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Millikan's Oil-Drop Experiment

This workbook uses an interactive simulation of Millikan and Fletcher's famous 1909 experiment, so you can balance your own oil drops and repeat their reasoning for yourself — from a single balanced drop to a full set of results that reveals the quantisation of charge.

By the end of this workbook you should be able to:
  • Explain what Millikan and Fletcher's experiment was designed to find out.
  • Explain the balance of forces on a stationary charged oil drop.
  • Use your own simulated data to show that electric charge is quantised.
  • Evaluate the experiment, including its assumptions and sources of uncertainty.

1. Why the experiment mattered

By the early 1900s, scientists knew that atoms contained charged particles. But nobody knew whether electric charge could take any value, or whether it was built up from smaller, indivisible packets — a bit like how matter is built from atoms.

In 1909, the American physicist Robert Millikan, working with his student Harvey Fletcher, designed an experiment to measure the charge on individual, microscopic drops of oil. Their results were remarkable: every single charge they measured turned out to be a whole-number multiple of the same tiny value, 1.60 × 10⁻¹⁹ C. Millikan was awarded the Nobel Prize in Physics in 1923, partly for this work.

The big idea this experiment proved: electric charge is quantised — it only ever exists in whole-number multiples of a smallest unit, the elementary charge, e.
Key word — quantised: a quantity that can only take certain fixed values, rather than any value at all.

In this workbook, you will use a simulation of the experiment to collect your own "measurements" and repeat Millikan's reasoning for yourself.

2. The apparatus

Millikan's apparatus sprayed a fine mist of oil droplets into the space above two horizontal parallel metal plates. As the oil was forced through the fine nozzle of the atomiser, friction gave some droplets a small electric charge. A few drops fell through a small hole in the top plate, into the space between the plates, where they could be watched through a microscope.

atomiser + − − microscope several thousand volts
Fig. 2.1 A simplified diagram of Millikan's apparatus: charged oil drops fall between two horizontal parallel plates and are observed with a microscope.

By adjusting the potential difference across the plates, Millikan could change the electric force on a drop until gravity and the electric force exactly balanced, leaving the drop hanging motionless. Because the field between two parallel plates is uniform, this balance condition is very simple to analyse — which is exactly the setup you met in the Electric & Magnetic Fields workbook (E = V/d).

It's a little more complex

Friction during spraying only charges a few drops, and only by a small, uncontrolled amount. To produce more useful charged droplets—and to change the charge on a droplet while it was still being observed—Millikan exposed the space between the plates to X-rays. The X-rays ionised the air, knocking electrons off some air molecules and producing free electrons and positive ions. When an oil droplet captured one of these charged particles (or occasionally lost an electron), its net charge changed by one or more elementary charges.

This turned out to be one of the most convincing aspects of the experiment. Instead of only comparing many different droplets (which might differ in unknown ways), Millikan could expose the chamber to X-rays during an observation and watch a single droplet's balancing voltage suddenly jump to a new value. Since the droplet's mass and every other property remained the same, the jump could only be explained by a change in its electric charge. The measured charges were always whole-number multiples of the same tiny value, never fractions of it. Seeing the same droplet change between exact integer multiples of e provided especially strong evidence that electric charge is quantised.

3. Balancing the forces

Two forces act vertically on an oil drop sitting between the plates:

  • Weight, mg — always acts downwards in this experimental setup.
  • Electric force, Eq = Vq/d — acts upwards if the plates and the sign of the charge are arranged correctly.
+ − F = Eq (up) mg (down)
Fig. 3.1 A stationary, balanced oil drop: the upward electric force exactly cancels the downward weight.

When the drop is perfectly balanced and stationary:

mg = Eq = Vq / d

Rearranging for the charge on the drop:

q = mgd / V
If you know the drop's mass, m, the plate separation, d, and the balancing voltage, V, you can calculate the charge, q, on that one drop.

In the real experiment, the mass of each tiny drop was found separately (from its size and the density of the oil, or from how fast it fell with the field switched off). In this simulation, the mass of each drop is given to you directly, so that you can focus on the balancing technique and on what the pattern of charges tells us.

Worked example 3.1

An oil drop of mass 2.4 × 10⁻¹⁴ kg is held stationary between plates 1.6 cm apart by a potential difference of 4900 V. Calculate the charge on the drop, and state how many excess or missing electrons this represents. (g = 9.81 m s⁻²)

Answer:
q = mgd/V = (2.4×10⁻¹⁴ × 9.81 × 0.016) / 4900 = 7.69×10⁻¹⁹ C
number of electrons = q/e = 7.69×10⁻¹⁹ / 1.60×10⁻¹⁹ ≈ 5 electrons

Check your understanding

1An oil drop of mass 3.6 × 10⁻¹⁴ kg is balanced by a p.d. of 8300 V across plates 2.0 cm apart. Calculate the charge on the drop and the number of electrons this represents.
Hint: start from the balance condition mg = Vq/d, and rearrange to make q the subject before substituting any numbers in. Once you have a value for q in coulombs, divide by e = 1.60×10⁻¹⁹ C and round to the nearest whole number of electrons.

4. Run the simulation

Below is a simulated version of Millikan's apparatus. Each time you click New drop, a droplet with a random (hidden) mass and a random (hidden) whole number of extra electrons appears between the plates. Your job is to adjust the voltage slider until the drop is balanced — hovering still — then lock in your reading; the simulation calculates the charge on the drop for you automatically.

Plate separation for every trial: d = 2.00 cm (this stays fixed — only V and the hidden drop change between trials).

Controls

Drop mass, m = — ×10⁻¹⁴ kg
Velocity = —
Status: falling
Hint: at balance, mg = Vq/d, so q = mgd/V — this is exactly how the q value in your results table below is worked out for you. It's the same rearrangement you need for the calculation questions elsewhere in this workbook.
Fig. 4.1 The simulated apparatus: drag the voltage slider until the drop stops moving, then record the balanced reading.

Your results

After you record a balanced reading, a new row appears below, with the charge on that drop already worked out for you (from q = mgd/V) in units of 10⁻¹⁹ C — so you can spend your time on the pattern in the data, rather than repeating the same calculation 10–15 times. Collect at least 10–15 trials, then use the Graph data button to plot mass against balancing voltage.

Trialm (×10⁻¹⁴ kg)d (cm)V (V)q (×10⁻¹⁹ C)

5. Finding e from your data

Before looking at your own results, try this warm-up puzzle — it uses exactly the same reasoning Millikan used.

Warm-up puzzle

Five sealed bags each contain a whole number of identical marbles. You are not allowed to open the bags, but you can weigh them. Their masses are: 108 g, 72 g, 144 g, 36 g, 180 g. What is the most likely mass of a single marble?

Every mass given is a multiple of 36 g (108 = 3×36, 72 = 2×36, 144 = 4×36, 36 = 1×36, 180 = 5×36), and 36 g is the largest number that divides all of them exactly. So the most likely mass of one marble is 36 g.

Millikan used exactly this idea, but with electric charge instead of marble bags: he looked for the largest number that divided all of his measured charges (within experimental uncertainty).

Check your understanding

2Copy your ten (or more) values of q from the table in Section 4 into the space below, in units of 10⁻¹⁹ C. Find the largest number that divides exactly into all of them. Unlike the marble-bag puzzle, this number will not itself be a whole number — each q is a whole number of electrons multiplied by e, and e is not a whole number of these units. This number is your simulated estimate of e, in units of 10⁻¹⁹ C.
3Compare your estimate of e with the accepted value, 1.60 × 10⁻¹⁹ C. Calculate the percentage difference between your estimate and the accepted value.
4Explain why using only one trial would not have been enough for Millikan to reach his conclusion, even if that single measurement was perfectly accurate.
A single charge value on its own could be any multiple of e — with only one data point you cannot tell whether it corresponds to 1 electron, 2 electrons, or some other number, so you cannot pin down e itself. It is only by comparing many different charges and finding their largest common factor that the size of the underlying "packet" of charge becomes clear — just like the marble bags puzzle needed several bags, not just one.

6. Evaluating the experiment

No experiment is perfect. Millikan's real apparatus had to deal with several complications that this simulation leaves out or simplifies.

Check your understanding

5This simulation ignores air resistance building up on the way to a steady speed, and evaporation of the oil drop over time. Suggest, for each of these two effects, whether it would make the measured charge come out too high, too low, or would not affect it, and explain your reasoning.
If a drop evaporates during the measurement, its mass decreases, so the true balancing voltage needed also decreases over time — using an outdated (too large) mass value would make a calculated charge too high. Reaching a steady speed is a separate practical concern (needing the observer to wait for the drop to settle) rather than a source of systematic error in q itself, provided the observer waits long enough before taking a reading.
6The apparatus also has to account for a small upward buoyancy force from the air, which was ignored in Worked Example 3.1. Explain why ignoring buoyancy would make a calculated charge slightly too large.
Buoyancy acts upwards, in the same direction as the electric force, helping to support the drop. If it is ignored, all of the drop's true weight appears to be supported by the electric force alone, so the calculated electric force — and therefore the calculated charge — comes out larger than it really is.
7Suggest one practical way a real experimenter could increase their confidence that they had found the true value of e, beyond just repeating the experiment more times.
Accept any reasonable, justified answer, for example: using drops with very different numbers of charges (not always similar n), so the common factor is easier to identify unambiguously; using a more precise method of timing/voltage measurement to reduce uncertainty in each individual value of q; or cross-checking against a completely independent method of measuring charge.
Stretch — quarks: protons and neutrons are themselves built from smaller particles called quarks, which carry charges of ±e/3 or ±2e/3. However, quarks are never found on their own — they are always bound together in combinations that add up to a whole multiple of e. So Millikan's conclusion, that any charge you can actually isolate and measure is a whole-number multiple of e, still holds.

7. Exam-style question

8 An oil drop of unknown mass is held stationary between two horizontal parallel plates, separated by 1.50 cm, when the potential difference across the plates is 6120 V.

a) State the two forces acting on the drop when it is stationary, and their directions. [2]
b) The drop carries a charge of 4 times the elementary charge. Calculate the mass of the drop. [3]
c) The same drop is then observed with the p.d. switched off. Describe and explain its motion. [2]
d) A second, different drop is found to be balanced by a p.d. of exactly half that in part (a), with exactly the same mass as the first drop. Determine the charge on this second drop, as a multiple of e. [2]

a) Weight (mg), acting vertically downward; electric force (Eq = Vq/d), acting vertically upward. Both must be present and equal in size for the drop to stay still.
b) q = 4 × 1.60×10⁻¹⁹ = 6.40×10⁻¹⁹ C. Balance: mg = Vq/d, so m = Vq/(gd) = (6120 × 6.40×10⁻¹⁹)/(9.81 × 0.0150) = 2.66×10⁻¹⁴ kg.
c) With the field off, only weight acts, so the drop accelerates downwards. In air, drag quickly increases until it balances the weight, so the drop reaches a constant (terminal) velocity rather than continuing to accelerate.
d) Same mass, so mg is unchanged; if V is halved but the balance condition mg = Vq/d must still hold with the same m, d and g, then q must double to compensate: q₂ = 2 × 4e = 8e.

Glossary

Balanced (drop)
The state in which the upward electric force on a drop exactly equals its downward weight, so it stays stationary.
Elementary charge, e
The smallest possible unit of electric charge, 1.60 × 10⁻¹⁹ C.
Quantised
Restricted to certain fixed values only — for charge, whole-number multiples of e.
Terminal velocity
The constant speed reached by an object falling through a fluid, once resistive (drag) forces balance the driving force.
Uniform electric field
A field with the same strength and direction everywhere, such as the field between two parallel charged plates.