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2 November, 10:08

Two parallel plates are separated by 0.1 mm. A 10 V potential difference is maintained between those plates. If a proton is released from the positive plate, calculate the kinetic energy of the proton when it reaches the negative plate.

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  1. 2 November, 12:53
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    K. E = 1.6 x 10⁻¹⁸ J

    Explanation:

    given,

    thickness between two plate = t = 0.1 mm

    Voltage difference between two plate = 10 V

    charge of proton = q = 1.6 x 10⁻¹⁹ C

    When the charge moves from positive to negative the potential energy reduces to kinetic energy

    K. E = Δ PE

    K. E = q Δ V

    K. E = 1.6 x 10⁻¹⁹ x 10

    K. E = 1.6 x 10⁻¹⁸ J

    so, the kinetic energy of the proton when it reaches negative plate is equal to K. E = 1.6 x 10⁻¹⁸ J
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