Question 25

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

*E*q = *m*g

*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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Aqeel Ahmad
2019-05-02 18:31:25

Nice work

Aqeel Ahmad
2019-05-02 18:31:09

The question looks so difficult

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