Question

In the Millikan oil-drop experiment, consider a droplet having terminal velocity to fall 0.240 cm in 18 s with the E field deactivated. Find the radius of the droplet for = 891 kg/m3, and =1.80 X 10-5 kg/m.s.

Answer 1

$\frac 43 \rho \pi r^3=\frac{6\pi r\eta v}g,\implies$

$r=\sqrt{\frac{9\eta v}{2\rho g}}=1.1\cdot 10^{-6}~m.$