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27 January, 04:25

A capacitor with plates separated by a distance d is charged to a potential difference ΔVC. All wires and batteries are disconnected, then the two plates are pulled further apart to a new separation distance 2d. What happens to the potential difference across the capacitor, ΔV, and the charge on the capacitor, Q, as a result of this change?

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  1. 27 January, 05:41
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    Initial separation of plate = d

    final separation = 2d

    The capacitance of the capacitor will reduce from C to C/2 because

    capacitance = ε A / d

    d is distance between plates.

    As the batteries are disconnected, charge on the capacitor becomes fixed.

    Initial charge on the capacitor

    = Capacitance x potential difference

    Q = C ΔV

    Final charge will remain unchanged

    Final charge = C ΔV

    Final capacitance = C/2

    Final potential difference = charge / capacitance

    = C ΔV / C/2

    = 2 ΔV

    Potential difference is doubled after the pates are further separated.
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