Revise Right
Hold tight Hi there!

✓ Revise Right course

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

Stokes' law gives the viscous drag force on a small sphere moving slowly through a fluid: F = 6πηrv. Notes, worked examples and 15 questions answered.

Stokes' Law and Significance of Millikan's Results is part of The discovery of the electron (A-level only) in A Level AQA Physics.

6 min

Reading time

21

Lessons in this topic

A Level AQA

Pathway

Reviewed by Revise Right

Content review

What this lesson covers

Stokes' law gives the viscous drag force on a small sphere moving slowly through a fluid: F = 6πηrv. Notes, worked examples and 15 questions answered.

  • State Stokes' Law and identify each variable with correct units
  • Apply Stokes' Law to calculate the radius of an oil droplet at terminal velocity
  • Describe the method and key measurements in Millikan's oil-drop experiment
  • Explain why Millikan's results provide evidence for the quantisation of electric charge

Key ideas to keep in view

Use this lesson to stay inside The discovery of the electron (A-level only) while connecting the detail back to the wider Turning points in physics (A-level only) topic.

  • Stokes' Law: An equation giving the viscous drag force on a small sphere moving at low speed through a viscous fluid: $F = 6\pi\eta r v$. Applied when the sphere has reached terminal velocity and flow is laminar (non-turbulent).
  • Terminal velocity: The constant velocity reached when the net force on a falling object is zero — drag force plus upthrust equals weight. Used in Millikan's experiment to determine droplet radius.
  • Quantisation of charge: The observation that electric charge only exists in discrete integer multiples of a fundamental unit $e$. Millikan's experiment demonstrated that every measured charge was a whole-number multiple of $1.6 \times 10^{-19}$ C.
  • Elementary charge: The magnitude of charge on a single electron (or proton), $e = 1.6 \times 10^{-19}$ C. Millikan's oil-drop experiment provided the first precise experimental measurement of this value.

Revision notes

Read the core explanations from Stokes' Law and Significance of Millikan's Results before testing yourself from memory.

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.

Frequently asked questions

What is Stokes' law?

Stokes' law gives the viscous drag force on a small sphere moving slowly through a fluid with laminar flow. The drag depends on the fluid's viscosity η and depends linearly on both the sphere's radius r and its speed v: double the radius or the speed and the drag doubles.

What is quantisation of charge?

Quantisation of charge means electric charge does not vary continuously but only exists in discrete integer multiples of the elementary charge e = 1.6 × 10⁻¹⁹ C, so Q = ne. Millikan never measured a charge such as 2.4 × 10⁻¹⁹ C, which would be 1.5 times the fundamental unit.

What is dynamic viscosity?

Dynamic viscosity, η, is a measure of a fluid's resistance to flow, in units of Pa s (pascal-seconds). A higher value of η means a thicker, more resistive fluid; air has η ≈ 1.8 × 10⁻⁵ Pa s at room temperature.

How do you calculate the radius of an oil drop using Stokes' law?

The radius of an oil droplet comes from its terminal velocity v: r = √(9ηv ÷ (2(ρ − ρ₀)g)), where ρ is the oil density and ρ₀ the air density. For v = 1.2 × 10⁻⁴ m s⁻¹, η = 1.8 × 10⁻⁵ Pa s and ρ = 900 kg m⁻³, r ≈ 1.05 × 10⁻⁶ m.

What is the significance of Millikan's results?

Millikan's results gave the first direct measurement of the elementary charge e and proved that charge is quantised. Combined with Thomson's charge-to-mass ratio, his value of e allowed the electron mass, about 9.11 × 10⁻³¹ kg, to be calculated for the first time, and supported the foundations of quantum theory.

When is Stokes' law valid?

Stokes' law is only valid for small, slow spheres where the flow remains laminar, meaning the fluid moves in smooth layers with no turbulence. Millikan's oil droplets, with radii of order 10⁻⁶ m moving at fractions of a millimetre per second, this condition is comfortably satisfied.

How does Stokes' drag depend on radius and speed?

Stokes' drag depends linearly on both the radius and the speed of the sphere: doubling the radius doubles the drag, and doubling the speed doubles the drag. This linear relationship makes Stokes' law simple to use, but it only holds for small, slow spheres where the flow remains laminar.

What is laminar flow in Stokes' law?

Laminar flow is smooth, non-turbulent flow in which the fluid moves in layers. Stokes' law assumes the flow around the sphere is laminar, a condition comfortably satisfied by Millikan's oil droplets, with radii of order 10⁻⁶ m moving at fractions of a millimetre per second.

Why is upthrust included when finding the radius of an oil drop?

Upthrust from the air is included when finding the radius of an oil drop for accuracy, because at terminal velocity the weight balances the drag plus upthrust. It gives r = √(9ηv ÷ 2(ρ − ρ₀)g), though air density, 1.2 kg m⁻³, is much smaller than oil density, so the difference is close to 900 kg m⁻³.

How was the mass of the electron first calculated?

The mass of the electron was first calculated by combining Millikan's value of e with Thomson's earlier measurement of the charge-to-mass ratio e/m. Dividing e by e/m gives an electron mass of about 9.11 × 10⁻³¹ kg.

Why did Millikan never measure a charge of 2.4 × 10⁻¹⁹ C?

Millikan never measured a charge of 2.4 × 10⁻¹⁹ C because that would be 1.5 times the fundamental unit, and charge only comes in whole-number multiples of e = 1.6 × 10⁻¹⁹ C. Charges came in steps of 1e, 2e, 3e, 4e and so on, much as coins come in fixed units.

Does the stationary oil drop calculation use Stokes' law?

The stationary oil drop calculation, q = mgd ÷ V, does not use Stokes' law, because a motionless droplet has no drag force: the electric force and gravity are in perfect balance. Stokes' law was only needed earlier to find the radius, and hence the mass, from the terminal velocity.

Who discovered Stokes' law?

Stokes' law was found by George Gabriel Stokes in 1851. He showed that for a small sphere moving at low speed through a viscous fluid with laminar flow, the drag force is F = 6πηrv, depending linearly on the radius and the speed.

What does Q = ne mean?

Q = ne means any measured charge Q is a whole-number multiple of the elementary charge e = 1.6 × 10⁻¹⁹ C, where n is a positive or negative integer. Electric charge does not vary continuously; it only exists in discrete multiples of e, a fact shown by Millikan's results.

Why was Millikan's experiment important for quantum theory?

Millikan's experiment was important for quantum theory because 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. It also supported the picture of matter as made of discrete charged particles.

This CourseAll Revision NotesPast PapersPricing