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Physics (IAL)

Electric Fields

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Edexcel IAL Physics · Fields & Forces

Electric Fields

The Big Idea: any charged object creates an invisible "zone of influence" around itself — an electric field — and any other charge that enters that zone feels a push or a pull, whose strength depends on how much charge is involved and how far apart they are.

📋 Summary — What This Chapter Covers
  • Defining an electric field — a region where a charged particle feels a force, created by all charges around it.
  • Electric field strength (E) — force per unit charge on a positive test charge; a vector, points away from + and towards −.
  • Coulomb's Law — the force between two point charges: proportional to the charges, inversely proportional to the square of the distance.
  • Field due to a point charge — radial field, obeys an inverse-square law (1/r²).
  • Field & potential relationship — E is the gradient of the potential-distance graph.
  • Uniform fields between parallel plates — E = V/d, constant field strength everywhere between the plates.
  • Electric potential (V) in a radial field — work done per unit charge bringing a test charge from infinity; follows a 1/r relationship (not 1/r²).
  • Representing fields — field lines (direction + strength) and equipotential lines (surfaces of equal potential, always perpendicular to field lines).
1 · Defining an Electric Field

What actually is an electric field?

Think of any charged object — a balloon you've rubbed on your jumper, an electron, a proton — as constantly "broadcasting" its presence into the space around it. That broadcast is the electric field. Formally:

Definition An electric field is a region of space in which a charged particle experiences a force.

Crucially, the charged particle sitting in that field doesn't need to be moving — it feels the force whether it's stationary or in motion. This is a really important distinction from magnetic fields, where a charge only feels a force if it's .

Every charged particle creates its own field, and that field exerts an electrostatic force (FE) on any other charged particle that happens to be within range. The two golden rules of how charges interact:

  • Like charges repel (+ and +, or − and −) — the force pushes them apart.
  • Opposite charges attract (+ and −) — the force pulls them together.

And the size of that force isn't fixed — it changes with distance. Charges close together feel a much stronger push or pull than charges far apart. (We'll quantify exactly how much in Coulomb's Law below.)

Watch out Don't mix up electric and magnetic fields! A charge sitting still in an electric field still feels a force. A charge sitting still in a magnetic field feels nothing — it only feels a force once it starts moving.
Practice Question
A student says "electric fields only affect moving charges, just like magnetic fields." Explain why this statement is incorrect.
2 · Electric Field Strength (E)

Measuring "how strong" a field is at a point

Saying "there's an electric field here" is a bit like saying "it's windy outside" — useful, but not precise. To be precise, physicists define electric field strength, E, at a point as:

Definition The force per unit charge acting on a positive test charge placed at that point.

Why specifically a test charge? Because it fixes the direction convention. If we always imagine dropping a tiny positive charge into the field and asking "which way does it get pushed", then:

  • Field lines point away from a positive charge (it repels our positive test charge).
  • Field lines point towards a negative charge (it attracts our positive test charge).

E is a vector — it has both a size and a direction, and that direction is exactly the direction of the field lines.

Key Formula
E = F / Q
E = electric field strength (N C⁻¹)
F = electrostatic force on the charge (N)
Q = the charge experiencing the force (C)
Worked Example A charged particle sits in a field of strength 3.5 × 10⁴ N C⁻¹ and experiences a force of 0.3 N. Find its charge.

Step 1: Rearrange E = F/Q → Q = F/E
Step 2: Q = 0.3 ÷ (3.5 × 10⁴) = 8.571 × 10⁻⁶ C
Answer: Q ≈ 8.6 × 10⁻⁶ C (2 s.f.)
Watch out If you plug in a negative charge for Q, you'll get a negative value of E. That's not a mistake — it just means the field vector points in the opposite direction to what it would for a positive charge. Learn to read that sign, don't panic at it.
Practice Question
A test charge of +2.0 × 10⁻⁶ C is placed at a point in a field and experiences a force of 0.08 N directed to the right. Calculate the electric field strength at that point, including its direction.
3 · Coulomb's Law — Force Between Two Charges

Quantifying attraction and repulsion

We know like charges repel and opposite charges attract — but Coulomb's Law tells us that force is. Think of it like gravity's electric cousin: instead of masses attracting, it's charges pushing or pulling, and the same "inverse square" pattern shows up.

Coulomb's Law (in words) The electrostatic force between two point charges is proportional to the product of the charges, and inversely proportional to the square of their separation.
Key Formula
FE = Q₁Q₂ / (4πε₀r²)
FE = electrostatic force between the two charges (N)
Q₁, Q₂ = the two point charges (C)
ε₀ = permittivity of free space (a constant describing how well a vacuum "permits" electric fields)
r = distance between the of the charges (m)

Because of the r² in the denominator, this is called an inverse square law. It means if you double the separation between two charges, the force doesn't halve — it drops to one quarter (½² = ¼) of its original size. Triple the distance and the force falls to just 1/9th. Distance matters a lot.

The sign of FE tells you the nature of the interaction:

  • If Q₁ and Q₂ are opposite signs → FE comes out negativeattractive force.
  • If Q₁ and Q₂ are the same sign → FE comes out positiverepulsive force.
Handy fact A uniformly charged spherical conductor behaves — from the outside — exactly like a point charge sitting at its centre. This means you can use Coulomb's Law for charged spheres too, not just idealised points.
Worked Example An alpha particle sits 2.0 mm from a gold nucleus in a vacuum. Find the force between them. (Atomic number He = 2, Au = 79, charge of proton = 1.60 × 10⁻¹⁹ C)

Step 1: Q₁ (alpha) = 2 × 1.60×10⁻¹⁹ = +3.2×10⁻¹⁹ C
Step 2: Q₂ (gold nucleus) = 79 × 1.60×10⁻¹⁹ = +1.264×10⁻¹⁷ C
Step 3: r = 2.0 mm = 2.0×10⁻³ m
Step 4: FE = (3.2×10⁻¹⁹ × 1.264×10⁻¹⁷) / (4π × 8.85×10⁻¹² × (2.0×10⁻³)²)
Answer: FE9.1 × 10⁻²¹ N (repulsive, since both are positive)
Watch out Two classic slip-ups: (1) forgetting to square the distance r, and (2) forgetting to convert units — always check charge is in Coulombs (not nC or µC) and distance is in metres (not mm or cm) before you substitute into the equation.
Practice Question
Two point charges, +4.0 nC and −6.0 nC, are separated by 0.30 m in a vacuum. Calculate the magnitude and nature (attractive/repulsive) of the force between them. (ε₀ = 8.85 × 10⁻¹² F m⁻¹)
4 · Electric Field Due to a Point Charge

The radial field

A single point charge (or a charged sphere) produces a radial field — picture field lines spreading out symmetrically in every direction like sun rays, or sucking inward symmetrically for a negative charge.

Key Formula
E = Q / (4πε₀r²)
E = electric field strength at distance r (N C⁻¹)
Q = the point charge producing the field (C)
r = distance from the centre of the charge (m)
ε₀ = permittivity of free space (F m⁻¹)

Notice this is just Coulomb's Law with one of the two charges removed — because here we're not asking "what force acts between two charges", we're asking "how strong is the field itself, independent of what gets placed in it". This equation is only valid for the field around a single point charge (or equivalent sphere) — never use it for parallel plates.

Key features to remember:

  • E is not constant in a radial field — it changes with distance.
  • It follows an inverse square law (1/r²) — double the distance, and E drops to a quarter.
  • If the graph of E against r is plotted, the area under the graph equals the change in electric potential, ΔV.
  • For a negative charge, E works out negative — meaning the field vector points the charge.
Worked Example Calculate the field strength 2 m from an electron, and state its direction.

Q = −1.6×10⁻¹⁹ C, r = 2 m, ε₀ = 8.85×10⁻¹² F m⁻¹
E = (−1.6×10⁻¹⁹) / (4π × 8.85×10⁻¹² × 2²) = −3.6 × 10⁻¹⁰ N C⁻¹
The negative sign tells us the field points towards the electron (as expected for a negative charge).
Watch out Don't confuse this with Coulomb's Law! Coulomb's Law has two charges (Q₁ and Q₂) because it's about the force between them. This equation has only one Q — the charge producing the field — because we're describing the field itself, not a force on a second object.
Practice Question
The electric field strength at a distance r from a point charge is E. What is the field strength at a distance 4r from the same charge?
5 · Electric Field & Potential

Linking force-based and energy-based views

So far we've thought about electric fields in terms of . But there's a second, equally powerful way to think about them: in terms of . A positive test charge sitting in a field has electric potential energy due to its position — just like a mass has gravitational PE due to its height.

Moving a positive charge another positive charge takes work (you're fighting the repulsion). Moving it a negative charge also takes work (you're fighting the attraction, effectively "holding it back"). Either way, work done on the charge changes its potential energy.

Definition Electric potential at a point is the amount of work done per unit charge to bring a positive test charge to that point.

This links E and V beautifully:

The core relationship The electric field strength is proportional to the gradient of the electric potential.

In plain terms:

  • If potential changes rapidly with distance → field strength is large.
  • If potential changes gradually with distance → field strength is small.

So on a potential-distance (V–r) graph, E is just the gradient at that point. Steep graph = strong field. Flat graph = weak field.

Worked Example Two sets of charged plates, X and Y, have potential-vs-distance graphs plotted. Set X's line is steeper than Set Y's. Which set creates the larger electric field strength?

Since E is proportional to the gradient of the V–d graph, and Set X has the larger gradient, Set X creates the larger field strength.
Practice Question
On a potential–distance graph, the potential changes from +12 V to +4 V over a distance of 0.02 m. Estimate the magnitude of the electric field strength over this region.
6 · Electric Field Between Parallel Plates

The uniform field

Unlike the radial field around a point charge, the field between two oppositely charged parallel plates is uniform — same strength and same direction at every point between the plates (ignoring the fringing effects right at the edges).

Key Formula
E = V / d
E = electric field strength (V m⁻¹, equivalent to N C⁻¹)
V = potential difference between the plates (V)
d = separation between the plates (m)

What this tells us:

  • Bigger voltage between the plates → stronger field.
  • Bigger separation between the plates → weaker field.

The field direction runs from the plate connected to the positive terminal to the plate connected to the negative terminal. If one plate is earthed, treat its voltage as 0 V.

Watch out E = V/d is only for parallel plates (uniform fields). Never use it for a point charge — for that you need E = Q/(4πε₀r²). Mixing these two up is one of the most common exam errors.
Worked Example Two parallel plates are 3.5 cm apart with a p.d. of 7.9 kV. Find the force on a charged particle of 2.6 × 10⁻¹⁵ C between the plates.

Step 1: E = V/d = 7.9×10³ / 3.5×10⁻² = 2.257 × 10⁵ V m⁻¹
Step 2: F = QE = 2.6×10⁻¹⁵ × 2.257×10⁵ = 5.87×10⁻¹⁰ N
Answer: F ≈ 5.9 × 10⁻¹⁰ N (2 s.f.)
Practice Question
Two parallel plates are separated by 5.0 cm and produce a uniform field of strength 2.0 × 10⁴ V m⁻¹. Calculate the potential difference between the plates.
7 · Electric Potential for a Radial Field

Potential around a point charge

Electric potential (V) is a scalar — no direction — but it does carry a positive or negative sign depending on the charge that's creating it:

  • Positive around an isolated positive charge.
  • Negative around an isolated negative charge.
  • Zero at infinity (this is the reference point).
Key Formula
V = Q / (4πε₀r)
V = electric potential (V)
Q = point charge producing the potential (C)
r = distance from the centre of the charge (m)
ε₀ = permittivity of free space (F m⁻¹)
The critical distinction Electric potential V follows a 1/r relationship (no square!).
Electric field strength E follows a 1/r² relationship.
These are genuinely different equations and different graph shapes — this trips up a huge number of students.

A handy way to remember the direction of change: potential always decreases in the same direction as the field lines, and increases in the opposite direction.

Worked Example The electric potential at distance r from a proton is V. What's the potential at 3r?

V' = Q/(4πε₀ · 3r) = (1/3) × Q/(4πε₀r) = V/3
Tripling the distance shrinks the potential to one third — because V is inversely proportional to r (not r²).
Practice Question
A negative point charge is fixed in place. As a positive test charge is moved closer to it, does the electric potential at the test charge's location increase or decrease? Explain why.
8 · Representing Radial & Uniform Electric Fields

Field lines and equipotentials

Diagrams are a huge part of this topic, and examiners love testing whether you can draw and interpret them correctly. Two types of lines matter:

  • Field lines — show the direction and relative strength of the field. Closer lines = stronger field. Always directed from + to −.
  • Equipotential lines/surfaces — join points at the same potential. Always drawn as dotted lines (no arrows — potential is a scalar, it has no direction), and always perpendicular to field lines.
Uniform field (parallel plates)
  • Field lines: equally spaced, parallel, straight — constant E everywhere between the plates.
  • Equipotential lines: horizontal, parallel, equally spaced straight lines.
Radial field (point charge / sphere)
  • Field lines: spread outward (positive) or inward (negative), density decreasing with distance — this decreasing density is what represents E falling off with 1/r².
  • Equipotential lines: concentric circles around the charge, spaced progressively further apart as you move outward (because V changes more slowly at greater distances).
Two charges together For two opposite charges: field lines run from + to −, and there's a central equipotential line at 0 V where the two opposing potentials cancel out.

For two like charges: field lines point away from both (or towards both), and there's a neutral point at the midpoint where the resultant field is exactly zero — no field lines pass through this point.
Watch out Field lines must never cross each other, must always be perpendicular to a conducting surface, and must have arrows showing direction. Equipotential lines, by contrast, must never have arrows — they represent a scalar quantity (potential), which has no direction.
Practice Question
Two identical positive point charges are placed near each other. Describe (a) what happens to the field lines at the midpoint between them, and (b) what this tells you about the field strength there.

📌 What to Memorise
Electric field strength
Force per unit positive charge at a point. A vector — direction follows field lines.
E = F / Q
Coulomb's Law
Force between two point charges — proportional to charges, inversely proportional to r².
F₀₀ = Q₁Q₂ / (4πε₀r²)
Field due to a point charge
Radial field strength at distance r from a point charge. Inverse square law.
E = Q / (4πε₀r²)
Field between parallel plates
Uniform field strength — only valid for parallel plates, never for point charges.
E = V / d
Electric potential (radial field)
Work done per unit charge bringing a test charge from infinity. Scalar, follows 1/r.
V = Q / (4πε₀r)
Field ↔ Potential relationship
E equals the gradient of a potential-distance (V–r) graph at any point.
E = ΔV / Δr
The 1/r vs 1/r² rule
This is the single most-confused pair in the whole chapter — burn it into memory:
V ∝ 1/r   ·   E ∝ 1/r²
Direction conventions
Field lines: away from + charges, towards − charges. Always perpendicular to a conductor's surface. Equipotential lines: always perpendicular to field lines, never have arrows.

✅ Concepts Checklist

🎯 Exam Tips & Common Mistakes
Mixing up the two field equations

The single biggest source of lost marks: using E = Q/(4πε₀r²) for parallel plates, or E = V/d for a point charge. Ask yourself first — "is this a radial field or a uniform field?" — before picking the formula.

Forgetting to square r in Coulomb's Law / point charge field

Both FE and E for a point charge have r² in the denominator. It's an easy term to drop under exam pressure — always double check before you calculate.

Unit prefix errors

Charges are often given in nC or µC, distances in mm or cm. Convert everything to base SI units (C and m) before substituting — examiners specifically design questions to catch this.

Confusing V ∝ 1/r with E ∝ 1/r²

These look similar but behave very differently as distance changes. If a question asks about potential at "3× the distance", the answer changes by a factor of 3 (not 9). If it asks about field strength, the answer changes by a factor of 9.

Drawing field lines and equipotentials incorrectly

Mark schemes check for: arrows on field lines only (never on equipotentials), lines never crossing, lines touching the charge/plate surface perpendicular to it, and equipotential lines always drawn perpendicular to field lines.

Sign errors with attraction/repulsion

A negative result from Coulomb's Law means attraction, a positive result means repulsion. Don't just report the magnitude — state clearly whether the force is attractive or repulsive when asked.

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