IB Physics HLTopic 4 — Force FieldsPaper 1 & 2q = mgd/V~15 min read
Millikan’s Oil-Drop Experiment
Last page we said that charge comes in lumps of 1.60 × 10−19 C. That is a bold claim. You cannot put an electron on a balance, and nobody has ever seen one. So how was that number ever measured? In 1909, in a basement in Chicago, Robert Millikan and Harvey Fletcher took a perfume spray, a microscope and a very large battery — and weighed a single droplet of oil using nothing but electricity.
📘 What you need to know
Millikan and Fletcher performed the experiment in 1909; it measured the elementary chargee
Oil, not water, because oil does not evaporate quickly — so the drop’s mass stays constant
Drops are charged by friction at the spray nozzle, or ionised by X-rays
With no field, a drop falls and reaches terminal velocity when drag = weight
With a field, the electric force F = qE acts on the drop; adjust the p.d. until the drop hangs still
A stationary drop means qE = mg, and since E = V/d, this gives q = mgd/V
Every measured charge came out as a whole-number multiple of 1.60 × 10−19 C
That is the evidence for the quantisation of charge
The idea in one sentence
Let a tiny charged droplet fall between two metal plates. Then turn on a voltage, and turn it up until the electric force pulling the drop up exactly cancels the weight pulling it down. When the drop freezes in mid-air, you know those two forces are equal — and one of them contains the charge you are hunting.
This is a null measurement, and that is the genius of it. Millikan never had to measure a force. He only had to notice when a drop stopped moving, which the human eye is extremely good at. Whenever an experiment asks you to balance two things and watch for nothing to happen, that is the design telling you it wants to be precise.
The apparatus
Oil is sprayed in, the drops pick up charge from friction at the nozzle, and one lucky drop is trapped in the uniform field between the plates.
Why oil, and not water?
Water would evaporate while you watched it. The drop would shrink, its mass would change, and the whole calculation — which leans entirely on m staying put — would collapse. Oil barely evaporates at all. Its mass is constant for as long as you need.
How the drops get charged
As the mist is forced out of the nozzle, the drops rub against it and are charged by friction
Some drops lose electrons and become positive; others gain electrons and become negative
Alternatively the air in the chamber can be ionised by X-rays, and the drops pick up the ions
The two situations
The drop is negative and the top plate is positive, so E points down while the force on the drop points up. That is what holds it there.
With no field: terminal velocity
Switch the plates off and the drop simply falls. As it speeds up, air resistance grows, until the upward drag exactly matches the downward weight. The resultant force is zero, the acceleration is zero, and it drifts down at a steady terminal velocity.
This stage is not decoration. Millikan timed the drop’s terminal velocity with the field off, and used it — through Stokes’ law — to work out the drop’s radius, and hence its mass. In an IB question the mass is usually handed to you, so you can skip straight to the balance. But that’s where m comes from. We’ll do Stokes’ law properly on its own page.
With a field: the balance
Apply a potential difference across the plates. The field between them is uniform, and the drop — being charged — feels an electric force.
Electric force on the dropF = qEE = field strength (N C−1) • q = charge on the drop (C)
Weight of the dropW = mg
Turn the voltage up and the drop slows, stops, then starts to rise. Catch it at the moment it hangs motionless and the two forces must be equal and opposite:
The balance conditionqE = mgand for parallel plates E = V/d , so qV/d = mgq = mgd / Vevery quantity on the right is measurable — so q is measurable
Spray oil, drops pick up charge
tune V until the drop stops
qE = mg
and E = V/d
q = mgd/V
What the results showed
Millikan did this thousands of times, with thousands of drops. Every drop gave a different charge — and yet the charges were not random at all. Sort them and a pattern jumps out: they are all whole-number multiples of one number.
No drop ever came back with 2.4 or 5.5 units. The smallest step between any two charges is the elementary charge.
Nothing forces this. If charge were a smooth, continuous fluid, drops would carry any old value and the dots would scatter along that line. They don’t. They sit on the ticks. That single stubborn fact is the whole conclusion:
Millikan’s conclusionthe charge on any object is always a multiple of 1.60 × 10−19 C — charge is quantised
Note carefully what Millikan did not do. He never saw an electron, never isolated one, never claimed to. He measured the charges on drops, and let the arithmetic do the talking. The electron’s charge is simply the size of the smallest step. When an examiner asks “how does this provide evidence for quantisation”, that step is the answer they want — not the apparatus.
🔬 Working a Millikan question
Is the drop stationary? If yes, the forces balance: qE = mg. Start there, every time.
Given V and d, not E? Use E = V/d, with d in metres.
Find the charge.q = mgd/V. Use g = 9.81 N kg−1.
How many electrons?N = q/e. It should come out very close to a whole number — round it.
Given a list of charges? Find the highest common factor, or the smallest difference between them. That is e.
WE 1
An oil drop of mass 4.08 × 10−15 kg is held stationary between two horizontal plates 6.0 mm apart, with a potential difference of 500 V across them. Determine (a) the charge on the drop and (b) the number of excess electrons it carries. (g = 9.81 N kg−1)
(a) Step 1 — the drop is stationary, so the forces balanceqE = mg and E = V/d → q = mgd / VStep 2 — convert d, then substituted = 6.0 mm = 6.0 × 10⁻³ mq = (4.08 × 10⁻¹⁵) × 9.81 × (6.0 × 10⁻³) / 500q = 4.80 × 10⁻¹⁹ C(b) Step 3 — charge is quantised, so use N = q / eN = (4.80 × 10⁻¹⁹) / (1.60 × 10⁻¹⁹)N = 3 electronsSanity check: the field is E = 500 / 0.0060 = 8.3 × 10⁴ N C⁻¹, and qE = 4.0 × 10⁻¹⁴ N, which is exactly mg. Also notice N came out as a clean whole number — if yours gives 3.7, you have made an arithmetic slip, because there is no such thing as 3.7 electrons.
WE 2
In a repeat of Millikan’s experiment, four drops are measured to carry charges of 3.2 × 10−19 C, 4.8 × 10−19 C, 6.4 × 10−19 C and 8.0 × 10−19 C. (a) Deduce a value for the elementary charge. (b) State the number of excess electrons on each drop. (c) Explain what these results tell you about electric charge.
(a) Step 1 — look at the gaps between successive charges4.8 − 3.2 = 1.6 | 6.4 − 4.8 = 1.6 | 8.0 − 6.4 = 1.6
The smallest step, and the highest common factor, is the same number.
e = 1.6 × 10⁻¹⁹ C(b) Step 2 — divide each charge by e3.2/1.6 = 2 | 4.8/1.6 = 3 | 6.4/1.6 = 4 | 8.0/1.6 = 52, 3, 4 and 5 electrons(c) Step 3 — say what it means
Every charge is a whole-number multiple of one smallest value.
No drop carries a fraction of it.
charge is quantisedDon’t say “the charges are all similar”. Say the magic words: every measured charge is an integer multiple of 1.6 × 10⁻¹⁹ C, so charge exists only in discrete lumps.
WE 3
(a) Explain why oil is used rather than water. (b) A negatively charged drop is held stationary between two horizontal plates. State, with a reason, which plate is at the higher potential. (c) The voltage is now increased slightly. Describe and explain the motion of the drop.
(a) why oil
Water would evaporate during the measurement.
Its mass would change, but the calculation assumes m is constant.
oil does not evaporate quickly, so m stays fixed(b) which plate is positive
The drop is negative, so the electric force must point up to balance the weight.
The force on a negative charge is opposite to the field, so E points down.
Field lines run from positive to negative.
the top plate is at the higher potential(c) increase the voltageE = V/d increases → qE increases
Now qE > mg, so there is a resultant upward force.
the drop accelerates upwardsIn part (c) it does not rise at constant speed. It accelerates, then drag builds up, and only after that does it settle to a new terminal velocity. Say “accelerates upwards” and you have the mark.
💡 Top tips
“Oil doesn’t evaporate, so the mass stays constant” — that exact reason is the mark.
Stationary drop ⇒ write qE = mg immediately. It’s the whole question.
Convert plate separation to metres before using E = V/d. Millimetres kill answers here.
Your N = q/e must land near a whole number. It’s a free check on your arithmetic.
Given a list of charges, hunt the smallest difference, not the smallest charge.
Quantisation is shown by integer multiples, not by “the numbers are small”.
Millikan measured the charge on drops, and deduced e. He never trapped an electron.
⚠ Common mistakes
Saying oil is used “because it is heavier” or “because it is easier to see”. It’s about evaporation
Leaving d in millimetres in E = V/d
Thinking the balanced drop has no forces on it. It has two, and they cancel
Forgetting the drop must be charged. An uncharged drop feels no electric force and just falls
Saying the drop “rises at constant speed” the instant the voltage goes up. It accelerates
Using q = mgV/d. Check by units, or by asking whether a wider gap should need more charge to hold the drop (it should)
Claiming Millikan “measured the charge of an electron directly”. He measured drops, and took the smallest step
Quick recap: Millikan and Fletcher (1909) sprayed charged oil drops — oil because it will not evaporate, so the mass stays constant — between two parallel plates. With no field a drop falls at terminal velocity. With a field, the electric force qE can be tuned until it exactly balances the weight mg, and since E = V/d, this gives q = mgd/V. Every drop’s charge turned out to be a whole-number multiple of 1.60 × 10−19 C. That is the experimental evidence that charge is quantised.
One detail we glossed over: the drops arrive already charged, straight out of the nozzle. Why? Because rubbing two surfaces together moves electrons — the same effect that makes a balloon stick to a wall and gives you a shock off a car door. It has a name, a set of rules, and three distinct mechanisms you need to be able to describe. Next page: Static Electricity.
Oil drops floating away from you?
Book a free meeting and we’ll work through the force balance, the q = mgd/V derivation and the “evidence for quantisation” answer.
