Electric field strength on a dust particle having charge equal to 8 × 10 - 19 when plates are separated by distance of 2 cm and have a potential difference of 5 kV is {{}}

Question

Quizplus · Accepted Answer

The electric field strength (E) between two plates is given by the potential difference (V) divided by the distance (d) between them, E = V/d. Here, V = 5 kV = 5000 V and d = 2 cm = 0.02 m. So, E = 5000 / 0.02 = 250,000 V/m. The force (F) on a charge (q) in an electric field is given by F = qE. Here, q = 8 × 10^-19 C. So, F = 8 × 10^-19 C × 250,000 V/m = 2.0 × 10^-13 N.

Electric Field Strength on a Dust Particle Having Charge Equal