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Every JEE aspirant uses e = 1.6 × 10⁻¹⁹ C, usually without asking where it came from. Robert Millikan measured it by watching single oil drops, a thousandth of a millimetre across, rise and fall between charged plates. The experiment is also a favourite source of exam questions on electric fields. This post explains how it worked, runs a full worked example, lists the mistakes students make, and ends with a lesson about bias that every scientist learns.
What did Millikan set out to show?
In his Nobel lecture, Millikan called the oil-drop experiment the place where "the most direct and unambiguous proof of the existence of the electron" would be found. He describes experiments "first performed in 1909" in which a drop's charge could be changed and the force on it watched. Charge, he found, could be added or removed only in amounts that changed the force by "definite integral multiples of the smallest observed force".
He received the 1923 Nobel Prize in Physics "for his work on the elementary charge of electricity and on the photoelectric effect".
How did the apparatus work?
The drop was "an individual oil-droplet about a thousandth of a millimeter in diameter blown out of an ordinary atomizer", as the lecture puts it, kept in air from which convection currents had been removed. It drifted between two horizontal metal plates connected to a battery through a switch, so the field could be turned on and off. Millikan stressed that the drop needed about a centimetre of path so its speed could be timed accurately.
With the field off, the drop falls under gravity at a steady speed set by air resistance. With the field on, the electric force can slow it, stop it or drive it upward.
Worked example: find the charge on a drop
Take the simplest version, where the field is adjusted until the drop hangs still. Then the electric force balances the weight:
qE = mg, with E = V/d
Data for one drop (illustrative): oil density ρ = 900 kg/m³, drop radius r = 1.0 μm = 1.0 × 10⁻⁶ m, plate separation d = 1.0 cm = 0.010 m, balancing voltage V = 1150 V, g = 9.8 m/s².
Volume: (4/3)πr³ = (4/3) × 3.1416 × (1.0 × 10⁻⁶)³ = 4.19 × 10⁻¹⁸ m³.
Mass: m = ρ × volume = 900 × 4.19 × 10⁻¹⁸ = 3.77 × 10⁻¹⁵ kg.
Weight: mg = 3.77 × 10⁻¹⁵ × 9.8 = 3.69 × 10⁻¹⁴ N.
Field: E = V/d = 1150 / 0.010 = 1.15 × 10⁵ V/m.
Charge: q = mg/E = 3.69 × 10⁻¹⁴ / 1.15 × 10⁵ = 3.21 × 10⁻¹⁹ C.
Now suppose two more drops give 4.82 × 10⁻¹⁹ C and 8.03 × 10⁻¹⁹ C. Look for the largest value that divides all three into whole numbers. Try 1.60 × 10⁻¹⁹ C: the ratios are 2.0, 3.0 and 5.0. The charges are 2e, 3e and 5e. That whole-number pattern, repeated over thousands of drops, is Millikan's evidence that charge is quantised.
How did Millikan know the mass of a drop he could not weigh?
He timed its fall. A small sphere falling through air reaches a terminal speed v when the drag from Stokes' law, 6πηrv, balances its weight (ignoring buoyancy). Setting 6πηrv = (4/3)πr³ρg gives r = √(9ηv / 2ρg). Measure v, and you get r, then the mass. The catch is η, the viscosity of air. Any error in η feeds straight into r, the mass and the charge.
How close was Millikan's value?
His lecture gives e = 4.774 (± 0.005) × 10⁻¹⁰ absolute electrostatic units. One coulomb is 2.998 × 10⁹ electrostatic units, a factor set by the speed of light. So:
4.774 × 10⁻¹⁰ ÷ 2.998 × 10⁹ = 1.592 × 10⁻¹⁹ C
Today the elementary charge has an exact defined value of 1.602 176 634 × 10⁻¹⁹ C. Millikan was low by (1.6022 − 1.5924) ÷ 1.6022 ≈ 0.6%.
Why did later measurements creep upward?
Richard Feynman used this in his 1974 Caltech commencement address "Cargo Cult Science". Millikan's answer, he said, was "a little bit off, because he had the incorrect value for the viscosity of air". Later results rose only gradually toward the correct value. Feynman's explanation: when experimenters got a number far above Millikan's, they looked for a mistake; when they got one close to it, "they didn't look so hard".
Historians have also studied which drops Millikan chose to publish, including Gerald Holton's study of the Millikan-Ehrenhaft dispute and Allan Franklin's paper on Millikan's published and unpublished data. A 1982 reanalysis of the 1913 data by Fairbank and Franklin found strong evidence for charge quantisation and no convincing evidence for fractional charges.
Common mistakes in oil-drop problems
Using V instead of E. The force is qE, and E = V/d. Forgetting to divide by d is the most common error.
Centimetres left in. Convert plate separation and radius to metres before calculating.
Wrong direction. For a negative drop to be pushed up, the upper plate must be positive. Check the sign before writing the force.
Cubing badly. (1.0 × 10⁻⁶)³ = 1.0 × 10⁻¹⁸. Slip a power here and the charge is off by orders of magnitude; a quick dimensional and order-of-magnitude check catches it.
Assuming n = 1. A drop usually carries several electrons' worth of charge. Divide by e and check you get close to a whole number.
A practice routine
Do three oil-drop problems this week: one balanced drop, one with the field off and a terminal speed given, and one where a drop changes speed after picking up charge. For each, write the force equation before any numbers. If you can follow the theory but stall on the numbers, our post on why you understand theory but fail numericals explains the gap.
When you get stuck partway, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer, so you can find exactly which step went wrong.
Start now: rework the example above with a balancing voltage of 767 V and check that you get 3e.
