Question

Determine the intensity of electric field which can balance a drop of oil mass of 0.001 mg carrying electronic charge

Answer 1

Fe = mg

q E = mg

E = mg/q

q = 10^-6gram = 10^-9KG

g = 10

q = 3e = 3 × 1.6 × 10^-19

Put Value

E = 2 × 10^10V/m