remain constant.

    The final energy stored in the capacitor at d = 2d₀ is

    U = 2d₀Q^2/2ε₀ A ... 2

    The factor U/U₀ can be derived by substituting equation 1 and 2

    U/U₀ = (2d₀Q^2/2ε₀ A) / (d₀Q^2/2ε₀ A)

    Simplifying we have;

    U/U₀ = 2

    U = 2U₀

    Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected.
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