Question

An oil drop of 12 excess electrons is held stationary under a constant electric field of $2.55\times {10}^{4}N{C}^{-1}$ in Millikans oil drop experiment. The density of the oil is $1.26gc{m}^{-3}$. Estimate the radius of the drop. [g = $9.8m{s}^{-2};e=$1.60$\times {10}^{-19}$ C].

Answer 1

Excess electrons on an oil drop, $n=12$

Electric field intensity, $E=2.55\times {10}^{4}N{C}^{-1}$

Density of oil, $\rho =1.26gm/c{m}^3=1.26\times {10}^{3}kg/m^3$

Acceleration due to gravity, $g=9.81m{s}^{-2}$

Charge on an electron, $e=1.6\times {10}^{-19}C$

Radius of the oil drop $=r$

Force (F) due to electric field E is equal to the weight of the oil drop (W).

$F=W$

$Eq=mg$

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

Where, $q=$ Net charge on the oil drop = ne

$m=$ Mass of the oil drop

$=$ Volume of the oil drop × Density of oil

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

$\therefore r=\sqrt[3]{3\frac{3E_{ne}}{4\pi \rho g}}$

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

$=\sqrt[3]{3946.09\times {10}^{-21}}$

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

$=9.82\times {10}^{-4}mm$

Therefore, the radius of the oil drop is $9.82\times {10}^{-4}mm$.