Stokes' Law and Significance of Millikan's Results Revision Notes | A Level AQA Physics

1. Why This Experiment Changed Physics

In 1909, Robert Millikan suspended tiny oil droplets between charged plates and measured something nobody had pinned down before: the exact charge on a single electron. That number — $1.6 \times 10^{-19}$ C — is now one of the most important constants in all of physics.

To get there, Millikan needed to know the size of each droplet precisely. That is where Stokes' Law comes in. Before we reach the significance of the results, we need to understand the physics that made the measurement possible.

2. Stokes' Law — The Viscous Drag on a Sphere

Picture a tiny sphere moving slowly through a thick fluid — honey, oil, or even air. The fluid resists the sphere's motion. George Gabriel Stokes showed in 1851 that, for a small sphere moving at low speed through a viscous fluid with laminar (smooth, non-turbulent) flow, the drag force has a beautifully simple form.

Notice that the drag force depends linearly on both $r$ and $v$. Double the radius and you double the drag; double the speed and you double the drag. This linear relationship is what makes Stokes' Law tractable for calculation — and it is only valid for small, slow spheres where the flow remains laminar.

The key assumption behind Stokes' Law is that the flow around the sphere is laminar — the fluid moves in smooth layers with no turbulence. For the microscopic oil droplets in Millikan's experiment (radii of order $10^{-6}$ m moving at fractions of a millimetre per second), this condition is comfortably satisfied.

3. Using Terminal Velocity to Find Droplet Radius

Here is the clever part. When an oil droplet falls through air with no electric field applied, it accelerates briefly and then reaches terminal velocity. At that point, the net force is zero: the downward weight exactly balances the upward drag plus upthrust.

Writing out the force balance at terminal velocity (and noting that upthrust from air is small but included for precision):

Rearranging for the droplet radius $r$ gives:

This is the critical link between Stokes' Law and Millikan's experiment. By timing how fast a droplet falls, Millikan could calculate its radius — and from the radius, its mass. Knowing the mass was essential for the next stage: measuring the charge.

Worked example: Calculating Oil Droplet Radius Using Stokes' Law

An oil droplet falls through air at terminal velocity $v_0 = 1.2 \times 10^{-4}$ m s$^{-1}$. The dynamic viscosity of air is $\eta = 1.8 \times 10^{-5}$ Pa s, the density of the oil is $\rho = 900$ kg m$^{-3}$, and the density of air is $\rho_0 = 1.2$ kg m$^{-3}$. Calculate the radius of the droplet.

1. $$r = \sqrt{\frac{9\eta v_0}{2(\rho - \rho_0)g}}$$ — This comes from equating Stokes' drag to the net downward force (weight minus upthrust) at terminal velocity. [M1]
2. $\rho - \rho_0 = 900 - 1.2 = 898.8 \approx 899$ kg m$^{-3}$ — Air density is much smaller than oil density, so this is close to 900, but include it for accuracy.
3. $9 \times (1.8 \times 10^{-5}) \times (1.2 \times 10^{-4}) = 9 \times 1.8 \times 1.2 \times 10^{-9}$ $= 9 \times 2.16 \times 10^{-9} = 1.944 \times 10^{-8}$ — Multiply the constants first, then combine the powers of ten. $9 \times 2.16 = 19.44$, giving $1.944 \times 10^{-8}$.
4. $2 \times 899 \times 9.8 = 2 \times 8810.2 = 17620.4$ — $899 \times 9.8$: compute $900 \times 9.8 = 8820$, subtract $1 \times 9.8 = 9.8$, giving $8810.2$. Then multiply by 2.
5. $\frac{1.944 \times 10^{-8}}{17620.4} = 1.103 \times 10^{-12}$ m$^2$ — $1.944 \div 17620.4 \approx 1.103 \times 10^{-4}$, and $10^{-8} \div 10^{4} = 10^{-12}$. [M1]
6. $r = \sqrt{1.103 \times 10^{-12}} = \sqrt{1.103} \times 10^{-6} \approx 1.05 \times 10^{-6}$ m — $\sqrt{1.103} \approx 1.050$. The radius is approximately $1.05 \times 10^{-6}$ m, or about 1 micrometre — consistent with a fine oil mist. [A1]

Answer: $r \approx 1.05 \times 10^{-6}$ m (1.05 μm)

4. Millikan's Oil-Drop Experiment

Now that we can find the droplet radius from its terminal velocity, we have everything we need to understand Millikan's full method. The experiment has two distinct phases: measuring the droplet's mass (using Stokes' Law as above), and then measuring the charge on the droplet by applying an electric field.

The elegance of the stationary-droplet condition is worth pausing on. When the droplet hangs motionless, there is no drag force at all — the electric force and gravity are in perfect balance. This means the charge calculation $q = mgd/V$ does not require Stokes' Law at this stage; Stokes' Law was only needed earlier to find $r$ and hence $m$.

5. Quantisation of Electric Charge

After repeating his measurements on hundreds of droplets, Millikan noticed a striking pattern. The charges he measured were never random. Every single value was a whole-number multiple of the same tiny quantity: $1.6 \times 10^{-19}$ C.

He never measured a charge of $2.4 \times 10^{-19}$ C, for instance — that would be 1.5 times the fundamental unit, and it did not appear. Charges came in steps: $1e$, $2e$, $3e$, $4e$, and so on, where $e = 1.6 \times 10^{-19}$ C.

Think of it like money. You can have 10p, 20p, or 30p — but not 15p, because coins come in fixed denominations. Charge works the same way: it comes in fixed units of $e$, and you cannot have a fraction of one.

This was a profound result. Before Millikan, the idea that charge might be quantised was theoretical — J.J. Thomson had discovered the electron in 1897 and proposed it carried a fixed charge, but nobody had measured that charge directly on individual particles. Millikan's experiment turned a hypothesis into a measured fact.

6. Significance of Millikan's Results

The significance of Millikan's experiment extends well beyond a single number. It represents a turning point in physics for several interconnected reasons, and AQA expects you to be able to evaluate these reasons with precision.

• First direct measurement of the elementary charge $e$: Millikan obtained $e = 1.6 \times 10^{-19}$ C, a value that remains accurate to within the precision of his apparatus.
• Experimental proof of charge quantisation: The data showed unambiguously that charge comes in discrete units — it is not a continuous quantity. This confirmed a key prediction of atomic theory.
• Determination of electron mass: Combined with Thomson's earlier measurement of the charge-to-mass ratio $e/m$ for the electron, Millikan's value of $e$ allowed the electron mass to be calculated for the first time: $m_e = e \div (e/m) \approx 9.11 \times 10^{-31}$ kg.
• Foundation for quantum theory: The quantisation of charge was an early and direct demonstration that nature operates in discrete steps at the atomic scale — a principle that underpins all of quantum mechanics.
• Validation of the atomic model: The result supported the picture of matter as composed of discrete charged particles, strengthening the case for the nuclear model of the atom.

When evaluating significance in an exam, the strongest answers connect multiple consequences. Millikan did not merely measure a constant — he demonstrated a fundamental principle (quantisation) and enabled a cascade of further measurements (electron mass) that shaped twentieth-century physics.

Questions on this topic tend to fall into three categories: Stokes' Law calculations, description of Millikan's method, and evaluation of significance. Each has its own pitfalls.

7. Exam Technique and Common Mistakes

Stokes' Law and Millikan's experiment together illustrate how a single, carefully designed measurement can unlock multiple fundamental constants and principles. The radius formula connects macroscopic observations (a falling droplet's speed) to microscopic quantities (the charge on a single electron) — a chain of reasoning that exemplifies the power of physics at its best.