Year 11 Physics Module 4 ⏱ ~35 min 5 MC · 3 Short Answer Lesson 2 of 17

Electric Fields & Field Lines

Sydney Transgrid 330 kV transmission lines (Wolseley–Kemps Creek, 2022): conductors separated 8 m apart at 330,000 V; average field between conductors E = V/d = 330,000/8 = 41,250 V/m. Near the conductor surface, E ≈ 100 kV/m, sufficient to ionise air at E > 3,000 kV/m if conductors come closer. The corona discharge wastes ~1 MW of power along the 170 km line.

Today's hook: Sydney Transgrid's 330 kV Wolseley–Kemps Creek transmission line (2022): conductors 8 m apart, average electric field E = V/d = 330,000/8 = 41,250 V/m between them, and corona discharge wastes ~1 MW of power along the 170 km line. An electric field this strong exists between the conductors with nothing physical there, just altered space. What is an electric field, and how does field line density reveal its strength and direction?
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Warm up first

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Before you read, predict

A photocopier uses a rotating drum coated with a light-sensitive semiconductor. In the dark, the drum is electrically charged. Toner particles are attracted to the charged regions and transferred to paper.

Predict 1: How could an electric field cause toner particles to move toward the drum without any physical contact?

Predict 2: Why must the electric field be uniform across the drum surface for a clear, consistent image?

Warm-up, the unit of electric field strength N C⁻¹ is equivalent to:

Learning Intentions
goals

Know

  • Electric field is defined as force per unit positive test charge: $E = F/q$
  • Field line conventions: direction, density, never crossing
  • The formulas $E = F/q$, $E = kq/r^2$, and $E = V/d$

Understand

  • Why field line density represents field strength
  • Why the field inside parallel plates is uniform
  • How positive and negative charges move differently in a uniform field

Can Do

  • Draw and interpret field line diagrams for point charges and parallel plates
  • Calculate electric field strength using all three formulas
  • Predict the motion of a charged particle in a uniform electric field
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Scan these before reading

Scan these before reading
vocab
Electric field ($E$)The region around a charge where another charge experiences a force. Defined as force per unit positive test charge: $E = F/q$.
Field linesVisual representations showing the direction and relative strength of an electric field. Begin on positive charges, end on negative charges.
Uniform fieldA field with constant magnitude and direction everywhere, produced between parallel plates: $E = V/d$.
Test chargeA small positive charge used to probe the field without disturbing it. The field direction is the direction of force on this charge.
Cross-lesson links: L01 introduced force between charges. L02 introduces the electric field (E = F/q), the region of influence around a charge. Field lines give a visual map of force direction and strength. Understanding them is essential for L03 (potential energy), L04 (particle motion), and the entire electromagnetism strand.
Misconceptions to fix
✗ Wrong: Electric field lines can cross each other.
✓ Right: Field lines never cross. At any point in space, the electric field has only one direction.
✗ Wrong: A negative charge in an electric field moves in the same direction as the field.
✓ Right: The field direction is for a positive test charge. A negative charge experiences force in the opposite direction to the field.
✗ Wrong: $E = V/d$ works for any arrangement of charges.
✓ Right: $E = V/d$ is valid only for the uniform field between two parallel plates. For point charges, use $E = kq/r^2$.
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What is an Electric Field?

Core Content

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What is an Electric Field?
+5 XP

A property of space altered by charge

Hold a charged plastic rod near a thin stream of water from a tap, the water bends toward the rod without touching it. Place a small charged piece of aluminium foil on a table and bring a charged balloon close, the foil slides across the table toward the balloon. In both cases, a force acts across empty space. The region around the charged object has been physically altered.

Electric field definition

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

Units: N C⁻¹ = V m⁻¹. Direction = force on a positive test charge.

The electric field is a property of space it exists whether or not a test charge is present to feel it.

An electric field is a region of space where a charge experiences a force. It is defined as $E = F/q$ (N C⁻¹ = V m⁻¹), with direction equal to the force on a positive test charge. The field exists in space independent of any test charge.

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

A test charge of $+2.0 \times 10^{-9}\ \text{C}$ experiences a force of $6.0 \times 10^{-5}\ \text{N}$ at a point. The electric field at that point is:

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Electric Field Lines

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Electric Field Lines
+5 XP

Visual maps of direction and strength

We just saw that an electric field exists in space and exerts a force on charges. That raises a question: how do we visually represent a field that is invisible? This card answers it → field lines map direction and relative strength across all of space.

Field lines are a visual tool for representing electric fields. They show both the direction and relative strength of the field at every point in space.

Rules for Drawing Field Lines

  • Direction: Lines point in the direction of force on a positive test charge.
  • Origin and termination: Lines begin on positive charges and end on negative charges (or extend to infinity).
  • Density = strength: Where lines are close together, the field is strong.
  • Never cross: Field lines cannot intersect, the field has only one direction at any point.
Four electric field patterns: arrows radiate outward from a positive point charge, point inward to a negative point charge, curve from positive to negative in a dipole, and run in equally spaced parallel lines from a positive plate to a negative plate.
Field arrows always show the force direction on a positive test charge: away from +, toward −, and from the positive plate to the negative plate.

Trace: Start at the tail of one arrow in each panel and follow it to its endpoint. Then compare the line spacing near a point charge with the equal spacing between the plates.

ConfigurationField Line Pattern
Single positive point chargeRadial lines pointing outward
Single negative point chargeRadial lines pointing inward
Electric dipole (+/−)Lines from + to −, curving between them
Two equal positive chargesLines repel outward; zero-field point midway
Parallel plates (opposite charges)Parallel, equally spaced lines from + to −
Real-World Anchor
A photocopier's drum is charged uniformly by a corona wire. The uniform electric field ensures toner particles are attracted evenly to charged regions. If the field were non-uniform, some regions would pull toner more strongly, producing a blotchy or faded copy.

Electric field lines start on positive charges and end on negative charges; they never cross. Line density is proportional to field strength. Between parallel plates the lines are parallel, equally spaced, and uniform, pointing from the positive plate to the negative plate.

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

Electric field lines can cross where the field is very strong.

More densely packed field lines indicate a stronger electric field.

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Calculating Electric Field Strength

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Calculating Electric Field Strength
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Two formulas for two different field geometries

We just saw that field lines map direction and strength. That raises a question: how do we calculate the actual numerical value of field strength for a point charge versus parallel plates? This card answers it → two distinct formulas cover the two geometries.

Point Charge

Field of a point charge

$E = k\dfrac{|q|}{r^2}$   where $k = 8.99 \times 10^9\ \text{N m}^2\text{ C}^{-2}$

Inverse-square: double the distance, quarter the field. Points outward from positive $q$, inward toward negative $q$.

Uniform Field Between Parallel Plates

Uniform field between plates

$E = \dfrac{V}{d}$   where $V$ = potential difference, $d$ = plate separation

Constant in magnitude and direction between the plates. Points from the positive plate to the negative plate.

Worked example, Photocopier drum field

Parallel plates separated $d = 5.0\ \text{mm}$, potential difference $V = 800\ \text{V}$. A toner particle carries $q = -1.6 \times 10^{-15}\ \text{C}$. Find (a) field strength, (b) force on particle.

  1. $d = 5.0 \times 10^{-3}\ \text{m}$
  2. $E = V/d = 800/(5.0 \times 10^{-3}) = 1.6 \times 10^5\ \text{V m}^{-1}$
  3. $F = |q|E = (1.6 \times 10^{-15})(1.6 \times 10^5) = 2.56 \times 10^{-10}\ \text{N}$
  4. Direction: toner is negative → force opposite to field → toward positive plate.

For a point charge: $E = k|q|/r^2$ (inverse-square, N C⁻¹). For parallel plates: $E = V/d$ (uniform field only). Force on any charge in a field: $F = qE$, with direction depending on the sign of $q$.

Pause, write the highlighted formulas into your book before the check below.

Two parallel plates are separated by 4.0 cm and connected to a 600 V supply. The field strength between the plates is:

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Motion in a Uniform Electric Field

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Motion in a Uniform Electric Field
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Constant force → constant acceleration (like gravity)

We just saw that $E$ can be calculated for both point charges and parallel plates. That raises a question: what happens to a charged particle placed in a uniform electric field? This card answers it → constant force gives constant acceleration, analogous to projectile motion.

Because the field between parallel plates is uniform, a charged particle placed there experiences a constant force and therefore a constant acceleration.

Force and acceleration in uniform field

$F = qE$   and   $a = \dfrac{qE}{m}$

  • Positive charge: accelerates in the direction of the field (toward negative plate).
  • Negative charge: accelerates opposite to the field (toward positive plate).
  • If the particle enters perpendicular to the field: follows a parabolic trajectory (constant $v$ parallel to plates, constant $a$ perpendicular).
HSC Exam Connection
The motion of charged particles in uniform electric fields is the foundation for cathode ray tubes (Lesson 4) and, in Year 12, particle accelerators and mass spectrometers. Master this now, it appears in almost every HSC electromagnetism question.

In a uniform electric field, a charge experiences constant force $F = qE$ and constant acceleration $a = qE/m$. Positive charges accelerate with the field; negative charges opposite to it. A particle entering perpendicular follows a parabolic trajectory.

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

Three of these are correct about electric field lines. Pick the odd one out.

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Activity 1, Analyse and Connect: Field Diagram Interpretation

Activity 1, Analyse and Connect: Field Diagram Interpretation
ApplyBand 3

A field line diagram shows lines emerging from point A and curving toward point B, with denser lines near A.

  1. Identify the sign of charge on A and B. Justify using field line conventions.
  2. At which point is the field stronger, near A or near B? What feature tells you this?
  3. Sketch the path a positive test charge would follow if released from rest at a point on the perpendicular bisector between A and B.

Two parallel plates are 2.0 cm apart with a potential difference of 400 V. The field strength is _____ V m⁻¹.

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Activity 2, Analyse and Connect: Parallel Plate Problem

Activity 2, Analyse and Connect: Parallel Plate Problem
ApplyBand 4

Two parallel plates separated by 4.0 cm. Top plate at +600 V, bottom at 0 V.

  1. Calculate the electric field strength and state its direction.
  2. A proton ($m_p = 1.67 \times 10^{-27}\ \text{kg}$, $q_p = +1.60 \times 10^{-19}\ \text{C}$) is released from the positive plate. Calculate the force and initial acceleration.
  3. An electron is released from the negative plate simultaneously. Compare accelerations of proton and electron and explain any difference.
Wrap-up, Key Definitions

Electric field definition

$E = F/q$ (N C⁻¹ = V m⁻¹). Direction = force on positive test charge. Field exists in space independent of test charge.

Field line rules

Start on +, end on −. Density ∝ strength. Never cross. Parallel plates: uniform, equally spaced.

Point charge field

$E = k|q|/r^2$. Inverse-square. Outward for +$q$, inward for −$q$.

Uniform field

$E = V/d$. Valid only between parallel plates. +charge accelerates with field; −charge opposite.

✓

Show what you have learned

Quick recall, electric fields & field lines
+5 XP

5 questions drawn from the lesson bank, feedback shown immediately.

Short Answer, 10 marks
+5 XP

UnderstandBand 3(3 marks) 3. Define electric field strength and explain why the direction of the electric field is defined as the direction of force on a positive test charge, not a negative one.

ApplyBand 4(3 marks) 4. Two parallel plates are separated by 3.0 cm and connected to a 900 V supply. A dust particle of mass $2.0 \times 10^{-6}\ \text{kg}$ and charge $+4.8 \times 10^{-12}\ \text{C}$ is placed between the plates. Calculate the electric field strength and the acceleration of the dust particle.

EvaluateBand 6(4 marks) 5. A student designs a particle separator using two parallel vertical plates. Charged ink droplets enter horizontally. Evaluate how changing plate separation $d$ and potential difference $V$ independently would affect the deflection, with reference to $E = V/d$ and $a = qE/m$.

Show all answers

Activity 1, Model Answers

  1. A is positive (field lines emerge from it); B is negative (field lines terminate on it).
  2. Field is stronger near A, the field lines are denser there. Field line density is proportional to field strength.
  3. A positive test charge released on the perpendicular bisector would accelerate toward B (the negative charge), curving along a field line.

Activity 2, Model Answers

  1. $E = V/d = 600/(4.0 \times 10^{-2}) = 1.5 \times 10^4\ \text{V m}^{-1}$ downward (from + to − plate).
  2. $F = qE = (1.60 \times 10^{-19})(1.5 \times 10^4) = 2.4 \times 10^{-15}\ \text{N}$. $a = F/m = (2.4 \times 10^{-15})/(1.67 \times 10^{-27}) = 1.44 \times 10^{12}\ \text{m s}^{-2}$ toward the negative plate.
  3. The electron has the same charge magnitude but ~1836× smaller mass, so its acceleration is ~1836× greater. The electron accelerates upward (toward positive plate); the proton downward.

Short Answer, Model Answers

Q3 (3 marks): Electric field strength is the force per unit positive test charge at a point: $E = F/q$. The direction is defined using a positive test charge to provide a single, unambiguous convention, if a negative charge were used, the force direction would be opposite to the field, creating confusion.

Q4 (3 marks): $E = V/d = 900/(3.0 \times 10^{-2}) = 3.0 \times 10^4\ \text{V m}^{-1}$. $F = qE = (4.8 \times 10^{-12})(3.0 \times 10^4) = 1.44 \times 10^{-7}\ \text{N}$. $a = F/m = (1.44 \times 10^{-7})/(2.0 \times 10^{-6}) = 7.2 \times 10^{-2}\ \text{m s}^{-2}$ toward negative plate.

Q5 (4 marks): Deflection depends on $a = qE/m = qV/(md)$. Increasing $d$ while $V$ constant decreases $E$, so $a$ and deflection decrease. Increasing $V$ while $d$ constant increases $E$ proportionally, increasing deflection. To maximise deflection, increase $V$ or decrease $d$. However, decreasing $d$ too much risks dielectric breakdown (sparking), so a practical lower limit exists for $d$.

✓

Retrieve, reflect and finish

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How did your thinking change?

The Sydney Transgrid 330 kV Wolseley–Kemps Creek line (2022) you read about in the hero has conductors 8 m apart and an average field of 41,250 V/m between them, enough to cause corona discharge wasting ~1 MW. That field exists in empty space, exerting forces on any charge that enters it.

Now look back at your Think First answers about the photocopier: the field exerts a force $F = qE$ on charged toner particles, the same principle as the Transgrid field acting on charged air molecules. A uniform field ($E = V/d$) ensures consistent force at every point.

← Electric Charge & Electrostatic Force Next: Electric Potential Energy & Work Done →

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