Stokes' law and terminal velocity in Millikan's experiment

Turning Points in Physics | AQA A-Level Physics

Key Definition

Stokes' law: The viscous drag force on a small sphere moving through a fluid at low speed is $F = 6\pi \eta r v$, where $\eta$ is the viscosity, $r$ is the radius, and $v$ is the speed.

Finding the mass of the drop

To use the stationary condition $Q = mgd/V$, Millikan needed to know the mass of each oil drop. The drops are far too small to weigh directly.
Instead, the electric field is switched off and the drop is allowed to fall freely through the air. It quickly reaches terminal velocityThe constant velocity reached when the driving force (here, gravity) is exactly balanced by resistive forces (here, viscous drag). At terminal velocity, acceleration is zero. $v_t$.
At terminal velocity, the weight equals the drag force:

$$mg = 6\pi \eta r v_t$$

Calculating the radius and mass

The mass of a spherical drop is $m = \frac{4}{3}\pi r^3 \rho$, where $\rho$ is the density of oil.
Substituting into the terminal velocity equation:

$$\frac{4}{3}\pi r^3 \rho g = 6\pi \eta r v_t$$

Solving for the radius:

$$r = \sqrt{\frac{9 \eta v_t}{2 \rho g}}$$

Once $r$ is known, the mass follows from $m = \frac{4}{3}\pi r^3 \rho$.
The terminal velocity $v_t$ is measured by timing the drop as it falls a known distance, observed through the microscope with a calibrated graticule.

Conditions for Stokes' law to apply

The sphere must be small and smooth.
The flow around it must be laminar (not turbulent).
The speed must be low. Oil drops in Millikan's experiment satisfy all three conditions.

Common Mistake

When deriving the radius, students often forget to cancel $\pi$ and $r$ from both sides before rearranging. Write out the cancellation explicitly: $\frac{4}{3}\cancel{\pi} r^{\cancel{3}\,2} \rho g = 6\cancel{\pi} \eta \cancel{r}\, v_t$. This avoids algebraic errors.