12/ph/c1/25

Question 1.25: An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ N C⁻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).

Answer :

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

 Ene 

Where,

q = Net charge on the oil drop = ne

m = Mass of the oil drop

= Volume of the oil drop × Density of oil

= 9.82 × 10⁻⁴ mm

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

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