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26 February, 17:56

The velocity of a particle (m = 10 mg, q = - 4.0 μC) at t = 0 is 20 m/s in the positive x direction. If the particle moves in a uniform electric field of 20 N/C in the positive x direction, what is the particle's speed at t = 5.0 s?

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  1. 26 February, 19:48
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    Vf = - 20 m/s (-ve sign shows that the particle is moving opposite to positive x - direction).

    Explanation:

    Given:

    Vi = 20 m/s, m = 10 mg = 1 * 10⁻⁵ kg, q = - 4.0 * 10⁻⁶ C, E = 20 N/C. t = 5.0 s

    first to find Electric Force

    F = Eq = 20 * - 4.0 10⁻⁶ C = - 8 * 10⁻⁵ N (-ve sign shows that the field will push the particle opposite to positive x - direction)

    We also have F=ma

    ⇒ a = F/m = - 8 * 10⁻⁵ N / 10 * 10⁻⁵ kg = - 8 m/s² (-ve sign shows that the particle is accelerated opposite to positive x - direction)

    Now according the first equation of Motion.

    Vf = Vi + at

    Vf = 20 m/s + - 8 m/s² * 5 s

    Vf = - 20 m/s (-ve sign shows that the particle is moving opposite to positive x - direction)
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