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13 December, 23:36

A 2.45 nF parallel-plate capacitor is charged to an initial potential difference of 56 V and then isolated. The dielectric material between the plates has a dielectric constant of 5.6. How much work is required to withdraw the dielectric sheet? Answer in units of J

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  1. 14 December, 00:04
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    The work required to withdraw the dielectric sheet = 3.84*10⁻⁶ J

    Explanation:

    Energy Stored in a capacitor = 1/2CV²

    E = 1/2CV² ... Equation 1

    Where E = Energy stored in the capacitor, C = capacitance of the capacitor, V = potential difference.

    Note: The work required to withdraw the dielectric sheet is equal to the energy stored in the capacitor

    Given: C = 2.45 nF = 2.45 * 10⁻⁹ F, V = 56 V

    Substituting these values into equation 1

    E = 1/2 (2.45*10⁻⁹) (56) ²

    E = 1/2 (2.45*10⁻⁹) (3136)

    E = 1568 (2.45*10⁻⁹)

    E = 3841.6*10⁻⁹

    E = 3.84*10⁻⁶ J.

    Thus the work required to withdraw the dielectric sheet = 3.84*10⁻⁶ J
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