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30 November, 16:23

A particle of charge and mass kg is released from rest in a region where there is a constant electric field of + 668 N/C. What is the displacement of the particle after a time of?

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  1. 30 November, 16:44
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    The displacement of the particle after a time of 3.64 * 10⁻² s = 0.366 m

    Explanation:

    Given q = 19.6 μC = 19.6 * 10^-6 C

    m = mass of the charge = 2.37 * 10^-5 kg

    E = 668 N/C

    We first calculate the force on the charge

    F = Electric field * Charge = Eq = 668 * 19.6 * 10⁻⁶ = 0.0131 N

    Then, using Newton's law of motion's F = ma

    We can obtain the acceleration of the particle

    0.0131 = 2.37 * 10⁻⁵ * a

    a = 0.0131 / (2.37 * 10⁻⁵) = 552.4 m/s²

    Then, using the equations of motion, we can obtain the displacement of the charge after t = 3.64 * 10⁻² = 0.0364 s

    u = initial velocity = 0 m/s (since the particle was initially at rest)

    t = 0.0364 s

    a = 552.4 m/s²

    d = displacement of the particle = ?

    d = ut + at²/2

    d = (0) (0.0364) + (552.4) (0.0364²) / 2 = 0 + 0.366 = 0.366 m
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