An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ NC⁻¹ in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm ⁻³. Estimate the radius of the drop. (g = 9.81 m s⁻²; e = 1.60 × 10⁻¹⁹ C).

Excess electrons on an oil drop, n = 12

Electric field intensity, E = 2.55 x 10⁴ N C^-1

Density of oil, ρ = 1.26 gm/cm³ = 1.26 x 10³ kg/m³

Acceleration due to gravity, g = 9.81 m s^-2

Charge on an electron, e = 1.6 x 10^-19 C

Radius of the oil drop = r

Force (F) due to electric field E is equal to the weight of the oil drop (W)

F = W => Eq = mg => Ene = 4/3 x π r³ x ρ x g

Where,

q = Net charge on the oil drop = ne

m = Mass of the oil drop = Volume of the drop x Density of oil =4/3 π r³ x ρ

r= [ 3Ene/4πρg]^1/3

= [ 3x 2.55 x 10⁴ x 12 x 1.6 x10^-19 /4 x 3.14 x 1.26 x 10³ x 9.81]^1/3

=[ 3946.09 x 10^–21]^1/3
=  9.82 x 10^–7 mm

Therefore, the radius of the oil drop is 9.82 x 10^–4 mm.