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Inquiry Question 2: How is it known that atoms are made up of protons, neutrons and electrons?

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

A focused answer to the HSC Physics Module 8 dot point on Millikan's oil drop experiment. Balancing gravity and electrical force on charged oil droplets between parallel plates, the equation mg = qE with E = V/d, the integer-multiple distribution of measured charges, and the value of the elementary charge e.

Generated by Claude Opus 4.814 min answerUpdated 2026-07-01

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context

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 $e = 1.602 \times 10^{-19}$ 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:

qE = mg, \quad E = V/d

So:

q = \frac{mgd}{V}

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:

q_n = n e, \quad n = 1, 2, 3, \dots

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 $m_e = 9.109 \times 10^{-31}$ 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.

Worked models

Model 1 - stationary balance. An oil drop of mass $5.72 \times 10^{-15}\ \text{kg}$ is held motionless between plates $6.00\ \text{mm}$ apart with a potential difference of $700\ \text{V}$ . Find the charge on the drop and the number of elementary charges it carries. ( $g = 9.8\ \text{m s}^{-2}$ , $e = 1.602 \times 10^{-19}\ \text{C}$ .)

State the physics first: the drop is stationary, so $qE = mg$ with $E = V/d$ , giving $q = \dfrac{mgd}{V}$ .

$E = \dfrac{V}{d} = \dfrac{700}{6.00 \times 10^{-3}} = 1.17 \times 10^{5}\ \text{V m}^{-1}$ .

$q = \dfrac{mg}{E} = \dfrac{(5.72 \times 10^{-15})(9.8)}{1.17 \times 10^{5}} = 4.80 \times 10^{-19}\ \text{C}$ .

$n = \dfrac{q}{e} = \dfrac{4.80 \times 10^{-19}}{1.602 \times 10^{-19}} = 3.0$ , so the drop carries $3$ elementary charges.

Marker's note: state $qE = mg$ before substituting numbers - this earns the method mark even if a later arithmetic step slips. Always finish by dividing by $e$ and rounding to the nearest whole number, since real charges must be integer multiples of $e$ .

Model 2 - mass from Stokes' law, then charge. A drop of radius $1.6 \times 10^{-6}\ \text{m}$ falls at terminal velocity $2.2 \times 10^{-4}\ \text{m s}^{-1}$ with the field off ( $\eta = 1.81 \times 10^{-5}\ \text{Pa s}$ for air). The field is then switched on, with plates $8.00\ \text{mm}$ apart and a potential difference of $1400\ \text{V}$ , and the drop is held stationary. Find its charge and the number of elementary charges it carries.

Stage 1 - mass. At terminal velocity, drag balances weight: $6\pi\eta r v_t = mg$ , so $m = \dfrac{6\pi\eta r v_t}{g}$ .

$m = \dfrac{6\pi (1.81 \times 10^{-5})(1.6 \times 10^{-6})(2.2 \times 10^{-4})}{9.8} = 1.23 \times 10^{-14}\ \text{kg}$ .

Stage 2 - charge. $E = \dfrac{V}{d} = \dfrac{1400}{8.00 \times 10^{-3}} = 1.75 \times 10^{5}\ \text{V m}^{-1}$ .

$q = \dfrac{mg}{E} = \dfrac{(1.23 \times 10^{-14})(9.8)}{1.75 \times 10^{5}} = 6.89 \times 10^{-19}\ \text{C}$ .

$n = \dfrac{q}{e} = \dfrac{6.89 \times 10^{-19}}{1.602 \times 10^{-19}} = 4.3$ , so the drop carries the nearest integer, $n = 4$ elementary charges.

Marker's note: this is the two-stage falling/stationary problem - do the field-off stage (Stokes' law for mass) completely before the field-on stage (force balance for charge). Marks are lost for trying to find $q$ without first establishing $m$ .

Common traps

Confusing $E$ and $V$: Between parallel plates, $E = V/d$ . The field is in V/m and is uniform between the plates; the voltage is the work per unit charge to move from one plate to the other.
Setting up the force balance with the wrong sign: The electric force must be opposite to gravity (upward, for a negative droplet) to balance it. Always check by drawing the free-body diagram.
Reporting a non-integer multiple of $e$ without comment: If your calculation gives $q/e = 2.4$ , you should say either that experimental error puts the actual value at 2 or 3, or that the problem is testing your ability to round. Real charges are integer multiples of $e$ .
Saying Millikan measured the mass of the electron directly: He measured the charge $e$ . The mass of the electron then follows from Thomson's $e/m$ .
Claiming the experiment proves the electron is the smallest charge in nature: It proves the smallest free charge is $e$ . Quarks have smaller charges but are not free.
Forgetting to convert millimetres to metres in $d$: A plate separation given in mm must be converted before it is substituted into $E = V/d$ , or the field is out by a factor of $1000$ .

Exam technique

State the relationship before the numbers: write " $qE = mg$ , so $q = mgd/V$ " first - a stated relationship earns a method mark even if the arithmetic slips.

Use the NESA data-sheet value $e = 1.602 \times 10^{-19}$ C and keep intermediate and final answers to two or three significant figures with the correct unit.

For a two-stage problem (find $m$ from Stokes' law with the field off, then find $q$ from the force balance with the field on), complete the mass stage fully before starting the charge stage - do not mix the two equations.

When reading a $q$ -vs- $n$ graph, the gradient is $e$ : take two well-separated points, compute $\Delta q / \Delta n$ , and quote the answer with units of coulombs, not an unlabelled number.

In plain English

Imagine a tiny, tiny water balloon hanging perfectly still in mid-air because a fan blowing up from below pushes it exactly as hard as gravity pulls it down - not too much, not too little. Millikan did the electrical version of this: he sprayed a fine mist of oil, let a stray charged droplet fall between two metal plates, and then dialled up the voltage on the plates until the electric push exactly cancelled the droplet's weight and it hung motionless, like that balloon. Because he knew the droplet's weight and the strength of the electric field, he could work out exactly how much charge was sitting on that one tiny drop. He did this for thousands of different drops - and every single time, the charge he measured was a whole-number multiple of one fixed, tiny amount, never a fraction of it. It is a bit like weighing lots of different stacks of coins and finding every stack weighs an exact whole number of five-cent pieces, never one-and-a-half: it tells you the coins themselves cannot be broken into smaller pieces. That fixed amount of charge is the electron's charge, $e$ , and the fact that it never comes in fractions is what physicists mean when they say electric charge is "quantised."

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.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 HSC4 marksAn oil drop of mass 3.20 x 10^-15 kg is held stationary between two parallel plates separated by 6.00 mm. The potential difference between the plates is 490 V. Calculate the charge on the drop and state how many elementary charges this represents. (g = 9.80 m/s^2, e = 1.60 x 10^-19 C.)

Show worked answer →

The drop is in equilibrium: electrical force up balances gravity down.

$qE = mg$ , with $E = V/d$ :

$q = \frac{mgd}{V} = \frac{3.20 \times 10^{-15} \times 9.80 \times 6.00 \times 10^{-3}}{490}$
$q = \frac{1.882 \times 10^{-16}}{490} = 3.84 \times 10^{-19}$ C.

In elementary charges:

$n = q/e = 3.84 \times 10^{-19} / 1.60 \times 10^{-19} = 2.4$ .

Rounding to the nearest integer, the drop carries 2 elementary charges, suggesting the experimentally rounded charge would be $2e = 3.20 \times 10^{-19}$ C. (The exam value of 2.4 likely indicates rounding in the question; either answer with $n = 2$ or commentary on the integer-multiples observation is acceptable.)

Markers reward $E = V/d$ , force balance, numerical answer for $q$ , and the explicit "integer multiple of $e$ " interpretation.

2018 HSC3 marksExplain how Millikan's experimental results demonstrated that electric charge is quantised.

Show worked answer →

Millikan measured the charge on each of many individual oil drops. He found that every measured value was an integer multiple of a single basic charge: $q = n e$ with $n = 1, 2, 3, \dots$ . No drop ever carried, say, 1.5 or 2.7 times that basic charge. Sometimes a single drop's charge changed (after exposure to X-rays, for example), but the new value was always an integer multiple of the same basic charge.

The natural explanation is that charge comes in discrete packets of size $e$ , the elementary charge, and macroscopic charges are integer multiples of these packets. The continuous-charge model of classical electromagnetism does not predict this clustering.

Markers reward the observation of integer multiples, no fractional charges, and the conclusion that charge is quantised in units of $e$ .

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksTwo horizontal parallel plates are separated by

8.0\ \text{mm}

with a potential difference of

250\ \text{V}

applied between them. Calculate the electric field strength between the plates.

Show worked solution →

Use $E = \dfrac{V}{d}$ , with $d$ converted to metres.

$E = \dfrac{250}{8.0 \times 10^{-3}} = 3.1 \times 10^{4}\ \text{V m}^{-1}$ .

Marks: one for the correct formula with the substituted values, one for the answer to two significant figures with the correct unit.

foundation4 marksAn oil drop of mass

4.09 \times 10^{-15}\ \text{kg}

is held stationary between two horizontal plates separated by

4.00\ \text{mm}

with a potential difference of

500\ \text{V}

. Calculate (a) the electric field between the plates and (b) the charge on the drop, stating

n

, the number of elementary charges it carries. (

g = 9.8\ \text{m s}^{-2}

e = 1.602 \times 10^{-19}\ \text{C}

Show worked solution →

(a) Field. $E = \dfrac{V}{d} = \dfrac{500}{4.00 \times 10^{-3}} = 1.25 \times 10^{5}\ \text{V m}^{-1}$ .

(b) Charge. The drop is stationary, so the electric force balances gravity: $qE = mg$ , so $q = \dfrac{mg}{E}$ .

$q = \dfrac{(4.09 \times 10^{-15})(9.8)}{1.25 \times 10^{5}} = 3.21 \times 10^{-19}\ \text{C}$ .

$n = \dfrac{q}{e} = \dfrac{3.21 \times 10^{-19}}{1.602 \times 10^{-19}} = 2.0$ , so the drop carries $n = 2$ elementary charges.

Marks: one for $E = 1.25 \times 10^{5}$ V m $^{-1}$ , one for the force-balance equation $qE = mg$ , one for $q = 3.21 \times 10^{-19}$ C with the unit, one for correctly stating $n = 2$ .

core5 marksAn oil drop of radius

1.35 \times 10^{-6}\ \text{m}

falls at a measured terminal velocity of

1.8 \times 10^{-4}\ \text{m s}^{-1}

with the field switched off (air viscosity

\eta = 1.81 \times 10^{-5}\ \text{Pa s}

). **(a)** Use Stokes' law to find the mass of the drop. **(b)** The field is switched on, with plates

5.00\ \text{mm}

apart and a potential difference of

430\ \text{V}

, and the drop is held stationary. Find the charge on the drop and the number of elementary charges it carries.

Show worked solution →

(a) Mass from Stokes' law. At terminal velocity the viscous drag balances weight, $6\pi\eta r v_t = mg$ , so $m = \dfrac{6\pi\eta r v_t}{g}$ .

$m = \dfrac{6\pi (1.81 \times 10^{-5})(1.35 \times 10^{-6})(1.8 \times 10^{-4})}{9.8} = 8.46 \times 10^{-15}\ \text{kg}$ .

(b) Charge from the force balance. $E = \dfrac{V}{d} = \dfrac{430}{5.00 \times 10^{-3}} = 8.6 \times 10^{4}\ \text{V m}^{-1}$ .

$q = \dfrac{mg}{E} = \dfrac{(8.46 \times 10^{-15})(9.8)}{8.6 \times 10^{4}} = 9.64 \times 10^{-19}\ \text{C}$ .

$n = \dfrac{q}{e} = \dfrac{9.64 \times 10^{-19}}{1.602 \times 10^{-19}} = 6.0$ , so the drop carries $6$ elementary charges.

Marks: one for Stokes' law rearranged for $m$ , one for $m = 8.46 \times 10^{-15}$ kg, one for $E = 8.6 \times 10^{4}$ V m $^{-1}$ , one for $q = 9.64 \times 10^{-19}$ C, one for correctly stating $n = 6$ .

core4 marksThe figure shows the charge

q

measured on five different oil drops plotted against the integer

n

that best fits each drop. **(a)** Describe the relationship shown by the graph. **(b)** Using the points

(n = 1,\ q = 1.60 \times 10^{-19}\ \text{C})

and

(n = 5,\ q = 8.01 \times 10^{-19}\ \text{C})

, calculate the gradient of the line. **(c)** State what the gradient represents and explain why this graph supports the quantisation of charge.

Show worked solution →

(a) The graph is a straight line through the origin: the charge on a drop is directly proportional to a small integer $n$ , i.e. $q = ne$ .

(b) Gradient $= \dfrac{\Delta q}{\Delta n} = \dfrac{(8.01 - 1.60) \times 10^{-19}\ \text{C}}{5 - 1} = \dfrac{6.41 \times 10^{-19}}{4} = 1.60 \times 10^{-19}\ \text{C}$ .

(c) The gradient equals the elementary charge $e$ . Because every measured drop's charge lies on this single straight line through the origin at an integer value of $n$ (never at $n = 1.5$ or $n = 2.7$ ), charge cannot take arbitrary values - it exists only in whole-number multiples of one fixed unit, $e$ . This is the direct evidence that electric charge is quantised.

Marks: one for identifying the direct proportionality $q = ne$ (line through the origin), one for a correctly calculated gradient with working shown, one for identifying the gradient as $e$ , one for linking the integer spacing of the data to the quantisation conclusion.

exam6 marksAssess the extent to which Millikan's oil drop experiment can be considered a reliable and valid method for determining the elementary charge.

Show worked solution →

Band-6 plan. (1) State what the experiment set out to measure and how (force balance, $q = mgd/V$ ). (2) Assess validity: does the method actually isolate $e$ , and what sources of systematic error existed (viscosity/Stokes' law approximation, drop evaporation, contact potential, buoyancy neglect)? (3) Assess reliability: repeatability across thousands of drops, whether independent drops gave consistent integer multiples. (4) Weigh both and reach an explicit judgement, using the closeness of Millikan's value ( $1.592 \times 10^{-19}$ C) to the modern value ( $1.602 \times 10^{-19}$ C, about $0.6\%$ low) as evidence.

Model answer. Millikan's method is valid in principle: balancing the known weight $mg$ of a drop against the electric force $qE = qV/d$ isolates the charge $q$ directly from measurable quantities ( $V$ , $d$ , and $m$ found from the drop's terminal velocity via Stokes' law), with no dependence on charge-to-mass ratio or any other prior assumption. This makes it a genuinely independent measurement of $e$ , distinct from Thomson's $e/m$ result.

However, the method carries systematic uncertainties that limit its validity. Stokes' law assumes a smooth, continuous fluid, which breaks down slightly for micron-sized drops where the air's molecular structure becomes significant (later corrected by the Cunningham slip-flow correction); Millikan's droplets could also evaporate slowly during observation, changing their mass; and stray contact potentials between dissimilar plate metals introduce a small unaccounted voltage. These effects explain why Millikan's original value of $e = 1.592 \times 10^{-19}$ C sits about $0.6\%$ below the modern accepted value of $1.602 \times 10^{-19}$ C - small, but systematic rather than random.

The experiment is highly reliable: Millikan measured thousands of individual, independent drops over several years, and every single one gave a charge that was, within experimental uncertainty, an integer multiple of the same base value. This consistency across a very large, independent sample is strong evidence the result is not a coincidence of a few drops but a genuine physical regularity.

Weighing these points, the experiment is judged highly reliable (the integer-multiple pattern is robust and repeatable across thousands of trials) and valid in its underlying logic, but with a small, identifiable systematic bias from the Stokes'-law approximation that later, more refined experiments corrected. It remains one of the most convincing single-experiment demonstrations in physics precisely because its central conclusion - quantisation - does not depend on removing that small systematic error.

Marker's note: the top band explicitly separates reliability (repeatability/consistency across many trials) from validity (whether the method truly isolates $e$ , and named sources of systematic error), cites the $0.6\%$ discrepancy from the modern value as evidence, and closes with an explicit judgement rather than simply listing pros and cons.

exam7 marksAnalyse how Millikan's oil drop experiment, taken together with Thomson's earlier cathode-ray experiments, established both the charge and the mass of the electron, and evaluate the significance of this result for the developing model of the atom.

Show worked solution →

Band-6 plan. (1) State what Thomson's experiment measured ( $e/m$ only) and why that was insufficient. (2) State what Millikan's experiment measured ( $e$ alone, via the force balance) and how it is combined with Thomson's ratio to give $m_e$ . (3) Give the resulting value and evaluate its significance for atomic structure (a real, quantised, massive constituent particle, feeding directly into Rutherford's and Bohr's models).

Model answer. Thomson's 1897 cathode-ray experiments used crossed electric and magnetic fields to measure the charge-to-mass ratio $e/m$ of the particles making up cathode rays, showing they were far lighter than any atom and were a universal constituent of matter - the electron. However, a ratio alone cannot separate charge from mass: the same $e/m$ could arise from a small charge on a light particle or a larger charge on a heavier one, so neither quantity was individually known.

Millikan's oil drop experiment (1909-1913) supplied the missing, independent measurement. By balancing the electric force $qE$ on a charged oil drop against its weight $mg$ (with the drop's own mass found separately from its terminal velocity via Stokes' law), Millikan measured the charge $q$ on many individual drops directly, with no reference to $e/m$ at all. Every measured charge turned out to be an integer multiple of a single value, $e = 1.602 \times 10^{-19}\ \text{C}$ (Millikan's own result, $1.592 \times 10^{-19}\ \text{C}$ , was close to this).

Combining the two results closed the loop: dividing Thomson's ratio $e/m$ by Millikan's directly measured $e$ gives the electron's mass, $m_e = 9.109 \times 10^{-31}\ \text{kg}$ , roughly $1/1836$ of a hydrogen atom. For the first time, both properties of a fundamental particle were pinned down independently and combined.

This was highly significant for the atomic model. It confirmed the electron as a real, discrete particle with a fixed, quantised unit of charge rather than a continuous fluid, which directly supported the idea that atoms have internal structure built from countable particles rather than being indivisible. This quantisation of charge, together with the electron's tiny mass, fed directly into Rutherford's nuclear model (light electrons orbiting a massive, compact nucleus) and later Bohr's model (which relied on electrons as discrete, countable particles occupying quantised orbits). Without a reliable value of $e$ and $m_e$ , neither model could have been quantitatively tested.

Marker's note: the top band explains WHY Thomson's result alone was insufficient (a ratio, not two separate quantities), correctly describes Millikan's method as an independent measurement of $q$ (not a repeat of Thomson's method), shows the combination $m_e = (e/m) \div e$ , and evaluates significance by linking quantised, discrete charge to the later nuclear and quantum atomic models rather than simply stating "it was important".