# An oil drop of 12 excess electrons is held stationary under a constant electric field of

Question:

An oil drop of 12 excess electrons is held stationary under a constant electric field of $2.55 \times 10^{4} \mathrm{~N} \mathrm{C}^{-1}$ in Millikan's oil drop experiment. The density of the oil is $1.26 \mathrm{~g} \mathrm{~cm}^{-3}$. Estimate the radius of the drop. $\left(\mathrm{g}=9.81 \mathrm{~m} \mathrm{~s}^{-2} ; e=1.60 \times 10^{-19} \mathrm{C}\right)$.

Solution:

Excess electrons on an oil drop, n = 12

Electric field intensity, E = 2.55 × 104 N C−1

Density of oil, ρ = 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 E is equal to the weight of the oil drop (W)

F = W

Eq mg

Ene $=\frac{4}{3} \pi r^{3} \times \rho \times \mathrm{g}$

Where,

q = Net charge on the oil drop = ne

= Mass of the oil drop

= Volume of the oil drop × Density of oil

$=\frac{4}{3} \pi r^{3} \times \rho$

$\therefore r=\sqrt[3]{\frac{3 E n e}{4 \pi \rho g}}$

$=\sqrt[3]{\frac{3 \times 2.55 \times 10^{4} \times 12 \times 1.6 \times 10^{-19}}{4 \times 3.14 \times 1.26 \times 10^{3} \times 9.81}}$

$=\sqrt[3]{946.09 \times 10^{-21}}$

$=9.82 \times 10^{-7} \mathrm{~m}$

= 9.82 × 10−4 mm

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