Year 12 Physics Module 6 ⏱ ~40 min 5 MC · 2 Short Answer Lesson 1 of 21

Uniform Electric Fields

In 1909, Robert Millikan at the University of Chicago suspended charged oil droplets motionless between two parallel metal plates separated by 16 mm, balancing gravity against a precisely tuned electric field. His measurement of the elementary charge (1.59 × 10⁻¹⁹ C, within 1% of today's accepted value) earned him the 1923 Nobel Prize, and it all depended on understanding the uniform field between parallel plates.

Today's hook: In Millikan's 1909 oil drop experiment at the University of Chicago, two horizontal plates 16 mm apart were connected to a high-voltage supply. A charged oil droplet of mass 1.6 × 10⁻¹⁴ kg, carrying a single elementary charge (1.6 × 10⁻¹⁹ C), hung perfectly still in mid-air. What voltage did Millikan need to apply across the plates to hold it there?
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Orient and predict

Recall what field direction means, then predict how the sign of a charge changes its force.

Worksheets

Practise this lesson

Use the available foundations worksheet for guided practice on this lesson.

Build Foundations & guided practice
Before you read, predict

Two large metal plates are connected to a 9 V battery: the top plate is positive, the bottom plate is negative. A tiny positive dust particle is released halfway between them.

Before reading on, answer:

  1. Which way will the dust particle move, toward the top plate, the bottom plate, or stay still? Why?
  2. If you replaced the dust particle with an electron, would it move the same way, the opposite way, or not at all?
  3. The plates are 3 cm apart. Make a rough guess: how strong is the electric field between them, in volts per metre?

Warm-up, electric field lines between parallel plates point from the positive plate to the negative plate. Where is the field strongest?

Learning Intentions
goals

Know, Field Definitions

  • An electric field is a region where a charged particle experiences a force
  • Between large parallel plates, away from their edges, $E = |\Delta V|/d$ is approximately uniform in magnitude and direction
  • The force on a charge is $F = qE$; direction depends on charge sign

Understand, Why Uniform?

  • Why the field between large parallel plates is uniform (not $1/r^2$ like a point charge)
  • Why a negative charge experiences force opposite to the field direction
  • How electric field strength relates to potential gradient

Can Do, Calculate & Predict

  • Calculate $E$ from plate voltage and separation
  • Calculate force and acceleration on charged particles in uniform fields
  • Predict direction of motion for positive and negative charges
Scan these before reading
vocab
Electric field ($\vec E$)Force per unit positive test charge at a point. Its direction is the force direction for a positive test charge; units are N/C or V/m.
Uniform electric fieldA field with constant magnitude and direction everywhere in the region.
Potential difference ($\Delta V$)Difference in electric potential between two points, measured in volts (V). In this lesson $V$ means its positive magnitude across the plates.
Electric field strength (parallel plates)$E = |\Delta V|/d$, where $d$ is the perpendicular distance between large plates, away from their edges.
Elementary charge ($e$)$1.60 \times 10^{-19}$ C. Magnitude of charge on a proton or electron.
Cross-lesson links: M6 opens with electric fields, the same physics you used in Year 11 (Coulomb's law) is now extended to uniform fields between parallel plates. The Millikan experiment is the first HSC application: identifying the charge on the electron by balancing forces.

Core Content

2

Model the uniform field

Read the field diagram, define the plate model and use $E=V/d$ with units.

1
Uniform Electric Fields Between Parallel Plates
+5 XP

Why the field is constant, and how to calculate its strength

When two large conducting plates are connected to a battery, one becomes positively charged and the other negatively charged. Near the centre, the charges arrange themselves so that the electric field between the plates is uniform, meaning it has the same magnitude and direction throughout that modelled region.

Higher potential, positive plate (+) Lower potential, negative plate (−) d Uniform field E = V/d V

Figure 1, Uniform electric field lines between two oppositely charged parallel plates

This is fundamentally different from the electric field around a point charge, which follows Coulomb's law and falls off as $1/r^2$. Between parallel plates, the field lines are straight, parallel, and equally spaced because the plates are large relative to their separation. Edge effects exist near the boundaries, but we ignore them in the ideal model.

Key Relationship, Uniform Field

$E = \dfrac{|\Delta V|}{d}$

$E$ = electric field strength (N/C or V/m)   ·   $V$ = potential difference between plates (V)   ·   $d$ = perpendicular separation (m)

Stop & Check

Two plates are 5.0 cm apart with 200 V across them. A student claims the field is stronger near the positive plate because "that's where the force starts." Is this correct? Explain using the diagram above.

For large parallel plates, away from the edges, the electric field is approximately uniform: $E = |\Delta V|/d$ (V/m or N/C). Its field lines are straight, parallel and equally spaced, unlike the radial $1/r^2$ field of a point charge.

Pause, copy the highlighted definition and formula into your book before moving on.

Two parallel plates are connected to a 12 V battery with separation 3.0 cm. What is the electric field strength between them?

3

Represent force as a vector

Separate the electric field from the force it produces on positive and negative charges.

2
Force on a Charged Particle
+5 XP

Direction matters: why positive and negative charges move differently

We just saw that the electric field between parallel plates is uniform and given by $E = V/d$. That raises a question: what force does this field exert on a charge placed inside it, and which way does it act? This card answers it → $F = qE$, with direction depending on charge sign.

A charged particle placed in an electric field experiences a force. The magnitude is simple: $F = qE$. The direction depends entirely on the sign of the charge, and getting this right is a key exam skill.

Force on a Charge

$\vec F = q\vec E$   and   $|\vec F|=|q|E$

$\vec F$ = force on the charge (N)   ·   $q$ = signed charge (C)   ·   $\vec E$ = electric field (V/m or N/C)

The direction of the force depends on the sign of the charge:

  • Positive charge ($q > 0$): Force is in the same direction as $\vec{E}$, toward the negative plate.
  • Negative charge ($q < 0$): Force is in the opposite direction to $\vec{E}$, toward the positive plate.
Positive plate Negative plate + F (down) − F (up) Positive q Negative q

Figure 2, A positive charge experiences force in the direction of E; a negative charge experiences force opposite to E

Misconceptions to fix
✗ Wrong: "The electron moves toward the negative plate because opposite charges attract."
✓ Right: The electron moves toward the positive plate because the electric field points from positive to negative, and the force on a negative charge is opposite to the field direction.
Stop & Check

An oxygen ion ($O^{2-}$) is placed between two horizontal plates where the field points upward. In which direction does the force act on the ion? Explain in one sentence.

The force on a charged particle in a uniform electric field is $F = qE$ (N). Positive charges experience force in the same direction as $\vec{E}$; negative charges (electrons) experience force opposite to $\vec{E}$, toward the positive plate.

Add the highlighted principle to your notes before the check below.

A negative charge placed in an electric field experiences a force in the same direction as the field.

The force on a proton and an electron placed in the same electric field have equal magnitude but opposite direction.

4

Calculate a particle response

Chain field, force and acceleration while keeping the electron’s direction separate from magnitude.

3
Worked Example, Electron Between Charged Plates
+5 XP

Calculate field strength, force, and acceleration for a charged particle

We just saw that $F = qE$ gives the force on a charge in a uniform field. That raises a question: how do you chain $E = V/d$, $F = qE$, and $a = F/m$ into a single calculation? This card answers it → a three-step worked example using an electron between charged plates.

Problem

Two horizontal parallel plates are separated by 2.0 cm and connected to a 100 V power supply. The top plate is positive and the bottom plate is negative. An electron is released from rest near the negative plate.

  1. (a) Calculate the electric field strength between the plates.
  2. (b) Calculate the magnitude of the electric force on the electron.
  3. (c) Calculate the acceleration of the electron.
Step 1, Electric field strength

Use $E = V/d$ with $V = 100$ V and $d = 0.020$ m.

$E = \dfrac{100}{0.020} = 5.0 \times 10^{3}$ V/m

Direction: From the positive plate toward the negative plate.

Step 2, Force on the electron

Use $F = qE$. For an electron, $q = -e = -1.60 \times 10^{-19}$ C.

$F = (1.60 \times 10^{-19})(5.0 \times 10^{3}) = 8.0 \times 10^{-16}$ N

Direction: The electron is negative, so the force is opposite to $\vec{E}$, upward, toward the positive plate.

Step 3, Acceleration

Use Newton's second law: $a = F/m$. Mass of electron $m_e = 9.11 \times 10^{-31}$ kg.

$a = \dfrac{8.0 \times 10^{-16}}{9.11 \times 10^{-31}} = 8.8 \times 10^{14}$ m/s$^2$

Context: This acceleration is roughly $10^{13}\,g$. The electric force on an electron completely dominates gravity, which is why we ignore gravity when analysing charged particles in electric fields.

Step order for plate problems: (1) choose an axis and identify $\vec E$, (2) use $\vec F=q\vec E$, (3) use $\vec a=\vec F/m$. In this electron example the electric acceleration is about $10^{13}$ times $g$, so gravity is negligible; do not discard gravity in a different situation unless the force comparison justifies it.

Pause, write the highlighted step sequence into your book before moving on.

Two parallel plates are 4.0 cm apart with 200 V across them. The electric field strength is _____ V/m.

5

Compare models and limits

Use the parallel-plate model where it applies, and distinguish it from a point-charge field.

4
From Coulomb's Law to Uniform Fields
+5 XP

Connecting point-charge fields to the parallel-plate approximation

We just saw how to calculate field, force and acceleration between parallel plates. That raises a question: how does the uniform $E = V/d$ model relate to Coulomb's radial field, and when should you use each? This card answers it → parallel plates superpose many point charges to produce a uniform field; always use $E = V/d$, not Coulomb's law, for plate problems.

In Year 11, you learned Coulomb's law for the force between two point charges. Understanding how that connects to, and differs from, the uniform parallel-plate model is an essential HSC skill.

Coulomb's Law

$F = k\dfrac{|q_1q_2|}{r^2}$

$k = 8.99 \times 10^{9}$ N m$^2$ C$^{-2}$   ·   A single point charge has $E = k|Q|/r^2$: outward for $+Q$, inward for $-Q$, and not uniform.

So why is the field between parallel plates uniform ($E = V/d$, independent of position)?

  • Each plate contains many charges distributed across its surface.
  • Near the centre of the plates, the contributions from all those charges add up to produce a field that is nearly constant in both magnitude and direction.
  • This is an idealisation valid when the plates are large compared to their separation and we ignore edge effects.
HSC Tip

For large parallel plates, use the uniform-field approximation away from the edges unless told otherwise. Use $E = |\Delta V|/d$, not Coulomb's law, for this plate model.

Stop & Check

A student says: "Since $E = V/d$, if I move the plates further apart while keeping the same battery connected, the field gets weaker. But if I use Coulomb's law, the field from each charge gets weaker with distance too. So both equations say the same thing." Is this reasoning valid? Explain the flaw.

Point charge field: magnitude $E = k|Q|/r^2$, radial and falling as $1/r^2$. Large parallel plates: $E = |\Delta V|/d$, approximately uniform away from edges because many surface-charge contributions superpose. Use the stated model, not both equations at once.

Add the highlighted distinction to your notes before the check below.

Three of these statements about uniform electric fields between parallel plates are correct. Pick the odd one out.

6

Apply and consolidate

Use the plate model quantitatively, then explain how charge sign and mass change a particle's response.

Activity 1, Parallel Plate Calculations
ApplyBand 3

Practise field strength, force and acceleration calculations

  1. Set the plates to 12 V and 4.0 cm apart. Calculate the electric field strength. Show your working: $E = V/d$.
  2. A positive charge of $3.2 \times 10^{-19}$ C is placed between those plates. Calculate the force on it and state its direction.
  3. Double the voltage to 24 V (keep $d = 4.0$ cm). By what factor does the force change? Explain why.
  4. Return to 12 V but halve the separation to 2.0 cm. By what factor does the force change? Explain the physical reason.

The potential difference between two parallel plates is doubled while their separation is halved. The electric field strength:

Activity 2, Concept Challenge
UnderstandBand 4

Explain force and acceleration for different particles

A proton and an electron are released from rest in the same uniform electric field. Which one hits the opposite plate first, or do they arrive simultaneously? Explain using $F = qE$ and $a = F/m$.

Synthesis, connect the ideas

In this lesson, three separate ideas lock together:

  • For large parallel plates, away from edges, the electric field is approximately uniform: $E = |\Delta V|/d$.
  • A charged particle in this field experiences $F = qE$, with direction depending on charge sign.
  • The acceleration $a = qE/m$ is enormous for electrons, which is why we ignore gravity in these problems.
7

Practise uniform fields

Retrieve the model, then calculate field, force, acceleration and direction.

Quick recall, electric fields & Coulomb's law
+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 5 marks
+5 XP

ApplyBand 4(2 marks) 1. Two identical metal plates are arranged horizontally, 4.0 cm apart, with the top plate at +80 V and the bottom plate at 0 V. Calculate (a) the electric field strength and (b) the magnitude of the electric force on an electron placed between the plates.

1 mark: correct $E$ with unit conversion · 1 mark: correct $F = qE$ with electron charge

EvaluateBand 5(3 marks) 2. An electron is released from rest near the negative plate of a parallel plate system with $E = 5.0 \times 10^3$ V/m. (a) Calculate the acceleration of the electron. (b) Explain why gravitational force on the electron can be ignored in this calculation. (c) State the direction of the electron's motion.

1 mark: correct $a = F/m$ · 1 mark: comparison of electric and gravitational forces · 1 mark: correct direction stated

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1 (2 marks): (a) $E = V/d = 80 / 0.040 = 2.0 \times 10^3$ V/m (1 mark, must convert cm to m). (b) $F = qE = (1.60 \times 10^{-19})(2.0 \times 10^3) = 3.2 \times 10^{-16}$ N (1 mark).

Q2 (3 marks): (a) $F = eE = (1.60 \times 10^{-19})(5.0 \times 10^3) = 8.0 \times 10^{-16}$ N. $a = F/m_e = 8.0 \times 10^{-16} / 9.11 \times 10^{-31} = 8.8 \times 10^{14}$ m/s² (1 mark). (b) The electric force ($8.0 \times 10^{-16}$ N) is approximately $10^{13}$ times larger than the gravitational force ($m_e g \approx 8.9 \times 10^{-30}$ N), so gravity is negligible (1 mark). (c) The electron, being negatively charged, experiences a force opposite to the field direction, it moves toward the positive plate (1 mark).

8

Review the model

Check what transferred, then return to the force-balance prediction.

Check what actually stuck
How did your thinking change?

At the start you were asked about Millikan's 1909 oil drop experiment, specifically what voltage he needed across 16 mm plates to hold a 1.6 × 10⁻¹⁴ kg droplet carrying one elementary charge stationary.

The answer: balance gives $qE = mg$, so $E = mg/q$. With $q = 1.6 \times 10^{-19}$ C (one elementary charge), $E = (1.6 \times 10^{-14} \times 9.8)/(1.6 \times 10^{-19}) \approx 9.8 \times 10^5$ V/m, and $V = Ed = 9.8 \times 10^5 \times 0.016 \approx 15{,}700$ V. This is exactly the scale of voltage Millikan used in his apparatus. He measured 1.59 × 10⁻¹⁹ C, within 1% of the modern value of 1.602 × 10⁻¹⁹ C.

The key insight: the uniform parallel-plate field makes the electric force constant throughout the gap, just like gravity, which is why Millikan could set up a perfect force balance. In a non-uniform field this method would not work.

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