Question

An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 N C−1 in Millikan’* 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).

Answer 1

Excess electrons on an oil drop, n = 12

Electric field intensity, E = 2.55 × 104 N C−1

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

Acceleration due to gravity, g = 9.81 m s−2

Charge on an electron, e = 1.6 × 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 or Ene = (4/3)πr3 x ρ x g

Where, q = Net charge on the oil drop = ne

m = Mass of the oil drop

= Volume of the oil drop × Density of oil = (4/3)πr3 x ρ

= 9.82 × 10−4 mm

Therefore, the radius of the oil drop is 9.82 × 10−4 mm.