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

#### Solution

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 = `4/3pi"r"^3 xx rho xx "g"`

Where,

q = Net charge on the oil drop = ne

m = Mass of the oil drop

= Volume of the oil drop × Density of oil

= `4/3 pi"r"^3 xx rho`

∴ `"r" = root(3)((3"Ene")/(4pirho"g"))`

= `root(3)((3 xx 2.55 xx 10^4 xx 12 xx 1.6 xx 10^-19)/(4 xx 3.14 xx 1.26 xx 10^3 xx 9.81))`

= `root(3)(946.09 xx 10^-21)`

= `9.82 xx 10^-7` m

= 9.82 × 10^{−4} mm

Therefore, the radius of the oil drop is 9.82 × 10^{−4 }mm.