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20 November, 11:07

What must the charge (sign and magnitude) of a particle of mass 1.49 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 670 n/c?

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  1. 20 November, 12:05
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    Answer;

    = - 2.18 * 10^-5 C

    Explanation;

    m = 1.49 * 10^-3 kg

    Take downward direction as positive.

    Fg = m g

    E = 670 N/C

    Fe = q E

    Fe + Fg = 0

    q E + m g = 0

    q = - m g/E

    = - 1.49 * 10^-3 * 9.81/670

    = - 2.18 * 10^-5 C

    = - 2.18 * 10^-5 C
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