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Question 25

An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 x 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 x 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, ÃÂ = 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 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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