TASC TCE 2022-style practice: In a Millikan-type experiment, a small plastic ball of density 941\ kg m^-3 and volume 3.08 × 10^-18\ m^3 is held stationary between two metal plates with a potential difference of 285\ V separated by 5.00\ mm. Calculate the weight of the sphere, the electric field strength between the plates and the charge on the sphere. If this is 3 elementary charges, find the experimental value of the elementary charge.

When the ball is stationary the upward electric force balances its weight: qE = mg. Mass: m = V = (941)(3.0810^-18) = 2.90 × 10^-15\ kg. Weight: W = mg = (2.9010^-15)(9.81) = 2.84 × 10^-14\ N. Field strength: E = Vd = 2850.005 = 5.70 × 10^4\ V m^-1. Charge: q = WE = 2.8410^-145.7010^4 = 4.99 × 10^-19\ C. If this is 3 elementary charges, e = q3 = 4.9910^-193 = 1.66 × 10^-19\ C. This is close to the accepted 1.60 × 10^-19\ C, illustrating that charge is quantised. Markers want qE = mg, E = Vd, then dividing by the number of charges.

§-Syllabus dot point

TASPhysicsSyllabus dot point

How was the charge on a single electron measured?

Explain how Millikan's oil drop experiment balanced electric and gravitational forces to find the elementary charge.

How Millikan balanced the electric force on a charged oil drop against its weight to measure the quantised elementary charge, and the calculation behind it.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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What this dot point is asking

This dot point is a classic application of balancing forces in fields, and it gives the experimental evidence that charge is quantised.

The apparatus

Millikan's apparatus had two horizontal parallel plates with a known separation, connected to a variable voltage. A fine mist of oil was sprayed above the top plate; the drops picked up charge through friction in the sprayer and from ionising radiation. A drop drifted through a small hole into the uniform field between the plates, where it could be watched through a microscope.

Between the plates the field is uniform, $E = \dfrac{V}{d}$ , so a charged drop feels a constant electric force $F = qE$ that can be directed upward to oppose gravity.

Balancing the forces

By adjusting the voltage, Millikan could bring a drop to rest, hovering motionless. At that moment the upward electric force exactly balanced the downward weight:

qE = mg \quad\Rightarrow\quad q\frac{V}{d} = mg.

Rearranging gives the charge on the drop:

q = \frac{mgd}{V}.

The mass of each tiny drop was found from its terminal velocity when falling without a field, using the viscosity of air (Stokes' law). Knowing $m$ , $g$ , $d$ and the balancing voltage $V$ gave the charge $q$ .

In practice the drop is rarely brought to a perfect standstill. Millikan often let it rise or fall at a steady (terminal) speed instead, where the electric force, the weight and the air drag balance. Because the drag depends on the same air viscosity used to find the mass, comparing the rising and falling speeds at a known voltage gives the charge without ever needing the drop to hover, which improved the precision of his result.

Why oil and not water

Millikan chose oil rather than water for a subtle but important reason: water evaporates quickly, which would change a drop's mass during the measurement and ruin the force balance. Oil has a very low vapour pressure, so a drop keeps a constant mass for the minutes needed to time its fall and balance it. The drops also had to be extremely small, around a micrometre across, so that the electric force from just a few electrons could balance the weight at a practical voltage.

The discovery of quantisation

Repeating the measurement for many drops, Millikan found that the charges were never random. Every value was a whole-number multiple of one smallest amount:

q = ne

where $n$ is an integer and $e = 1.6 \times 10^{-19}\ \text{C}$ . This showed that charge comes in indivisible packets, the elementary charge, carried by the electron. It was direct evidence that electric charge is quantised, one of the foundations of modern physics, and it let Millikan determine $e$ with good accuracy.

Worked example: charge on a balanced drop

An oil drop of mass $3.3 \times 10^{-15}\ \text{kg}$ hovers between plates $6.0\ \text{mm}$ apart when the voltage is $410\ \text{V}$ . Find the charge on the drop and the number of excess electrons.

Step 1: Apply the force-balance equation

When the drop hovers without moving, the upward electric force and the downward weight are exactly equal, giving the balance condition $qE = mg$ . Substituting $E = V/d$ and rearranging for charge yields $q = \dfrac{mgd}{V}$ , where every quantity on the right is known from the setup. Substituting in:

q = \frac{(3.3\times10^{-15})(9.8)(6.0\times10^{-3})}{410} = 4.7\times10^{-19}\ \text{C}.

Step 2: Find the number of electrons

Because charge is quantised in units of the elementary charge $e = 1.6 \times 10^{-19}\ \text{C}$ , the total charge on the drop must be a whole-number multiple of $e$ . Dividing the measured charge by $e$ gives that integer directly:

n = \frac{q}{e} = \frac{4.7\times10^{-19}}{1.6\times10^{-19}} \approx 3.

Final answer: The drop carries about $3$ excess electrons, with a total charge of $4.7 \times 10^{-19}\ \text{C}$ .

In the exam, set up the balance condition $qE = mg$ with $E = \dfrac{V}{d}$ , solve for $q$ , then divide by $1.6 \times 10^{-19}\ \text{C}$ to find the number of electrons. Round to the nearest whole number, since charge is quantised.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20226 marksIn a Millikan-type experiment, a small plastic ball of density

941\ \text{kg m}^{-3}

and volume

3.08 \times 10^{-18}\ \text{m}^3

is held stationary between two metal plates with a potential difference of

285\ \text{V}

separated by

5.00\ \text{mm}

. Calculate the weight of the sphere, the electric field strength between the plates and the charge on the sphere. If this is

3

elementary charges, find the experimental value of the elementary charge.

Show worked answer →

When the ball is stationary the upward electric force balances its weight: $qE = mg$ .

Mass: $m = \rho V = (941)(3.08\times10^{-18}) = 2.90 \times 10^{-15}\ \text{kg}$ .
Weight: $W = mg = (2.90\times10^{-15})(9.81) = 2.84 \times 10^{-14}\ \text{N}$ .

Field strength: $E = \dfrac{V}{d} = \dfrac{285}{0.005} = 5.70 \times 10^4\ \text{V m}^{-1}$ .

Charge: $q = \dfrac{W}{E} = \dfrac{2.84\times10^{-14}}{5.70\times10^4} = 4.99 \times 10^{-19}\ \text{C}$ .

If this is $3$ elementary charges, $e = \dfrac{q}{3} = \dfrac{4.99\times10^{-19}}{3} = 1.66 \times 10^{-19}\ \text{C}$ .

This is close to the accepted $1.60 \times 10^{-19}\ \text{C}$ , illustrating that charge is quantised. Markers want $qE = mg$ , $E = \dfrac{V}{d}$ , then dividing by the number of charges.

TCE 20193 marksMillikan measured the charge on many different oil drops and found values such as

1.6

3.2

4.8

and

6.4 \times 10^{-19}\ \text{C}

. Explain what this pattern shows about electric charge and why no drop was ever found with a charge of

2.4 \times 10^{-19}\ \text{C}

Show worked answer →

Every measured charge is a whole-number multiple of the same smallest value, $1.6 \times 10^{-19}\ \text{C}$ : the listed values are $1\times$ , $2\times$ , $3\times$ and $4\times$ this amount.

This shows that electric charge is quantised, that is, it comes only in indivisible packets of size $e = 1.6 \times 10^{-19}\ \text{C}$ , the elementary charge carried by the electron. A drop gains charge by gaining or losing whole electrons, so its total charge must be $q = ne$ for some integer $n$ .

A charge of $2.4 \times 10^{-19}\ \text{C}$ would be $1.5$ times the elementary charge, a non-integer multiple, so it cannot occur: you cannot transfer half an electron. Markers want the idea $q = ne$ with integer $n$ , identification of $e$ , and the explanation that fractional multiples are impossible.