Investigate, assess and model Millikan's oil drop experiment to determine the elementary charge and the quantisation of electric charge

What this dot point is asking

NESA wants you to describe Millikan's apparatus, explain the force balance on a charged oil drop between parallel plates ($qE = mg$ with $E = V/d$), use it to extract the charge on individual drops, and account for the observation that all measured charges are integer multiples of the elementary charge $e = 1.60 \times 10^{-19}$ C, with the conclusion that electric charge is quantised.

The answer

Why the experiment was needed

Thomson's 1897 measurement of $e/m$ for the electron was the charge-to-mass ratio, not the charge itself. To separate the two and find both the mass and charge of the electron, an independent measurement of $e$ alone was required.

The apparatus

Robert Millikan's 1909 experiment (refined through about 1913) used:

• A small chamber containing two horizontal parallel metal plates separated by distance $d$, with a small hole in the upper plate.
• A potential difference $V$ applied between the plates, creating a uniform vertical electric field $E = V/d$.
• An atomiser to spray tiny oil droplets above the upper plate. A few droplets fall through the hole into the space between the plates.
• A short-wavelength source (X-rays, or ionising radiation) to ionise some air molecules and so charge some droplets by attachment.
• A microscope to track individual droplets and a stopwatch to measure terminal velocities.

Two methods

Stationary method (the simplest to describe). Adjust the voltage until a chosen droplet hangs motionless. The electric force on the charge balances gravity:

So:

The mass $m$ of the droplet is found by switching off the field and measuring the terminal velocity of free fall through the air, then using Stokes' law (or, in modern presentations, treating the droplet density and radius separately).

Falling-and-rising method (Millikan's actual method). With the field off, the droplet falls at terminal velocity $v_g$ set by gravity vs viscous drag. With the field switched on (in the direction that drives the negative droplet upward), it rises at terminal velocity $v_E$ set by net electric force vs drag. Combining $v_g$ and $v_E$ eliminates the radius-dependent constants and gives the charge $q$ directly.

Results

Millikan measured thousands of drops over many years. Every measured charge was a positive integer multiple of a single value:

with $e \approx 1.60 \times 10^{-19}$ C. Drops with $n = 1$ (singly charged) were the most common, but $n = 2, 3, 4$ appeared often, and occasionally larger values. Sometimes a drop's charge would jump (after a momentary exposure to ionising radiation), but always to a different integer multiple of the same base unit.

The interpretation is direct: charge is quantised. The smallest unit of free charge in nature is $e$, and macroscopic charges are integer multiples of it.

Millikan's best value was $e = 1.592 \times 10^{-19}$ C, very close to the modern value $1.602 \times 10^{-19}$ C. Combined with Thomson's $e/m$, this fixed the electron mass at $m_e = 9.11 \times 10^{-31}$ kg.

Worked example: a heavier drop

A drop of mass $5.0 \times 10^{-15}$ kg is held stationary between plates 5.0 mm apart with potential difference 460 V. Find the charge on the drop.

Electric field: $E = V/d = 460 / 5.0 \times 10^{-3} = 9.2 \times 10^4$ V/m.

Force balance: $qE = mg$, so $q = mg/E = (5.0 \times 10^{-15})(9.80)/(9.2 \times 10^4) = 5.3 \times 10^{-19}$ C.

In elementary charges: $n = q/e = 5.3 \times 10^{-19} / 1.60 \times 10^{-19} \approx 3.3$.

The closest integer is 3, so the drop carries $3e = 4.8 \times 10^{-19}$ C. The 10% discrepancy in this textbook problem usually reflects measurement uncertainty rather than fractional charge.

Modern view

Charge quantisation in units of $e$ is observed in every macroscopic system. Quarks have charges of $\pm e/3$ and $\pm 2e/3$, but they are confined inside hadrons and cannot be isolated as free particles. The smallest free charge is the electron's $-e$ (or its antiparticle's $+e$), exactly the unit Millikan measured.

Try it: Electric field calculator for $E = V/d$ between parallel plates, and explore the force on a charged droplet between them.

Examples in context

Example 1. Replicating Millikan's experiment at a Sydney high-school open day. A student observes an oil drop of radius $r = 1.0 \times 10^{-6} \text{ m}$ falling at terminal speed $v_t = 1.2 \times 10^{-4} \text{ m/s}$ in air ($\eta = 1.81 \times 10^{-5} \text{ Pa s}$, oil $\rho = 920 \text{ kg/m}^3$). Stokes's law gives the drop mass $m = (6 \pi \eta r v_t) / g = 6 \pi \times 1.81 \times 10^{-5} \times 10^{-6} \times 1.2 \times 10^{-4} / 9.8 = 4.18 \times 10^{-15} \text{ kg}$. Switching on $V = 5500 \text{ V}$ across $d = 1.5 \text{ cm}$ plates (so $E = 3.67 \times 10^5 \text{ V/m}$) holds it stationary: $qE = mg$, so $q = mg/E = 4.18 \times 10^{-15} \times 9.8 / 3.67 \times 10^5 = 1.12 \times 10^{-19} \text{ C}$. This is $\sim 0.7 e$, so the student should round to $1 e$ within experimental error.

Example 2. Charge quantisation in a modern Lucas Heights ion-trap. Single-ion Paul traps at ANSTO Lucas Heights hold isolated $^{40}$Ca$^+$ ions with charge $q = +e = 1.60 \times 10^{-19} \text{ C}$. The trap measures changes in the ion's micromotion when it captures an extra electron, dropping $q$ to $0$. The minimum step is exactly $e$, just as Millikan found, but now resolved to $1$ part in $10^9$ rather than Millikan's $1\%$. CODATA 2019 fixed $e \equiv 1.602176634 \times 10^{-19} \text{ C}$ exactly as a defined SI constant - a direct lineage from Millikan's 1909-1913 oil drops to the modern definition of the ampere.

Try this

Q1. State Millikan's conclusion about electric charge from the oil-drop experiment. [2 marks]

• Cue. Electric charge is quantised; all observed charges are integer multiples of $e = 1.60 \times 10^{-19} \text{ C}$.

Q2. An oil drop of mass $3.2 \times 10^{-15} \text{ kg}$ is held stationary between parallel plates separated by $1.0 \text{ cm}$ with $V = 980 \text{ V}$ applied. Calculate the drop's charge and the number of excess electrons. [4 marks]

• Cue. $E = V/d = 9.8 \times 10^4 \text{ V/m}$; $qE = mg$, so $q = mg/E = 3.2 \times 10^{-15} \times 9.8 / 9.8 \times 10^4 = 3.2 \times 10^{-19} \text{ C}$; $n = q/e = 2$ electrons.

Q3. Millikan observed drops with the following charges (in units of $10^{-19} \text{ C}$): $1.6, 3.3, 4.7, 6.5, 8.0$. (a) Identify the common factor. (b) Explain why no drops had charge $2.4 \times 10^{-19} \text{ C}$. (c) Outline the role of Stokes's law in determining drop mass. [2+2+2 marks]

• Cue. (a) Each is approximately $n \times 1.6 \times 10^{-19} \text{ C}$ with $n = 1, 2, 3, 4, 5$. (b) Not an integer multiple of $e$. (c) Terminal velocity $v_t$ in still air gives radius then mass via $\rho$.