In Millikan's oil drop experiment, an oil drop carrying a charge Q is held stationary by a potential difference 2400 V between the plates. To keep a drop of half the radius stationary, the potential difference had to be made 600 V . What is the charge on the second drop?

Step-by-step solution

Force on charge, F=Q E or F= Q V d ( V=E d) For drop to be stationary, weight of drop = force due to charge i.e., m g= Q V d For two drops, Q₁ Q₂ V₁ V₂ = m₁ m₂ Q₁ Q₂ V₁ V₂ = 4 3 r₁^3 4 3 r₂^3 Q₂ Q₁ = r₂^3 r₁^3 V₁ V₂ Q₂ Q = (r / 2)^3 r^3 2400 600 Q₂= Q 2