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).
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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Thank u sir
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Nice work
nice solution
The answer is 9.89m sir
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