Addition of a metal slab of thickness “a” between the plates of a parallel plate capacitor of plate separation “d” is equivalent to introduc

Question

Addition of a metal slab of thickness “a” between the plates of a parallel plate capacitor of plate separation “d” is equivalent to introducing a dielectric with dielectric constant “K” between the plates. Find an expression for “K” in terms of “a” and “d”?

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Neala 4 years 2021-08-17T19:41:25+00:00 1 Answers 13 views 0

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  1. yenoanh
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    2021-08-17T19:43:21+00:00
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    Answer:

    K = \frac{d}{d+a}

    Explanation:

    The capacitance of a capacitor in terms of the dielectric constant, area of the plate and the distance separating the plate is given by:

    C = \frac{\epsilon A}{d}

    Where A = Area of the plate

    d = distance between the plates

    \epsilon = dielectric constant

    Case 1:

    When a meta slab of thickness, a, is added between the plates of the parallel plate capacitor , the effective separation between the plates becomes d+a

    Therefore the capacitance of the capacitor becomes:

      C = \frac{\epsilon A}{d + a} …………………..(1)

    Case 2:

    Introducing a dielectric with dielectric constant K between the plates, the capacitance of the capacitor becomes:

    C = \frac{K\epsilon A}{d}…………………….(2)

    Equating (1) and (2)

    \frac{K\epsilon A}{d} = \frac{\epsilon A}{d+a}\\\frac{K}{d} = \frac{1}{d+a} \\K = \frac{d}{d+a}

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