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 electric field of magnitude 670 n/c ?

Answer 1

Answer;

= -2.18 × 10^-5 C

Step-by-step 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