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Millikan's oil drop experiment

Thomson measured a ratio; Millikan pinned down the ingredient. By floating single droplets of oil between charged plates and timing them as they fell, he read off the charge on the electron and found nature's smallest permissible instalment.

Year 13AQA 3.12.1.4

Builds on Cathode rays and the electron and Coulomb's law and electric field strength.

IN THIS TOPIC

  • Use the balance condition QV/d = mg for a stationary charged droplet.
  • Use Stokes' law and terminal speed to find a droplet's radius and mass.
  • Explain the significance of Millikan's results: charge is quantised in units of e.

WHAT YOU PROBABLY THINK

Electric charge comes in any amount you like, like water from a tap.

Holding a droplet still

Thomson's ratio left a gap: e and m were only known combined. Robert Millikan's idea, perfected around 1913, was to measure the charge alone, one droplet at a time. An atomiser sprays a mist of oil above a pair of horizontal plates; friction in the spray leaves some droplets charged, a few drift through a small hole, and a microscope watches them in the illuminated space between the plates.

With a pd V across plates a distance d apart, the uniform field E = V/d exerts a force QE on a droplet of charge Q. Adjust V until one droplet hangs perfectly still, and the electric force exactly supports the weight:

QVd = mgNOT ON THE DATA SHEET — LEARN IT
Millikan's balance condition: a charged oil droplet hangs stationary between charged plates when the electric force QV over d equals its weight mg+−electric force = QV/dweight = mga charged oil droplet, watched through a microscopestationary: adjust V until the electric pull exactly holds the weight
FIG. 1The balance condition. A charged droplet hangs stationary when the electric force QV over d exactly equals its weight, and the voltage dial reads off the balance.

WORKED EXAMPLE

Reading a droplet's charge

A droplet of mass 3.69 × 10−15 kg hangs stationary between plates 5.0 mm apart when the pd is 565 V. Find its charge.

Q = mgd/V = (3.69 × 10−15 × 9.81 × 5.0 × 10−3) / 565.

Q = 3.2 × 10−19 C: exactly twice 1.6 × 10−19 C. This droplet carries two electrons' worth of charge.

The equation is a see-saw: heavier droplets or wider gaps need more volts, and the microgram-scale masses are why the voltages stay in the hundreds rather than the millions.

Weighing the invisible

The balance equation has a hole in it: the droplet's mass, far too small for any scale. Millikan's solution is the part examiners love. Switch the field off and watch the droplet fall. Within milliseconds it reaches terminal speed, where the viscous drag of the air balances its weight, and for a small sphere that drag is given by Stokes' law:

F = 6π ηrvNOT ON THE DATA SHEET — LEARN IT
With the field off the droplet falls at terminal speed: the viscous drag of Stokes' law balances the weight, and the measured speed hands over the droplet's radiusfalls at steady speed vdrag = 6πηrvweight = mgterminal speed: measure v under the microscope,and Stokes' law hands you the radius, then the mass
FIG. 2Field off: the droplet falls at a steady terminal speed, drag balancing weight. The measured speed is the only unknown besides the radius, so the radius follows.

with η the viscosity of air and r the droplet's radius. At terminal speed, 6πηrv equals the weight, and the weight itself is (4/3)πr3ρg with ρ the oil's density. One measured speed therefore pins down r, and from r the mass. Timing a speck across the microscope's graduations weighs it.

YOUR TURN

From a stopwatch to a mass

A droplet falls at a steady 2.4 × 10−4 m s−1 with the field off. Using η = 1.8 × 10−5 Pa s and oil density 880 kg m−3, find its radius and mass, before opening the working.

Show the working

Setting 6πηrv = (4/3)πr3ρg and cancelling gives r = √9ηv/2ρg = √(9 × 1.8 × 10−5 × 2.4 × 10−4 / (2 × 880 × 9.81)) = 1.5 × 10−6 m.

m = (4/3)πr3ρ = 1.2 × 10−14 kg. A speck three micrometres across, weighed with nothing but a stopwatch and a viscosity table.

The smallest instalment

Millikan ran the measurement on droplet after droplet, hundreds of them, charged by chance in the spray. The charges that came out were not spread smoothly. Every single droplet carried a whole-number multiple of one value, 1.6 × 10−19 C, and nothing ever landed in between.

Millikan's result: measured droplet charges cluster at whole-number multiples of one value, 1.6 times ten to the minus nineteen coulombs, and never in betweene2e3e4emeasured droplet chargecharges land on the rungs, never between themcharge is quantised: e is the indivisible unit
FIG. 3The result that mattered: droplet charges cluster at e, 2e, 3e and 4e, and the spaces between the rungs stay empty. Charge cannot be subdivided below e.

So much for charge flowing like water from a tap. Charge is quantised: it exists only in instalments of the electronic charge e, nature's indivisible unit, and any measured charge is some whole number of electrons added or missing. Combined with Thomson's ratio, the electron was finally complete: e measured directly, and the mass following at once from e divided by e/m.

TRY IT UNSEEN

Completing the electron

Using Millikan's e = 1.60 × 10−19 C and Thomson's e/m = 1.76 × 1011 C kg−1, find the mass of the electron.

Show the working

m = e ÷ (e/m) = 1.60 × 10−19 / 1.76 × 1011 = 9.1 × 10−31 kg.

Two experiments, sixteen years apart, and a division: the first measured mass of a fundamental particle, about one two-thousandth of a hydrogen atom, exactly as Thomson's comparison predicted.

THE EXAM BIT

  • The balance condition QV/d = mg is the anchor: name every symbol, and remember V/d is the field strength between parallel plates, an idea imported from the electric fields unit.
  • The field-off measurement exists to find m: terminal speed, Stokes' drag 6πηrv equal to weight, radius from the speed, mass from the radius. Tell it as that chain.
  • Stokes' law needs its conditions: a small sphere moving slowly through a fluid. At a droplet's size and speed both hold, which is why the method works.
  • "Explain the significance of Millikan's results" wants quantisation stated in full: every measured charge is a whole-number multiple of 1.6 × 10−19 C, so charge comes only in units of e.
  • The classic follow-up pairs the experiments: Millikan's e with Thomson's e/m gives the electron's mass. Practise that one-line division.

CHECK YOURSELF

A droplet weighing 3.84 × 10−14 N hangs stationary between plates 6.0 mm apart with 480 V across them. Find the droplet's charge, and state how many electrons' worth it carries.

Show a hint

Rearrange QV/d = mg for Q; the weight mg is given whole.

Show the answer

Q = (weight × d)/V = (3.84 × 10−14 × 6.0 × 10−3) / 480 = 4.8 × 10−19 C.

4.8 × 10−19 / 1.6 × 10−19 = 3: the droplet carries three electrons' worth of charge.

A non-integer answer here means an arithmetic slip, and that check, whole numbers or nothing, is Millikan's discovery in miniature.

Hold the droplet still: QV/d = mg. Let it fall: Stokes' law weighs it.

Charge only comes in whole multiples of e; there is no smaller instalment.

WORKBOOK

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Question workbook

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Question workbook

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Answer book

PDF · 52 KB · fully worked · navy

CHECK YOUR PROGRESS

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  • Use the balance condition QV/d = mg for a stationary charged droplet.
  • Use Stokes' law and terminal speed to find a droplet's radius and mass.
  • Explain the significance of Millikan's results: charge is quantised in units of e.

Open the full revision checklist to track your progress across the whole unit.

No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.