Question 25

An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10^{4} 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 × 10^{4} N C^{−1}

Density of oil, ρ = 1.26 gm/cm^{3} = 1.26 × 10^{3} kg/m^{3}

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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Rupesh
2019-06-15 17:23:02

Thank u sir

Ujwal
2019-06-11 21:54:12

Nice work

Ujwal
2019-06-11 21:53:53

Nice work

Atul Gupta
2019-05-15 10:47:37

nice solution

Sudeep
2019-04-08 15:36:35

The answer is 9.89m sir

Lolu
2017-11-04 16:23:35

Lol...nice one...Rofl

Sivani Gupta
2017-08-23 08:27:12

Good

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