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1 November, 05:05

An electron is released from rest in a uniform electric field. The electron accelerates, travelling 5.50 m in 4.00 µs after it is released. What is the magnitude of the electric field in N/C?

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  1. 1 November, 06:52
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    3.91 N/C

    Explanation:

    u = 0, s = 5.50 m, t = 4 us = 4 x 10^-6 s

    Let a be the acceleration.

    Use second equation of motion

    s = u t + 1/2 a t^2

    5.5 = 0 + 1/2 a (4 x 10^-6) ^2

    a = 6.875 x 10^11 m/s^2

    F = m a

    The electrostatic force, Fe = q E

    Where E be the strength of electric field.

    So, q E = m a

    E = m a / q

    E = (9.1 x 10^-31 x 6.875 x 10^11) / (1.6 x 10^-19)

    E = 3.91 N/C
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