Q4 Buoyancy, Terminal speed, Stokes’ law

Electric Fields, Charged Oil Drops, Terminal Speed, and Quantized Charge

IB Physics HL • Paper 2 • Long-Answer Question • Electric Fields, Forces, Viscosity, and Millikan Oil-Drop Physics

Lesson Overview

In this Paper 2 long-answer question, we study the electric field between two oppositely charged parallel plates and then apply this idea to a charged oil drop suspended between the plates. The question develops into force balance, buoyancy, electric force, terminal speed, viscous drag, and finally the quantization of electric charge.

This is a classic IB Physics context because it links fields, mechanics, and particle ideas in one experiment. To score well, you should identify the forces clearly, write the equilibrium condition carefully, and explain how the equations connect to the physical motion of the drop.

⭐ Key Concepts

The electric field strength between parallel plates is

$E=\frac{V}{d}$

The electric force on a charged particle in an electric field is

$F=qE$

The weight of an object is

$W=mg$

The mass of a spherical drop can be written as

$m=\rho V$

The buoyancy force is the weight of displaced air.
If the density of the oil is much larger than the density of air, the buoyancy force is much smaller than the weight of the oil drop.
At terminal speed, the resultant force is zero.
For a small sphere moving through a fluid, the viscous drag force is

$F_D=6\pi \eta rv$

Electric charge is quantized in integer multiples of the elementary charge

$e=1.60\times10^{-19}\text{ C}$

📘 Clear IB-Style Explanation

This question is based on the physics behind the Millikan oil-drop experiment. The central idea is that a tiny oil drop can be held stationary when the upward electric force balances its weight.

A strong IB response should distinguish clearly between:

the electric field between the plates,
the electric force on the drop,
the weight of the drop,
and the drag force when the drop falls at terminal speed.

The final part of the question is especially important conceptually. Even though the drop splits into two parts, the charge on each part must still be an integer multiple of the elementary charge. This is strong evidence that charge is quantized.

📌 Worked Example 1 — Electric Field Between Parallel Plates

Two parallel conducting plates are oppositely charged.

(a)(i) Draw the electric field lines due to the charged plates.

(a)(ii) The potential difference between the plates is $960\text{ V}$ and the distance between them is $8.0\text{ mm}$ . Calculate the electric field strength $E$ between the plates.

Solution

(a)(i) Electric field lines

The electric field lines between parallel plates are:

straight,
parallel,
equally spaced,
directed from the positive plate to the negative plate.

Small edge effects may also be shown near the sides of the plates.

📊 Marks: A1 A1

(a)(ii) Electric field strength

For parallel plates,

$E=\frac{V}{d}$

Convert the distance into metres:

$d=8.0\text{ mm}=8.0\times10^{-3}\text{ m}$

Substitute:

$E=\frac{960}{8.0\times10^{-3}}$

$E=1.2\times10^5\text{ V m}^{-1}$

Therefore, the electric field strength is

$1.2\times10^5\text{ N C}^{-1}$

📊 Marks: M1 A1

📌 Worked Example 2 — Why the Oil Drop Becomes Charged

An oil drop is introduced into the space between the plates through a small hole in the upper plate. The oil drop moves through air in a tube before falling between the plates.

(b) Explain why the oil drop becomes charged as it falls through the tube.

Solution

As the oil drop moves through the tube, it collides and interacts with air molecules or with the walls of the tube.

These interactions can transfer electrons to or from the drop by friction.

Therefore, the oil drop becomes electrically charged.

📊 Marks: A1

📌 Worked Example 3 — Why Buoyancy Is Much Smaller Than the Weight

The oil drop is observed to be stationary in the space between the plates. Buoyancy is one of the forces acting on the drop.

The density of oil is $730$ times greater than that of air.

(c)(i) Show that the buoyancy force is much smaller than the weight.

Solution

The weight of the oil drop is

$W_{\text{oil}}=\rho_{\text{oil}}Vg$

The buoyancy force equals the weight of displaced air:

$F_B=\rho_{\text{air}}Vg$

So the ratio is

$\frac{F_B}{W_{\text{oil}}}=\frac{\rho_{\text{air}}Vg}{\rho_{\text{oil}}Vg}$

$=\frac{\rho_{\text{air}}}{\rho_{\text{oil}}}$

Since the oil is $730$ times denser than air,

$\frac{\rho_{\text{air}}}{\rho_{\text{oil}}}=\frac{1}{730}$

$\approx1.4\times10^{-3}$

This ratio is much less than

$1$

so the buoyancy force is much smaller than the weight.

📊 Marks: M1 M1 A1

📌 Worked Example 4 — Forces on the Stationary Oil Drop

(c)(ii) Draw the forces acting on the oil drop, ignoring the buoyancy force.

Solution

Ignoring buoyancy, the two forces on the stationary oil drop are:

weight, acting vertically downward,
electric force, acting vertically upward.

Because the drop is stationary, these two forces are equal in magnitude.

📊 Marks: A1 A1

📌 Worked Example 5 — Deriving the Charge on the Oil Drop

(c)(iii) Show that the electric charge on the oil drop is given by

$q=\frac{\rho V g}{E}$

where $\rho$ is the density of oil and $V$ is the volume of the oil drop.

(c)(iv) State the sign of the charge on the oil drop.

Solution

(c)(iii) Derivation

Since the drop is stationary, the electric force equals the weight:

$qE=mg$

The mass of the drop is

$m=\rho V$

Substitute into the force balance:

$qE=\rho V g$

$q=\frac{\rho V g}{E}$

Hence the required expression is shown.

📊 Marks: M1 A1

(c)(iv) Sign of the charge

The electric force on the drop is upward, while the electric field is downward.

For the force to act opposite to the field direction, the charge must be

$\text{negative}$

📊 Marks: A1

📌 Worked Example 6 — Why the Drag Force Equals $6\pi \eta rv$

The electric field is turned off. The oil drop falls vertically, reaching a constant speed $v$ .

(d)(i) Outline why, for this drop, the drag force is

$6\pi \eta rv$

where $\eta$ is the viscosity of air and $r$ is the radius of the oil drop.

Solution

When the drop reaches constant speed, it is moving at terminal speed.

At terminal speed, the resultant force on the drop is zero, so the acceleration is zero.

Therefore, the downward weight is balanced by the upward drag force.

For a small sphere moving through a fluid, the viscous drag force is given by Stokes’ law:

$F_D=6\pi \eta rv$

So for this oil drop, the drag force has this form.

📊 Marks: A1 A1

📌 Worked Example 7 — Showing the Charge Is About $3\times10^{-19}\text{ C}$

The following data for the oil drop are available:

$\eta=1.8\times10^{-5}\text{ Pa s}$

$r=8.2\times10^{-7}\text{ m}$

$v=1.4\times10^{-4}\text{ m s}^{-1}$

$\rho=920\text{ kg m}^{-3}$

(d)(ii) Show that the charge on the oil drop is about $3\times10^{-19}\text{ C}$ .

Solution

At terminal speed, the drag force equals the weight of the drop:

$6\pi \eta rv=\rho V g$

For a spherical drop,

$V=\frac43\pi r^3$

So the mass, and hence the weight, can be found from the radius. Equivalently, we first calculate the radius from the terminal-speed balance and then substitute into the electric-force balance.

Using the given values:

$6\pi \eta rv=6\pi(1.8\times10^{-5})(8.2\times10^{-7})(1.4\times10^{-4})$

$=3.89\times10^{-14}\text{ N}$

This is the weight of the drop to a very good approximation.

From the earlier force balance when the drop was stationary between the plates,

$qE=W$

with

$E=1.2\times10^5\text{ N C}^{-1}$

$q=\frac{W}{E}$

$q=\frac{3.89\times10^{-14}}{1.2\times10^5}$

$q=3.24\times10^{-19}\text{ C}$

Therefore, the charge on the oil drop is about

$3.2\times10^{-19}\text{ C}$

which is approximately

$3\times10^{-19}\text{ C}$

📊 Marks: M1 M1 A1

📌 Worked Example 8 — Charge Quantization When the Drop Splits

The oil drop splits into two parts of equal mass. Both are charged.

(d)(iii) Deduce the net charge on each part.

Solution

The original charge is about

$3.2\times10^{-19}\text{ C}$

This is approximately

$2e$

since

$e=1.60\times10^{-19}\text{ C}$

Electric charge is quantized, so the two smaller drops must each have a net charge equal to an integer multiple of $e$ .

The only possible split of a total charge of

$2e$

between two charged parts is

$e\text{ and }e$

$2e\text{ and }0$

But since both parts are charged, the second option is not allowed.

Therefore, each part must have net charge

$1e=1.60\times10^{-19}\text{ C}$

📊 Marks: A1 A1

⚠ Common Mistakes

Drawing electric field lines with unequal spacing between parallel plates.
Forgetting to convert $8.0\text{ mm}$ into metres before calculating the electric field strength.
Including buoyancy as comparable to the weight, even though the density ratio shows it is much smaller.
Getting the sign of the oil-drop charge wrong by not comparing the field direction with the force direction.
For terminal speed, forgetting that the resultant force is zero.
Using Stokes’ law incorrectly by leaving out one of the factors in

$6\pi \eta rv$

Assuming charge can split continuously, instead of recognizing that charge is quantized in integer multiples of $e$ .

📘 IB Exam Tips

In parallel-plate questions, always think first about field direction and force direction.
When the drop is stationary, start with a clear force balance.
When the drop is falling at constant speed, say explicitly that it is moving at terminal speed.
Use density arguments carefully when deciding whether buoyancy can be ignored.
For quantization questions, compare the charge found with integer multiples of

$e$

Always finish with a direct sentence that answers the question clearly.

🧪 Challenge Problem

Two parallel plates are separated by $6.0\text{ mm}$ and have a potential difference of $720\text{ V}$ . A small charged oil drop is stationary between the plates.

(a) Calculate the electric field strength between the plates.

(b) Explain why the oil drop can become charged as it moves through air.

(c) State the two main forces on the stationary drop if buoyancy is ignored.

(d) If the weight of the drop is $2.4\times10^{-14}\text{ N}$ , calculate the magnitude of the charge on the drop.

✅ Self-Practice

An oil drop of radius $7.0\times10^{-7}\text{ m}$ falls through air and reaches terminal speed $1.2\times10^{-4}\text{ m s}^{-1}$ . The viscosity of air is $1.8\times10^{-5}\text{ Pa s}$ .

(a) Calculate the drag force on the oil drop at terminal speed.

(b) Explain why this drag force is equal in magnitude to the weight of the drop.

(c) The oil drop is later held stationary in an electric field of strength $1.0\times10^5\text{ N C}^{-1}$ . Calculate the charge on the drop.

(d) Deduce how many elementary charges this corresponds to.

Show Solutions

Challenge Problem

(a)

$E=\frac{V}{d}=\frac{720}{6.0\times10^{-3}}$

$E=1.2\times10^5\text{ N C}^{-1}$

(b)

The oil drop becomes charged because collisions and friction with air molecules or the tube can transfer electrons to or from the drop.

(c)

The two main forces are the weight downward and the electric force upward.

(d)

$q=\frac{W}{E}$

$q=\frac{2.4\times10^{-14}}{1.2\times10^5}$

$q=2.0\times10^{-19}\text{ C}$

Self-Practice

(a)

$F_D=6\pi \eta rv$

$F_D=6\pi(1.8\times10^{-5})(7.0\times10^{-7})(1.2\times10^{-4})$

$F_D=2.85\times10^{-14}\text{ N}$

(b)

At terminal speed the acceleration is zero, so the resultant force is zero. Therefore, the drag force equals the weight in magnitude.

(c)

$q=\frac{W}{E}=\frac{2.85\times10^{-14}}{1.0\times10^5}$

$q=2.85\times10^{-19}\text{ C}$

(d)

$\frac{q}{e}=\frac{2.85\times10^{-19}}{1.60\times10^{-19}}\approx1.78$

So this would correspond experimentally to about

$2e$

since charge is quantized.

Electric Fields, Charged Oil Drops, Terminal Speed, and Quantized Charge

Lesson Overview

⭐ Key Concepts

📘 Clear IB-Style Explanation

📌 Worked Example 1 — Electric Field Between Parallel Plates

Solution

📌 Worked Example 2 — Why the Oil Drop Becomes Charged

Solution

📌 Worked Example 3 — Why Buoyancy Is Much Smaller Than the Weight

Solution

📌 Worked Example 4 — Forces on the Stationary Oil Drop

Solution

📌 Worked Example 5 — Deriving the Charge on the Oil Drop

Solution

📌 Worked Example 6 — Why the Drag Force Equals $6\pi \eta rv$

Solution

📌 Worked Example 7 — Showing the Charge Is About $3\times10^{-19}\text{ C}$

Solution

📌 Worked Example 8 — Charge Quantization When the Drop Splits

Solution

⚠ Common Mistakes

📘 IB Exam Tips

🧪 Challenge Problem

✅ Self-Practice

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Q4 Buoyancy, Terminal speed, Stokes’ law

Electric Fields, Charged Oil Drops, Terminal Speed, and Quantized Charge

Lesson Overview

⭐ Key Concepts

📘 Clear IB-Style Explanation

📌 Worked Example 1 — Electric Field Between Parallel Plates

Solution

📌 Worked Example 2 — Why the Oil Drop Becomes Charged

Solution

📌 Worked Example 3 — Why Buoyancy Is Much Smaller Than the Weight

Solution

📌 Worked Example 4 — Forces on the Stationary Oil Drop

Solution

📌 Worked Example 5 — Deriving the Charge on the Oil Drop

Solution

📌 Worked Example 6 — Why the Drag Force Equals

Solution

📌 Worked Example 7 — Showing the Charge Is About

Solution

📌 Worked Example 8 — Charge Quantization When the Drop Splits

Solution

⚠ Common Mistakes

📘 IB Exam Tips

🧪 Challenge Problem

✅ Self-Practice

Related

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📌 Worked Example 6 — Why the Drag Force Equals $6\pi \eta rv$

📌 Worked Example 7 — Showing the Charge Is About $3\times10^{-19}\text{ C}$