Electric force on a dust particle having charge equal to 8X10-19 C when plates are separated by a distance of 2cm and have a potential difference of 5 kV is

8×10⁻¹⁷ N

Explanation:

from the question, Electric force is given as

F = QV/r Equation 1

Where F = Electric Force,  Q = Charge, V = Electric potential, r = distance.

Given: Q = 8×10⁻¹⁹ C, V = 5 kV = 5000 V, r = 2 cm = 0.02 m.

Substitute into equation 1

F = 8×10⁻¹⁹(5000)(0.02)

F = 8×10⁻¹⁷ N

Hence the electric force on the dust particle is  8×10⁻¹⁷ N

answer: a ball

explanation: because it will bounce

i'm just gonna leave this comment here, so i can get a challange done-

explanation: