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Question
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 N C−1 (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).
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Solution
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 = `4/3pi"r"^3 xx rho xx "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 pi"r"^3 xx rho`
∴ `"r" = root(3)((3"Ene")/(4pirho"g"))`
= `root(3)((3 xx 2.55 xx 10^4 xx 12 xx 1.6 xx 10^-19)/(4 xx 3.14 xx 1.26 xx 10^3 xx 9.81))`
= `root(3)(946.09 xx 10^-21)`
= `9.82 xx 10^-7` m
= 9.82 × 10−4 mm
Therefore, the radius of the oil drop is 9.82 × 10−4 mm.
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