Millikan's Oil Drop Experiment: Equilibrium and Forces Revision Notes | A Level AQA Physics

1. Why Millikan's Experiment Changed Physics

Picture a tiny droplet of oil, invisible to the naked eye, suspended perfectly still in mid-air between two metal plates. No string, no shelf — held up entirely by an electric field. That image is the heart of one of the most elegant experiments in physics history.

Robert Millikan performed this experiment between 1909 and 1913, and it gave physicists the first precise measurement of the elementary charge $e$. More than that, it proved that charge is quantised — it comes in fixed packets, not a continuous smear. Every value he measured was an integer multiple of $1.6 \times 10^{-19}$ C.

To understand how the experiment works, you need to analyse the forces acting on a droplet in two distinct situations: falling freely under gravity, and held stationary by an electric field. Those two situations together unlock both the mass and the charge of the droplet.

2. The Stationary Droplet: Electrostatic Equilibrium

Think about what it takes to hold something still in mid-air. The net force must be zero. For a charged oil droplet between two parallel plates, there are two forces in play: gravity pulling it down, and the electric force pushing it up.

The electric field between two parallel plates separated by distance $d$ with potential difference $V$ across them is uniform, with magnitude $E = V/d$. Substituting this into the force balance gives the key equation for this experiment.

Notice that if you know $m$, $g$, $V$, and $d$, you can calculate $Q$ directly. The challenge Millikan faced was finding $m$ — the droplets are far too small to weigh on a balance. That is where the falling droplet analysis comes in.

3. The Falling Droplet: Finding the Mass

With the electric field switched off, the droplet falls under gravity. It accelerates initially, but very quickly reaches a constant speed — terminal speed. At terminal speed, the net force is zero: weight downward equals viscous drag upward.

The drag force on a small sphere moving through a viscous fluid is given by Stokes' law: $F_{drag} = 6\pi\eta r v$, where $\eta$ is the viscosity of air, $r$ is the droplet radius, and $v$ is the speed. At terminal speed $v_0$, this drag force equals the weight of the droplet.

Once you know $r$ from this equation, you can calculate the mass using $m = \frac{4}{3}\pi r^3 \rho$, where $\rho$ is the density of the oil. With $m$ now known, you can return to the equilibrium equation $QV/d = mg$ and solve for $Q$.

When the electric field is switched back on and adjusted so the droplet rises at a new terminal speed $v_1$, the force balance changes: the net upward force (electric force minus weight) equals the drag force. This gives a second route to $Q$ and provides a consistency check on the measurement.

4. Comparing Motion With and Without the Electric Field

It helps to compare the two scenarios side by side. The forces change when the field is applied, and so does the motion — but the underlying physics of terminal speed applies in both cases.

The stationary condition is the cleanest measurement because there is no drag force to account for — the droplet is not moving. This is why Millikan preferred to adjust the voltage until the droplet was perfectly still, rather than measuring a rising speed.

A key practical point: the droplets are charged by exposing them to ionising radiation (such as X-rays or radiation from a radioactive source). The charge on a given droplet can change if it captures or loses an ion, and Millikan observed that whenever the charge changed, it always changed by an integer multiple of $1.6 \times 10^{-19}$ C — direct evidence for quantisation.

5. Worked Calculation: Finding Charge from Equilibrium

The equilibrium equation $QV/d = mg$ looks straightforward, but exam questions often give you the mass in non-SI units or ask you to rearrange for $V$ or $d$. The worked example below walks through a standard calculation with full mark annotations.

Worked example: Calculating the charge on a stationary oil droplet

An oil droplet of mass $4.8 \times 10^{-15}$ kg is held stationary between two parallel plates separated by $d = 5.0 \times 10^{-3}$ m. The potential difference across the plates is $V = 600$ V. Calculate the charge $Q$ on the droplet. [3 marks]

1. $$\frac{QV}{d} = mg$$ — [M1] Stating the correct force balance earns the method mark. The upward electric force equals the downward weight.
2. $$Q = \frac{mgd}{V}$$ — Multiply both sides by $d$ and divide both sides by $V$. Keep the rearrangement as a separate line — examiners award a method mark for correct rearrangement.
3. $$Q = \frac{(4.8 \times 10^{-15}) \times 9.8 \times (5.0 \times 10^{-3})}{600}$$ — [M1] Substituting all values correctly with consistent SI units. Note: mass is already in kg, distance in m, voltage in V.
4. $$\text{Numerator} = 4.8 \times 10^{-15} \times 9.8 \times 5.0 \times 10^{-3}$$\n$$= 4.8 \times 9.8 \times 5.0 \times 10^{-15} \times 10^{-3}$$\n$$= 235.2 \times 10^{-18} = 2.352 \times 10^{-16}$$ — Work through the powers of ten separately: $10^{-15} \times 10^{-3} = 10^{-18}$. Then $4.8 \times 9.8 = 47.04$ and $47.04 \times 5.0 = 235.2$.
5. $$Q = \frac{2.352 \times 10^{-16}}{600} = 3.92 \times 10^{-19} \text{ C}$$ — $2.352 \div 600 = 3.92 \times 10^{-3}$, so $2.352 \times 10^{-16} \div 600 = 3.92 \times 10^{-19}$ C. [A1] for the correct answer with units.
6. $$\frac{Q}{e} = \frac{3.92 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.45$$ — This result is approximately midway between 2 and 3 elementary charges, which in a real Millikan experiment would indicate a measurement error — genuine results should be very close to an integer multiple of $e$. The example values here are chosen for calculation practice; in practice, Millikan's measured charges were always within experimental uncertainty of an exact integer multiple of $e$.

Answer: $Q = 3.92 \times 10^{-19}$ C. Note: $Q/e \approx 2.45$, which is not close to an integer — in a real experiment this would suggest a measurement error. Millikan's actual results always gave values very close to integer multiples of $e$.

6. Worked Calculation: Using Terminal Speed to Find Mass

Before you can use the equilibrium equation to find $Q$, you need $m$. Here is how terminal speed measurements feed into that calculation. The key relationship is $mg = 6\pi\eta r v_0$ at terminal speed with no field.

Worked example: Determining droplet mass from terminal speed

An oil droplet falls at terminal speed $v_0 = 1.2 \times 10^{-4}$ m s⁻¹ with no electric field. The viscosity of air is $\eta = 1.8 \times 10^{-5}$ N s m⁻², and the density of the oil is $\rho = 900$ kg m⁻³. Find the radius and mass of the droplet. (Take $g = 9.8$ N kg⁻¹.)

1. $$mg = 6\pi\eta r v_0$$ — At terminal speed with no field, weight equals Stokes drag. This is the starting equation.
2. $$m = \frac{4}{3}\pi r^3 \rho$$ — The droplet is a sphere of radius $r$ and density $\rho$. Substituting this expression for $m$ allows us to solve for $r$.
3. $$\frac{4}{3}\pi r^3 \rho g = 6\pi\eta r v_0$$ $$\frac{4}{3} r^2 \rho g = 6\eta v_0$$ $$r^2 = \frac{6\eta v_0 \times 3}{4 \rho g} = \frac{18\eta v_0}{4 \rho g} = \frac{9\eta v_0}{2 \rho g}$$ — Divide both sides by $\pi r$ (valid since $r eq 0$), then rearrange. The factor of $\pi$ cancels cleanly.
4. $$r^2 = \frac{9 \times (1.8 \times 10^{-5}) \times (1.2 \times 10^{-4})}{2 \times 900 \times 9.8}$$ — Substitute $\eta = 1.8 \times 10^{-5}$ N s m⁻², $v_0 = 1.2 \times 10^{-4}$ m s⁻¹, $\rho = 900$ kg m⁻³, $g = 9.8$ N kg⁻¹.
5. $$\text{Numerator} = 9 \times 1.8 \times 10^{-5} \times 1.2 \times 10^{-4}$$ $$= 9 \times 1.8 \times 1.2 \times 10^{-9} = 9 \times 2.16 \times 10^{-9} = 19.44 \times 10^{-9}$$ $$\text{Denominator} = 2 \times 900 \times 9.8 = 17640$$ — $1.8 \times 1.2 = 2.16$; $9 \times 2.16 = 19.44$; powers: $10^{-5} \times 10^{-4} = 10^{-9}$. Denominator: $2 \times 900 = 1800$; $1800 \times 9.8 = 17640$.
6. $$r^2 = \frac{19.44 \times 10^{-9}}{17640} = 1.102 \times 10^{-12} \text{ m}^2$$ $$r = \sqrt{1.102 \times 10^{-12}} = 1.05 \times 10^{-6} \text{ m}$$ — $\sqrt{1.102 \times 10^{-12}} = \sqrt{1.102} \times 10^{-6} \approx 1.05 \times 10^{-6}$ m. The droplet radius is approximately 1 micrometre — consistent with a real Millikan droplet.
7. $$m = \frac{4}{3}\pi r^3 \rho = \frac{4}{3} \times \pi \times (1.05 \times 10^{-6})^3 \times 900$$ $$(1.05)^3 = 1.05 \times 1.05 \times 1.05 = 1.1025 \times 1.05 = 1.157$$ $$r^3 = 1.157 \times 10^{-18} \text{ m}^3$$ $$m = \frac{4}{3} \times 3.1416 \times 1.157 \times 10^{-18} \times 900$$ $$= 4.189 \times 1.157 \times 10^{-18} \times 900 = 4.366 \times 10^{-15} \text{ kg}$$ — Expand $(1.05)^3$ longhand: $1.05 \times 1.05 = 1.1025$; $1.1025 \times 1.05 = 1.157$. Then $\frac{4}{3}\pi \approx 4.189$; $4.189 \times 1.157 = 4.847$; $4.847 \times 10^{-18} \times 900 = 4.36 \times 10^{-15}$ kg.

Answer: $r \approx 1.05 \times 10^{-6}$ m; $m \approx 4.4 \times 10^{-15}$ kg

With $m$ now determined from the terminal speed measurement, you can substitute directly into $Q = mgd/V$ using the voltage required to hold the droplet stationary. That two-step process — terminal speed gives mass, equilibrium gives charge — is the complete Millikan method.

7. Quantisation of Charge: The Bigger Picture

Millikan did not measure just one droplet. He repeated the experiment hundreds of times with different droplets and different charges. Every single value of $Q$ he obtained was an integer multiple of $1.6 \times 10^{-19}$ C — never a fraction of it.

This was the experimental proof that charge is quantised. The smallest freely observable charge is $e = 1.6 \times 10^{-19}$ C, and all freely observable charges are $ne$ where $n$ is a non-zero integer. A droplet carrying three excess electrons has charge $Q = 3e = 4.8 \times 10^{-19}$ C. (Note: quarks carry fractional charges such as $\frac{1}{3}e$ or $\frac{2}{3}e$, but these are never observed in isolation.)

The experiment also gave the first accurate value of $e$ itself. Combined with the charge-to-mass ratio of the electron (measured by J.J. Thomson in 1897), Millikan's value of $e$ allowed physicists to calculate the electron mass for the first time. These two experiments together — Thomson's and Millikan's — form the foundation of the AQA 'Turning Points' topic.

8. Exam Technique and Common Mistakes

AQA questions on Millikan's experiment tend to test three things: stating the equilibrium condition correctly, performing the two-step mass-then-charge calculation, and explaining what the results tell us about quantisation. Each of these has its own set of traps.

For extended-response questions asking you to 'explain how Millikan determined the charge on an electron', structure your answer in two clear stages: the field-off measurement to find mass, then the field-on equilibrium to find charge. Finish with a statement about quantisation — that all measured charges were integer multiples of $e$ — to access the higher marks.

Millikan's experiment remains one of the clearest demonstrations of quantisation in physics: by balancing macroscopic forces on a microscopic droplet, it revealed a fundamental property of nature — that charge comes in indivisible packets — and gave us the precise value of $e$ that underpins all of modern atomic physics.