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17 July, 13:37

Each plate of a parallel-plate capacator is a square with side length r, and the plates are separated by a distance d. The capacitor is connected to a source of voltage, V. A plastic slab of thickness d and dielectric constant K is inserted slowly between the plates over the time period Δt until the slab is squarely between the plates. While the slab is being inserted, a current runs through the battery/capacitor circuit.

Assuming that the dielectric is inserted at a constant rate, find the current I as the slab is inserted.

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  1. 17 July, 17:12
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    Before the dialectic was inserted the capacitor is Co

    When the slab is inserted,

    The capacitor becomes

    C=kCo

    The charge Q is given as

    Q=CV

    Then, when C=Co

    Qo=CoV

    Then, when C=kCo

    Q=kCoV

    Then, the change in charges is given as

    Q-Qo = kCoV - CoV

    ∆Q = kCoV - CoV

    Current is given as

    I=dQ/dt

    I = (kCoV - CoV) / dt

    I=Co (kV-V) / dt

    Note Co is the value capacitor

    So, Capacitance of parallel plates capacitor is given as

    Co=εoA/d

    Then,

    I=εoA (kV-V) / d•dt

    I=VεoA (k-1) / d•dt

    Where A=πr²

    I = V•εo•πr²• (k-1) / d•dt

    This is the required expression for current is in the required term
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