Determination of charge of an electron - Millikan's oil drop experiment

Once the specific charge $e/m$ of the electron was known from Thomson's experiment, the next step was to measure the electron's actual charge $e$ itself. This was accomplished by R.A. Millikan using his celebrated oil drop experiment.

Apparatus. Two horizontal circular metal plates A and B, each about 20 cm in diameter and 1.5 cm apart, are enclosed in a glass-walled chamber and connected to a potential difference of about 10 kV, so the electric field between them points vertically. An atomizer sprays a fine mist of a highly viscous oil (such as glycerine) through a small hole in the upper plate; some droplets pick up a small negative charge from friction with air or from exposure to X-rays. The chamber is lit from the side, and a microscope perpendicular to the light lets an observer track a single chosen drop.

Step 1 - radius of the drop (field off). With the electric field switched off, the drop falls under gravity, opposed by buoyancy and by Stokes' viscous drag, quickly reaching a constant terminal velocity $v$. Balancing forces, $F_g=F_b+F_v$, i.e. $\tfrac{4}{3}\pi r^3\rho g=\tfrac{4}{3}\pi r^3\sigma g+6\pi r\eta v$ (where $\rho$ is the oil's density, $\sigma$ the density of air and $\eta$ the viscosity of air), gives the droplet's radius:

Step 2 - charge of the drop (field on). The field is switched on and adjusted so the same drop hovers stationary in the field of view, meaning there is no viscous force and the net upward electric force exactly balances the drop's weight minus buoyancy: $qE=\tfrac{4}{3}\pi r^3(\rho-\sigma)g$, so

Substituting the radius found in Step 1 gives the working formula

$q=18\pi\left[\frac{\eta^3v^3}{2g(\rho-\sigma)}\right]^{1/2}\frac{1}{E}$ …