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An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s2. What are the magnitude and direction of the electric field.

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2 answers

  1. E=(ma/q)
    Mass of electron=9.109e-31kg
    Charge of electron=-1.602e-19C
    so
    E=(9.109e-31kg*115m/s^2)/-1.602e-19C
    E=-6.539e-10
    formated answer
    E=6.539e-10 N/C South

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  2. Assuming I did it correctly:

    F=ma

    mass = the mass of the electron = 9.11 * 10^-31 kg

    a= 115 m/s

    plug in your known variables
    F will equal 1.05 * 10^-28 Newtons (N)

    Next, F=qE

    q equals the charge of the electron, or "point charge".
    q=1.6 * 10^-19 C

    Solve for E, so E=F/q

    plug in your values, and E will equal 6.56 *10^-10 N/C

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