Virtual laboratory/Physics/Electrostatics/Millikan's oil-drop experiment — the elementary charge

Millikan's oil-drop experiment — the elementary charge

An extension of the topic of charge and the electric field. A charged oil drop hangs between the plates when the electric force balances its weight (qE = mg). The measured charges of successive drops are always a multiple of the elementary charge e — proof that charge is quantised (Nobel Prize 1923).

Millikan's oil-drop experiment — the elementary charge
Physics · Electrostatics · Millikan oil-drop experiment (FIZ.ELEKTROSTAT.PP.05)
Voltage across the plates UU200 V
The measured charges of many different drops are…
voltage U
200 V
field E
33.3 kV/m
force qE
1.6·10⁻¹⁴ N
charge q
3·e
🎯 Challenge: balance the drop
streak: 0

I will knock the drop out of balance — your task is to find the voltage at which it hangs still again (qE = mg).

What just happened?

An oil drop between the capacitor plates carries a charge and is subject to two forces: its weight mgmg (downward) and the electric force qEqE (upward). By adjusting the voltage you find the value of UU at which the forces balance — the drop hangs motionless.

Balance condition:

qE=mg,E=UdqE = mg,\qquad E = \dfrac{U}{d}

— the field between the plates depends on the voltage and the plate spacing. Substituting, for a suspended drop:

q Ud=mg⇒q=mgdUq\,\dfrac{U}{d} = mg \quad\Rightarrow\quad q = \dfrac{mgd}{U}

What did Millikan discover?

Measuring the charge of hundreds of different drops, he always obtained a multiple of the same portion: q=n⋅eq = n\cdot e, where n=1,2,3,…n = 1, 2, 3, \dots Electric charge is quantised — it does not come in arbitrary amounts, but in portions of:

e≈1.6⋅10−19 Ce \approx 1{.}6\cdot10^{-19}\ \text{C}

Why you need this for the exam:

It is the landmark proof that the elementary charge exists. The trick: at balance set qE=mgqE=mg, substitute E=U/dE=U/d and solve for qq — and the result you get can always be written as n⋅en\cdot e.

Remember: at balance qE = mg, so q = mgd/U — and every measured charge is a multiple of the elementary charge, q = n·e.

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Theory & formulas

Equilibrium of the drop

A charged drop hangs motionless when the electric force balances its weight: qE=mg,E=UdqE = mg, \qquad E = \dfrac{U}{d}

Quantisation of charge

The measured charges of successive drops are always a multiple of the elementary charge: q=n e,e≈1.6⋅10−19 Cq = n\,e, \qquad e\approx1{.}6\cdot10^{-19}\ \mathrm{C} Proof that charge is quantised (Millikan, Nobel Prize 1923).

Symbols: qq — charge of the drop, E=U/dE=U/d — field strength, UU — voltage, dd — plate separation, mm — mass of the drop, gg — acceleration, nn — a whole number, ee — elementary charge.

Frequently asked questions

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What does Millikan's experiment determine?

The elementary charge e — the smallest portion of charge (e≈1.6·10⁻¹⁹ C).

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Why are the charges multiples of e?

Because charge is quantised — the drop carries a whole number of excess electrons, q=n·e.

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When does the drop hang motionless?

When the electric force balances the weight: qE=mg, with E=U/d.

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