Ask Question
8 August, 00:00

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?

+1
Answers (1)
  1. 8 August, 00:41
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers