D. Gravitational, Electric, and Magnetic Fields · Gravitational, Electric and Magnetic FieldsMillikan's Oil Drop Experiment and the Elementary Charge
D2.5 D2.3 D2.1About 70 minutes
Learning goals- I can describe Millikan's oil drop apparatus and explain how a drop can be balanced in a uniform electric field.
- I can use qE = mg and E = V/d to find the charge on a drop, q = mgd/V.
- I can analyse a set of measured charges to show that charge is quantized and estimate e.
- I can evaluate the experiment's design, sources of error and its place in the history of physics.
Words to know
quantized
Coming only in whole-number multiples of a smallest unit. Electric charge is quantized in units of e.
elementary charge (e)
The smallest free charge, the size of the charge on an electron or proton: 1.60 × 10⁻¹⁹ C.
atomizer
A device that sprays a fine mist of tiny droplets. Millikan's oil drops picked up charge by friction as they were sprayed.
balancing voltage
The potential difference across the plates that makes the electric force on a particular drop equal its weight, so the drop stays still.
terminal velocity
The constant speed a falling object reaches when air resistance balances its weight. Millikan used it to find each drop's size and mass.
Weighing a single electron's charge
By 1900, physicists knew electrons existed (Thomson had measured their charge-to-mass ratio), but nobody knew the charge of one electron. Was electric charge a smooth fluid that could come in any amount, or did it come in tiny identical lumps?
Between 1909 and 1913, Robert Millikan, working at the University of Chicago with his graduate student Harvey Fletcher, answered the question with an experiment so simple in idea that you can follow every step with this unit's physics: hold a tiny charged oil drop still in a uniform electric field. Millikan received the 1923 Nobel Prize in Physics, in part for this work.
Remember from Uniform Electric Fields Between Parallel Plates: between plates, E = V/d is uniform, and a charge feels F = qE anywhere in the gap.
The apparatus
- An atomizer sprays a fine mist of oil above two horizontal metal plates. (Oil does not evaporate the way water would, so the drop's mass stays steady.) Spraying rubs charge onto some drops.
- A few drops fall through a small hole in the top plate into the space between the plates.
- A light shines on the drops, and the experimenter watches them through a microscope with a scale in the eyepiece. The drops look like tiny stars.
- A variable power supply sets the voltage V across the plates, separated by d.
The experimenter picks one drop and adjusts V until the drop hangs motionless.
The physics: balancing a drop
When the drop is still, the net force is zero. The upward electric force balances the downward gravitational force:
qE = mg, and with E = V/d: q = mgd/V
If the top plate is positive, E points down, so the drop must be negative for the electric force to point up.
The hardest part is finding m: the drop is only about a micrometre across, far too small to weigh. Millikan switched the field off, let the drop fall, and timed it between two marks in the eyepiece. A tiny drop reaches terminal velocity almost at once, and that speed depends on its radius in a known way (Stokes's law for air resistance). From the radius and the oil's density came the mass.
Millikan also changed a drop's charge (in later work, using X-rays to ionize the air) and watched the same drop again and again. Every change in charge was a whole-number multiple of the same amount.
Worked exampleExample 1: Charge on a drop
An oil drop of mass 2.5 × 10⁻¹⁵ kg is balanced between plates 1.0 cm apart when the voltage is 510 V. Find the charge on the drop and the number of extra electrons.
- q = mgd/V = (2.5 × 10⁻¹⁵)(9.8)(0.010)/510.
- q = 2.45 × 10⁻¹⁶/510 = 4.80 × 10⁻¹⁹ C.
- N = q/e = 4.80 × 10⁻¹⁹/1.60 × 10⁻¹⁹ = 3.0.
q = 4.8 × 10⁻¹⁹ C, which is 3 extra electrons (the drop is negative if the top plate is positive).
Worked exampleExample 2: Finding the mass from the size
A drop has radius 1.0 × 10⁻⁶ m, and the oil's density is 880 kg/m³. The plates are 8.0 mm apart. It is balanced at 903 V. Find its charge.
- Volume = (4/3)πr³ = (4/3)π(1.0 × 10⁻⁶)³ = 4.19 × 10⁻¹⁸ m³.
- m = density × volume = 880 × 4.19 × 10⁻¹⁸ = 3.69 × 10⁻¹⁵ kg.
- q = mgd/V = (3.69 × 10⁻¹⁵)(9.8)(0.0080)/903 = 3.20 × 10⁻¹⁹ C.
Worked exampleExample 3: Finding e from a set of measurements
A class measures these charges (in units of 10⁻¹⁹ C): 3.2, 4.8, 4.8, 6.4, 8.0, 9.6, 9.6, 11.2, 3.2, 6.4. What is the most likely value of the elementary charge?
- Look at the smallest values and the differences between values: 4.8 − 3.2 = 1.6, 6.4 − 4.8 = 1.6, 8.0 − 6.4 = 1.6, 9.6 − 8.0 = 1.6, 11.2 − 9.6 = 1.6.
- Every value is a whole-number multiple of 1.6: 3.2 = 2 × 1.6, 4.8 = 3 × 1.6, 11.2 = 7 × 1.6, and so on.
- No measured value falls between the multiples, and no smaller step appears.
e ≈ 1.6 × 10⁻¹⁹ C. Notice that no drop in the set has a charge of 1e. The largest common factor of all the charges is what reveals e, not the smallest charge measured.
Measured charges from Example 3, stacked on a line. They cluster only at whole-number multiples of e = 1.6 × 10⁻¹⁹ C, never in between.Why it matters, and a lesson about science
The experiment showed that charge is quantized and gave the first accurate value of e. With e and Thomson's charge-to-mass ratio, physicists could calculate the mass of the electron, and with Faraday's constant from chemistry, Avogadro's number.
Millikan's value was slightly too low, mainly because he used a slightly wrong value for the viscosity of air. Later experimenters' results crept up toward the modern value over several years. Historians have also studied his notebooks and found that he did not publish all the drops he observed; he judged some runs unreliable. Scientists still debate whether his choices were reasonable. It is a good reminder of why modern science expects researchers to report all their data and explain how they selected it.
Today e is one of the constants used to define the SI units: since 2019 the elementary charge is fixed at exactly 1.602 176 634 × 10⁻¹⁹ C. (We round it to 1.60 × 10⁻¹⁹ C in this course.)
If the drop moves
If a drop is not balanced, it accelerates. Use Newton's second law with both forces: F_net = qE − mg (taking up as positive). In real air, a drop reaches a small terminal speed almost at once, so in Millikan's actual method speeds were measured, not accelerations.
Just for funMillikan's oil drops were so tiny that a few dozen of them side by side would span the width of a human hair. He still managed to count the electrons on them one by one.
Hands-on activityFind the "elementary mass" (a Millikan model)
This classic model needs a grown-up to prepare it, so the student does not see inside.
Grown-up prepares: 8 small opaque envelopes or paper bags, numbered 1 to 8, and a pile of identical small objects such as metal washers, coins of one kind (for example nickels) or paper clips. Put a different whole number of objects (between 1 and 9) in each bag; a few bags can have the same number. Seal them. Write down the numbers secretly. Also prepare one empty bag of the same kind.
Student:
- Weigh the empty bag on a kitchen scale (to 1 g, or 0.1 g if you have a jewellery scale).
- Weigh each sealed bag and subtract the empty bag's mass. These are your "measured charges".
- Sort the values from smallest to largest. Find the differences between neighbouring values.
- Find the largest value that divides every measurement into (nearly) a whole number. That is your "elementary mass".
- Predict how many objects are in each bag. Then open them and check.
What to look for: the masses cluster at whole-number multiples of one object's mass (a Canadian nickel is about 3.95 g; check the mass of one of your objects at the end). Small differences come from the scale's precision and slight differences in bag mass, just as Millikan's drops had measurement uncertainty. If no bag has exactly 1 object, you still find the unit from the differences, the same as with e.
Your turn
1Why must the oil drop be negative if the top plate is positive and the drop is held still?
E points from the + plate to the − plate: down. To get an upward force F = qE, q must be negative.
2A drop of mass 4.9 × 10⁻¹⁵ kg carries 3 extra electrons. The plates are 1.2 cm apart. What voltage will balance it?
q = 3e = 4.8 × 10⁻¹⁹ C. V = mgd/q = (4.9 × 10⁻¹⁵)(9.8)(0.012)/(4.8 × 10⁻¹⁹) = 1.2 × 10³ V (1200 V).
3A drop of mass 1.6 × 10⁻¹⁵ kg has one extra electron. What field strength balances it?
E = mg/q = (1.6 × 10⁻¹⁵)(9.8)/(1.60 × 10⁻¹⁹) = 9.8 × 10⁴ N/C, pointing down.
4A student reports a drop charge of 2.4 × 10⁻¹⁹ C. Explain why this result must contain an error.
2.4 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 1.5. Charge must be a whole-number multiple of e, so 1.5e is impossible; the mass, voltage or separation must have been measured incorrectly.
5A drop (m = 3.0 × 10⁻¹⁵ kg, q = −4.8 × 10⁻¹⁹ C) is between plates with an 8.0 × 10⁴ N/C downward field. Find the net force and the initial acceleration (ignore air resistance).
F_e = qE = (4.8 × 10⁻¹⁹)(8.0 × 10⁴) = 3.84 × 10⁻¹⁴ N up. mg = 2.94 × 10⁻¹⁴ N down. Net = 9.0 × 10⁻¹⁵ N up. a = F/m = 3.0 m/s² up. (In real air, drag quickly limits its speed.)
6Measured charges (× 10⁻¹⁹ C): 6.4, 9.6, 16.0, 12.8, 4.8. Estimate e.
Differences: 9.6 − 6.4 = 3.2, 6.4 − 4.8 = 1.6, 12.8 − 9.6 = 3.2, 16.0 − 12.8 = 3.2. All values are multiples of 1.6 × 10⁻¹⁹ C (4, 6, 10, 8 and 3 times).
More practice: warm-up
1Which two forces balance on a still oil drop?
The electric force (qE) up and the gravitational force (mg) down.
2Write the formula Millikan used for the charge on a balanced drop.
3What does it mean that charge is quantized?
Charge comes only in whole-number multiples of e.
4Why did Millikan use oil instead of water?
Oil does not evaporate quickly, so the drop's mass stays constant during a measurement.
5How many electrons make up −1.0 × 10⁻¹⁸ C?
N = 1.0 × 10⁻¹⁸/1.60 × 10⁻¹⁹ = 6.25, which is not a whole number, so this charge cannot exist exactly; the nearest possible charges are 6e or 7e.
More practice: on your own
1A drop of radius 1.0 × 10⁻⁶ m and density 880 kg/m³ is balanced between plates 1.5 cm apart. If it carries 2 extra electrons, what voltage balances it?
m = 3.69 × 10⁻¹⁵ kg (as in Example 2). V = mgd/q = (3.69 × 10⁻¹⁵)(9.8)(0.015)/(3.2 × 10⁻¹⁹) = 1.7 × 10³ V.
2A balanced drop suddenly starts to rise while V is unchanged. What could have happened?
Its charge increased in size (it picked up more electrons, for example from ionized air), so qE became larger than mg. (Losing mass by evaporation would also do it, which is one reason oil is used.)
3If the plate separation is doubled but V is kept the same, what happens to a drop that was balanced?
E = V/d is halved, so the electric force is halved and the drop falls (it accelerates downward until air drag balances).
4Why did Millikan not simply measure the mass of each drop on a balance?
The drops are about a micrometre across, with masses around 10⁻¹⁵ kg, far too small to weigh directly. He found the radius from the terminal speed of the falling drop, then calculated the mass from the density.
5Identify two sources of error in a Millikan experiment and say how each would affect q.
Examples: (1) an error in the air viscosity used to find the radius changes m and therefore q (this is what made Millikan's e slightly low); (2) the voltage reading or plate spacing error changes E directly, so q changes in proportion to d and inversely to V; (3) air currents or evaporation make the drop drift, giving a wrong balancing voltage.
Challenge
1A drop is balanced at 600 V. After a puff of X-rays changes its charge, it is balanced at 400 V. Plates are 1.0 cm apart and the drop's mass is 2.0 × 10⁻¹⁵ kg. Find both charges and the change in charge. Is the change consistent with quantization?
q₁ = mgd/V₁ = (2.0 × 10⁻¹⁵)(9.8)(0.010)/600 = 3.27 × 10⁻¹⁹ C (≈ 2e). q₂ = (1.96 × 10⁻¹⁶)/400 = 4.90 × 10⁻¹⁹ C (≈ 3e). Change = 1.63 × 10⁻¹⁹ C ≈ 1e. Yes, consistent (the small differences from 1.60 × 10⁻¹⁹ come from the rounded data).
2Show that for the same drop at a fixed plate spacing, the charge is inversely proportional to the balancing voltage, and use this to explain how Millikan could detect a one-electron change without knowing m precisely.
q = (mgd)/V, and mgd is constant for one drop, so q ∝ 1/V. The ratios of charges for the same drop equal the inverse ratios of the voltages, which are easy to measure precisely. A set of voltages whose inverses are in ratios like 2 : 3 : 4 shows quantization even if m is only roughly known.
3Explain why the result "no drop had a charge smaller than e" does not by itself prove that e is the smallest unit, and what additional evidence the experiment gave.
Small charges might just be rare in the sample. The stronger evidence was that every measured charge, and every change in charge on a single drop, was a whole-number multiple of the same e. A smaller unit would have shown up as charges or changes between those multiples. (Quarks have charges of ±e/3 and ±2e/3, but they are never found alone, so the smallest free charge is e.)
Quiz
1. In Millikan's experiment, a drop hangs motionless when
2. A balanced drop's charge is found from
3. Which measured drop charge is impossible?
4. Millikan's experiment showed that electric charge is
5. To find the mass of each tiny drop, Millikan
6. If the voltage across the plates is increased while a drop is balanced, the negative drop will
Big idea (D2.5): Millikan balanced an electric force against gravity for tiny oil drops (q = mgd/V) and found that every charge is a whole-number multiple of e = 1.60 × 10⁻¹⁹ C.
How to help: The envelope activity is a great family activity: you prepare the sealed bags, and your teen discovers the "unit mass" from the data without opening them. Ask afterwards how this is like finding e.
Common difficulties:
- Thinking the smallest measured charge must be e. The common factor of all charges is what matters.
- Forgetting to convert mm to m for d.
- Getting the sign of the drop wrong (with the top plate positive, a held-up drop is negative).
- Rounding too early, which can make a value like 3.0e look like 2.9e.
Note on history: the lesson mentions the debate about how Millikan selected data. This is a good opening to talk about honesty and transparency in science.