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

Since the drop is stationary, force $\left(F_E\right)$ due to electric field $E$ is
equal to the weight of the oil drop $\left(W\right)$
Let the radius of the oil drop be $r$
Mass of the oil drop $\left(m\right)$= Volume of the drop $\times$ Density of oil

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

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

Electrostatic Force $F_E=qE$

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

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

$q$= Net charge on the oil drop =$ne$= $12\times 1.6\times {10}^{-19}C=19.2\times {10}^{-19}C$

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

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

Acceleration due to gravity, $g=9.81m/s^2$

Putting the values in

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

$\Rightarrow r=\sqrt[3]{3\frac{3Ene}{4\pi \rho g}}$

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

$=\sqrt[3]{3946.09\times {10}^{-21}}=9.81\times {10}^{-7}m$