IB Physics SLTopic 4 — Electric & Magnetic FieldsPaper 1 & 2Eq = mg~7 min read
Millikan’s Oil-Drop Experiment
Last page we said charge comes in indivisible lumps of e. But how do you measure something that small? In 1909 Robert Millikan found a gorgeously simple trick: float a tiny charged oil drop in mid-air, perfectly balanced between gravity pulling down and an electric force pushing up — and from that balance, read off the charge on a single drop.
📘 What you need to know
Millikan (with Fletcher, 1909) used charged oil drops to measure the elementary chargee = 1.60 × 10⁻¹⁹ C
Oil is used, not water, because it barely evaporates — so each drop’s mass stays constant during the measurement
Drops are charged as they leave the spray nozzle (by friction, or ionised by X-rays), then watched through a microscope between two horizontal parallel plates
With no field, a drop falls and reaches terminal velocity, where air resistance = weight
Applying a p.d. makes an electric field; with the right voltage the drop hovers, so the upward electric force balances the weight: Eq = mg
Rearranging gives the drop’s charge, q = mg ÷ E
Every measured charge came out as a whole-number multiple of 1.60 × 10⁻¹⁹ C — direct evidence that charge is quantised
The Big Idea
The previous page claimed charge only exists in lumps of a smallest unit, e. Millikan’s experiment is the evidence. By measuring the charge on hundreds of individual oil drops, he found they were never just any value — every single one was a whole-number multiple of one tiny amount, 1.60 × 10⁻¹⁹ C. That smallest step is the elementary charge, the charge on one electron.
The Apparatus
The set-up is a box of clever simplicity. A spray (atomiser) puffs a fine mist of oil into a chamber. As the droplets squeeze out of the nozzle they pick up charge by friction (some lose electrons and go positive, some gain electrons and go negative). A few drift down through a small hole into the gap between two horizontal metal plates, where an experimenter watches a single drop through a microscope.
Oil is sprayed in and charged by friction, a few drops fall through a hole into the uniform field between two parallel plates, and one is tracked through a microscope while the p.d. is tuned.
Two Forces, One Drop
Field off — falling at terminal velocity
With no voltage across the plates, a drop simply falls under gravity. As it speeds up, air resistance grows until it matches the drop’s weight. The forces now cancel, so the drop stops accelerating and drifts down at a steady terminal velocity.
Field on — making the drop hover
Now switch on a p.d. across the plates to create an electric field. The field pushes on the drop’s charge with an electric force:
Electric force on a chargeF = Eq
Where F is the electric force (N), E is the electric field strength (N C⁻¹) and q is the drop’s charge (C). Tune the voltage carefully and this upward electric force can be made to exactly cancel the weight. The drop then hangs motionless — the whole point of the experiment.
Left: with no field the drop settles to a terminal velocity where air resistance balances weight. Right: with the field on and the voltage tuned, the upward electric force Eq exactly balances the weight mg, so the drop hangs still.
A hovering drop means the up and down forces are equal. Set the electric force equal to the weight and rearrange for the charge:
electric force up
balances
weight down
Eq = mg
q = mg ÷ E
Knowing the field strength E, the drop’s mass m, and g, you get the charge q on that one drop. Repeat for drop after drop.
Why It Proves Charge Is Quantised
Here’s the payoff. When Millikan collected the charges from hundreds of drops, they didn’t scatter randomly. Every value was a whole-number multiple of the same tiny amount — 1e, 2e, 3e, … of 1.60 × 10⁻¹⁹ C, but never 1.5e or 2.3e. Charge comes in indivisible packets, and the size of one packet is the charge on a single electron. That’s exactly what “charge is quantised” means, measured directly.
🧭 Solving a balanced-drop problem
Draw the two forces — weight mg down, electric force Eq up
Hovering means balanced — set them equal: Eq = mg
Rearrange for what’s asked — usually q = mg ÷ E
Find the number of electrons with N = q ÷ e — it should come out (near) a whole number
Sanity-check — if N isn’t close to an integer, re-check your powers of ten and units
Quick recap: charged oil drops (oil so the mass stays fixed) are balanced in a field between parallel plates. Hovering means Eq = mg, so q = mg ÷ E; every drop’s charge is a whole multiple of e = 1.60 × 10⁻¹⁹ C, proving charge is quantised.
WE 1
(a) Explain why oil droplets are used rather than water droplets. (b) The p.d. is adjusted until a charged drop hangs motionless. Explain what this tells you about the forces on the drop, and how it gives the drop’s charge.
Part (a) — why oil
Oil barely evaporates, so the drop’s mass stays constant while it’s measured
Water would evaporate, shrinking the drop and changing mg mid-experiment.Part (b) — what “motionless” means
No motion → no acceleration → resultant force = 0
So the upward electric force must exactly balance the weight:
Eq = mgq = mg ÷ EWith E, m and g known, this gives q. Millikan found every q was a whole multiple of 1.60 × 10⁻¹⁹ C.
WE 2
An oil drop of mass 4.9 × 10⁻¹⁵ kg is held stationary between charged plates where the electric field strength is 3.0 × 10⁴ N C⁻¹. (a) Calculate the charge on the drop. (b) How many excess electrons does it carry? (Take g = 9.8 N kg⁻¹, e = 1.60 × 10⁻¹⁹ C.)
Part (a) — balance the forces: Eq = mgq = mg ÷ E = (4.9 × 10⁻¹⁵ × 9.8) ÷ (3.0 × 10⁴)q = 4.8 × 10⁻¹⁴ ÷ 3.0 × 10⁴q = 1.6 × 10⁻¹⁸ CPart (b) — number of electrons: N = q ÷ eN = (1.6 × 10⁻¹⁸) ÷ (1.60 × 10⁻¹⁹)N = 10 electronsA clean whole number — exactly what “charge is quantised” predicts.
💡 Top tips
Oil, not water: the reason is always evaporation — oil keeps the drop’s mass constant during the measurement
“Stationary” is the trigger word: it means the forces balance, so write Eq = mg straight away
Expect a whole number of electrons. If q ÷ e isn’t near an integer, hunt for a powers-of-ten slip
What it measured: the experiment finds the chargee and shows charge is quantised — not the electron’s mass
⚠ Common mistakes
Forgetting that a stationary drop has balanced forces — writing an F = ma equation with a leftover acceleration
Mixing up the force directions — weight is always down; the electric force is the one tuned to point up and cancel it
Saying oil is used because it’s “heavier” or “denser” — the real reason is it doesn’t evaporate
Claiming the experiment measures the electron’s mass — it measures its charge
One balanced drop, and the charge of the electron falls out. Up next: Static Electricity — how everyday objects pick up charge in the first place (friction, induction and contact), the sparks it can cause, and why we earth things to stay safe.
Want this to actually click before the exam?
Book a free meeting and let’s work through the tricky bits together.