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-

Question

4 years ago

15

directed electric field of magnitude 670 n/c ?

1 answer:

Kipish [7] · Answer 1 · 2022-01-08T20:32:11+03:00

Kipish [7]4 years ago

6 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