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Question

An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55×104NC1 in Millikans oil drop experiment. The density of the oil is 1.26 g cm3. Estimate the radius of the drop. [g = 9.8 ms2; e=1.60×1019 C].

Solution
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Excess electrons on an oil drop, n=12

Electric field intensity, E=2.55×104NC1

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

Acceleration due to gravity, g=9.81ms2

Charge on an electron, e=1.6×1019C

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=43πr3×ρ×g

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

m= Mass of the oil drop

= Volume of the oil drop × Density of oil

=43πr3×ρ

r=33Ene4πρg

=33×2.55×104×12×1.6×10194×3.14×1.26×103×9.81

=3946.09×1021

=9.82×107m

=9.82×104mm

Therefore, the radius of the oil drop is 9.82×104mm.

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