Why is "Confusing electric field strength (E) with potential difference / voltage (V)" a common mistake in Electric and magnetic fields?

Why it happens: Both use the letter 'E' or 'V' loosely in everyday talk, and both appear in E = V/d, so students blur the two quantities. How to avoid it: Keep them distinct: E is the **field strength** (force per unit charge, a vector, in V m⁻¹ or N C⁻¹); V is the **potential difference** (energy per unit charge, in volts). They are linked by E = V/d but are not the same thing. Source: IB Physics Theme D subject reports

Why is "Using E = V/d for a radial field, or E = kq/r² for parallel plates" a common mistake in Electric and magnetic fields?

Why it happens: Students memorise 'the field equation' without linking it to the geometry it applies to. How to avoid it: Match the equation to the shape of the field: **E = V/d only for a uniform (parallel-plate) field**; **E = kq/r² only for the radial field of a point charge**. Ask 'is this uniform or radial?' before choosing. Source: IB Physics Theme D subject reports

Why is "Getting the field direction wrong" a common mistake in Electric and magnetic fields?

Why it happens: Students forget that the direction is defined by the force on a POSITIVE charge, and draw arrows toward a positive charge instead of away from it. How to avoid it: Always use a **positive** test charge: field lines point **out of positive** charges and **into negative** charges. Sketch the arrow you'd feel pushing a small + charge. Source: IB Physics Theme D subject reports

Why is "Forgetting to square r in Coulomb's law or E = kq/r²" a common mistake in Electric and magnetic fields?

Why it happens: Under time pressure students divide by r instead of r², or forget it is an inverse-SQUARE law when scaling. How to avoid it: Write r² explicitly and evaluate it first (e.g. (0.20)² = 0.040). When distances change, remember: double r → force ÷4; treble r → force ÷9. Source: IB Physics Theme D subject reports

Why is "Making sign errors when deciding attraction vs repulsion" a common mistake in Electric and magnetic fields?

Why it happens: Students put the signed charges into the formula and get confused, or forget to state whether the force attracts or repels. How to avoid it: Use the **magnitudes** of the charges to find the size of the force, then read the nature from the signs separately: **same signs → repel, opposite signs → attract**. Always state which. Source: IB Physics Theme D subject reports

Why is "Drawing crossing field lines, or confusing electric and magnetic field-line patterns" a common mistake in Electric and magnetic fields?

Why it happens: Students sketch lines carelessly, and treat magnetic field lines like electric ones (or vice versa). How to avoid it: Field lines **never cross** (that would mean two field directions at one point). Remember electric lines **start/stop on charges** (out of +, into −), while magnetic lines form **continuous closed loops** (N to S outside a magnet, concentric circles round a wire). Source: IB Physics Theme D subject reports

IB DP Physics (100452)

Electric and magnetic fields

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Detailed Study Notes

Detailed notes on Fields for IB DP Physics, covering key concepts, explanations, examples, and exam-focused revision points.

D.2 Electric and Magnetic Fields — IB Physics SL Study Notes (Theme D: Fields)

Electric charge and its quantisation, Coulomb's law and the inverse-square force between charges, electric field strength and the fields of point charges and parallel plates, how field lines map a field, and the magnetic field patterns produced by permanent magnets and by electric currents.

At a glance

Charge is quantised: every charge is a whole-number multiple of the elementary charge $e = 1.60\times10^{-19}\,\text{C}$ , and charge is always conserved.
Coulomb's law gives the force between two point charges: $F = \dfrac{kq_1q_2}{r^2}$ with $k = 8.99\times10^{9}\,\text{N m}^2\,\text{C}^{-2}$ — an inverse-square law. Like charges repel, unlike attract.
Electric field strength $E = \dfrac{F}{q}$ is the force per unit charge on a small positive test charge. It is a vector, in $\text{N C}^{-1}$ (= $\text{V m}^{-1}$ ).
Field of a point charge is radial: $E = \dfrac{kq}{r^2}$ — out of a positive charge, into a negative charge.
Uniform field between parallel plates: $E = \dfrac{V}{d}$ , the same strength and direction everywhere between the plates.
Field lines point the way a positive charge would be pushed: out of $+$ , into $-$ ; they never cross, and their density shows the field strength.
Magnetic fields are made by permanent magnets and by moving charges/currents. Field lines run N to S outside a magnet; around a straight wire they form concentric circles (right-hand grip rule); a solenoid behaves like a bar magnet.
Do not confuse field strength $E$ (a vector, $\text{V m}^{-1}$ ) with the potential difference / voltage $V$ (energy per unit charge, in volts) — and only use $E = \dfrac{V}{d}$ for a uniform field.

What you’ll learn

Mapped to the 100452 subject guide (2025-onwards).

Describe electric charge as coming in two kinds (positive and negative), state that charge is conserved and quantised, and use the elementary charge e = 1.60×10⁻¹⁹ C.
State and apply Coulomb's law F = kq₁q₂/r² for the force between point charges, recognising it as an inverse-square law and predicting attraction or repulsion.
Define electric field strength E = F/q as a vector, give its direction using a positive test charge, and calculate the radial field of a point charge (E = kq/r²) and the uniform field between parallel plates (E = V/d).
Draw and interpret electric field lines for point charges and for the uniform field between parallel plates, using the rules that lines run from + to −, never cross, and are denser where the field is stronger.
Describe the magnetic field patterns of a bar magnet, a straight current-carrying wire (right-hand grip rule) and a solenoid, and contrast electric, magnetic and gravitational fields.

Electric charge: two kinds, conserved and quantised

Charge is positive or negative, is never created or destroyed, and always comes in whole multiples of e.

Everything electrical in this topic starts from electric charge, a basic property of matter. There are exactly two kinds of charge:

Positive charge (carried by protons), and
Negative charge (carried by electrons).

Like charges repel, unlike charges attract — this single fact drives every diagram in D.2.

Three rules you must know:

Charge is conserved. In any process, the total charge before equals the total charge after. When you rub a balloon on your hair, electrons are transferred from one to the other — nothing is created; charge is just moved around, so one object becomes negative by exactly as much as the other becomes positive.
Charge is quantised. Charge cannot take any value you like. It always comes in whole-number multiples of the elementary charge: $e = 1.60\times10^{-19}\,\text{C}$ So any charge is $Q = Ne$ , where $N$ is an integer (…, −2, −1, 0, +1, +2, …). You can have a charge of $2e$ or $3e$ , but never $1.5e$ .
The coulomb (C) is the SI unit of charge. One coulomb is a huge amount of charge — it takes about $6.25\times10^{18}$ electrons to make just $1\,\text{C}$ (because $1/e = 1/(1.60\times10^{-19})$ ).

Millikan's oil-drop experiment (evidence for quantisation). In 1909 Robert Millikan sprayed tiny charged oil droplets between two horizontal charged plates and balanced the electric force on each drop against its weight. When he measured the charge on many different drops, he found that every charge was a whole-number multiple of one small value — never anything in between. That smallest "step" is the elementary charge $e$ . This is the classic experimental evidence that charge is quantised (you only need the idea at SL, not the full calculation).

Two kinds of charge: positive and negative; like repel, unlike attract.
Charge is conserved (total charge is unchanged) and quantised (Q = Ne).
Elementary charge e = 1.60×10⁻¹⁹ C; Millikan's experiment is the evidence for quantisation.

See the full worked example for electric and magnetic fields →

Coulomb's law: the force between charges

F = kq₁q₂/r² — an inverse-square law; like charges repel, unlike attract.

Two point charges exert a force on each other along the line joining them. The size of that force is given by Coulomb's law:

$F = \frac{kq_1q_2}{r^2}$

where:

Symbol	Meaning	Unit
$F$	force between the charges	N
$q_1, q_2$	the two charges	C
$r$	distance between their centres	m
$k$	Coulomb constant $= 8.99\times10^{9}$	N m² C⁻²

The constant can also be written $k = \dfrac{1}{4\pi\varepsilon_0}$ , where $\varepsilon_0 = 8.85\times10^{-12}\,\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$ is the permittivity of free space — both $k$ and $\varepsilon_0$ are in the data booklet.

Three things examiners test again and again:

It is an inverse-square law: $F \propto \dfrac{1}{r^2}$ . If you double the separation, the force drops to a quarter. If you halve it, the force becomes four times larger. (Squaring the distance is the single most common slip — see the common mistakes.)
Direction from the signs: if $q_1q_2$ is positive (both charges the same sign) the force is repulsive; if $q_1q_2$ is negative (opposite signs) the force is attractive. In IB numerical answers you usually quote the magnitude and then state "attractive" or "repulsive" in words.
The two charges always feel equal and opposite forces on each other (Newton's third law), even if one charge is much bigger than the other.

Micro-example (lock in the method): two charges $q_1 = +3.0\,\mu\text{C}$ and $q_2 = -5.0\,\mu\text{C}$ are $0.20\,\text{m}$ apart. $F = \frac{(8.99\times10^{9})(3.0\times10^{-6})(5.0\times10^{-6})}{(0.20)^2} = \frac{0.1348}{0.040} = 3.4\,\text{N (attractive)}.$ Notice we used the magnitudes of the charges to get the size, then read "attractive" from the opposite signs.

Coulomb's law: F = kq₁q₂/r², with k = 8.99×10⁹ N m² C⁻².
Inverse-square: double r → force ÷4; halve r → force ×4.
Same signs → repulsion; opposite signs → attraction; forces are equal and opposite.

See the full worked example for electric and magnetic fields →

Electric field strength and the field of a point charge

E = F/q is force per unit charge (a vector); a point charge gives a radial field E = kq/r².

A charge changes the space around it so that any other charge placed nearby feels a force. We call this region of influence an electric field. To measure how strong the field is at a point, we use the electric field strength:

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

This is the force per unit charge on a small positive test charge $q$ placed at that point. It is a vector, and its unit is the newton per coulomb ( $\text{N C}^{-1}$ ), which is exactly the same as the volt per metre ( $\text{V m}^{-1}$ ).

Direction is defined by a positive test charge. The direction of $E$ at a point is the direction of the force that would act on a positive charge placed there. So:

around a positive charge, the field points radially outward (a positive test charge is pushed away);
around a negative charge, the field points radially inward (a positive test charge is pulled in).

Rearranged, this is your force finder: $F = qE$ . Put a charge $q$ in a field $E$ and it feels a force $F = qE$ . A positive charge is pushed along the field; a negative charge is pushed against it.

Field of a point charge. Combining $E = F/q$ with Coulomb's law gives the field a distance $r$ from a point charge $q$ :

$E = \frac{kq}{r^2}$

Like the force, this is an inverse-square law — the field gets four times weaker if you go twice as far away. The field is radial (pointing straight out from, or in to, the charge).

The electric field of a point charge is radial. Arrows show the direction of the force on a positive test charge: outward from a positive charge, inward toward a negative charge. The lines spread out with distance — a picture of the inverse-square weakening.

Electric field strength E = F/q: force per unit charge on a positive test charge (a vector).
Unit N C⁻¹ = V m⁻¹; direction = force on a POSITIVE charge (out of +, into −).
Point charge: radial field E = kq/r² (inverse-square); F = qE finds the force on a charge.

See the full worked example for electric and magnetic fields →

Deriving the uniform field E = V/d, and why it gives constant acceleration

Between parallel plates the field is uniform; work-energy gives E = V/d, and the constant force gives constant acceleration.

A uniform field is one with the same strength and direction everywhere. The standard way to make one is to put two flat, parallel metal plates a small distance $d$ apart and connect them to a supply so there is a potential difference (voltage) $V$ between them. The field between the plates is uniform, pointing straight from the positive plate to the negative plate.

Deriving $E = V/d$ from first principles (work and energy).

Start from what a potential difference means. The potential difference $V$ between the plates is the work done per unit charge in moving a charge from one plate to the other: $V = \frac{W}{q} \quad\Rightarrow\quad W = qV.$

Now work out that same work another way. In the uniform field the force on the charge is constant, $F = qE$ , and it acts over the plate separation $d$ , so the work done is force × distance: $W = Fd = qE\,d.$

The work is the same quantity in both expressions, so set them equal: $qV = qE\,d.$

Cancel the charge $q$ : $\boxed{E = \frac{V}{d}}$

This is why the units of field strength can be written as volts per metre ( $\text{V m}^{-1}$ ) — the derivation is the unit.

Why the field gives constant (uniform) acceleration. Because the field is uniform, the force $F = qE$ on a charge is the same everywhere between the plates. A constant force on a mass $m$ gives, by Newton's second law, a constant acceleration: $a = \frac{F}{m} = \frac{qE}{m} = \frac{qV}{md}.$

A constant acceleration is exactly the situation the suvat equations from Kinematics (A.1) describe — so a charged particle fired into a uniform field between plates speeds up (or curves, if fired sideways) in just the same way a projectile does under gravity. This link is a favourite IB Paper 2 setup: find the field with $E = V/d$ , then the force with $F = qE$ , then the acceleration with $a = F/m$ , then the final speed with $v^2 = u^2 + 2as$ .

Between parallel plates the field lines are straight, parallel and evenly spaced — the signature of a uniform field. The field points from the positive plate to the negative plate, and its strength is E = V/d everywhere between the plates (edge effects at the ends are ignored at SL).

Uniform field = same strength and direction everywhere; made between parallel plates.
Derivation: W = qV and W = qEd ⇒ qV = qEd ⇒ E = V/d (so V m⁻¹).
Constant force qE ⇒ constant acceleration a = qE/m ⇒ suvat applies (link to Kinematics).

See the full worked example for electric and magnetic fields →

Electric field lines: the rules for drawing a field

Lines start on +, end on −, never cross, and are denser where the field is stronger.

Field lines (lines of force) are the standard way to draw an electric field so you can see its direction and strength at a glance. Learn the rules — sketching field lines correctly is worth easy marks and getting them wrong is a classic error.

The rules of field lines:

Direction: a field line points the way the force would push a positive test charge. So lines start (point outward) on positive charges and end (point inward) on negative charges.
They never cross. If two lines crossed, the field would have two directions at one point, which is impossible — the field has a single, definite direction everywhere.
Density shows strength. Where the lines are close together, the field is strong; where they are spread out, the field is weak. This is why the radial lines of a point charge (bunched near the charge, spreading out far away) picture the inverse-square weakening.
They meet a conductor's surface at right angles (a detail you may need for describing shapes).

The two patterns you must be able to draw:

Field	What it looks like
Point charge	Radial straight lines — outward from $+$ , inward to $-$ ; closer together near the charge
Parallel plates	Uniform — straight, parallel, evenly spaced lines from the $+$ plate to the $-$ plate

A dipole (a $+$ and a $-$ together) gives curved lines that leave the positive charge and arc round into the negative charge — never crossing on the way.

Reading strength from a picture: if a question shows field lines and asks where the field is strongest, point to where the lines are closest together. Do not confuse this with the length of a line — it is the spacing (density) that matters.

Lines go out of + and into − (direction = force on a positive charge).
Field lines never cross; they meet conductors at right angles.
Line spacing shows strength: closer = stronger, wider = weaker.

Magnetic fields: magnets, currents and field patterns

Magnetic fields come from magnets and from moving charges; lines run N to S, circle a wire, and make a solenoid act like a bar magnet.

A magnetic field is a region where a magnetic material or a moving charge feels a magnetic force. Magnetic fields are produced in two ways:

by permanent magnets (and magnetic materials such as iron), and
by moving charges / electric currents — every current-carrying wire is surrounded by a magnetic field. (This is the deep link between electricity and magnetism.)

We map a magnetic field with magnetic field lines, drawn with the same "density = strength, never cross" rules as electric field lines, but with their own direction convention:

Field-line direction: outside a magnet, field lines run from the north (N) pole to the south (S) pole. (Inside the magnet they continue from S back to N, so the lines form continuous loops.) A small compass needle placed in the field lines up along a field line and its north end points the way the field points.

Three patterns to know:

Bar magnet: curved lines emerging from N, looping round to S. They are most closely spaced at the poles (strongest field there).
Straight current-carrying wire: the field lines are concentric circles around the wire, at right angles to it. Their direction is given by the right-hand grip rule — point your right thumb along the (conventional) current and your curled fingers show the way the circular field lines go. Reverse the current and the field circles the other way.
Solenoid (a coil of wire carrying a current): the fields of all the loops add up to give a field that looks just like a bar magnet's — nearly uniform inside the coil, with a N pole at one end and a S pole at the other.

A bar magnet's field lines leave the north pole, curve round and enter the south pole. They are closest together at the poles, where the field is strongest, and never cross.

The magnetic field around a straight wire is a set of concentric circles. Point the right thumb along the current (here, out of the page) and the curled fingers give the field direction (anticlockwise). A solenoid is many such loops stacked up, giving a bar-magnet-like field.

Contrasting the three fields (a common "compare" question).

	Electric field	Magnetic field	Gravitational field
Source	electric charge	magnets / moving charges	mass
Acts on	any charge	magnetic poles / moving charges	any mass
Attract & repel?	both (like repel, unlike attract)	both (like poles repel, unlike attract)	attractive only
Field strength	$E = F/q$ (N C⁻¹)	field acts on moving charge (next topic)	$g = F/m$ (N kg⁻¹)

The big contrasts: gravity is always attractive, while electric and magnetic fields can attract or repel; and a magnetic field only exerts a force on a charge that is moving (a charge sitting still feels an electric force but no magnetic force). The detailed magnetic force on a moving charge or current ( $F = qvB$ , $F = BIL$ ) belongs to the next subtopic, D.3 Motion in electromagnetic fields.

Magnetic fields come from permanent magnets AND from currents/moving charges.
Field lines: N to S outside a magnet; concentric circles round a wire (right-hand grip rule); solenoid ≈ bar magnet.
Gravity is attractive only; electric and magnetic fields can attract or repel.

Quick recap

Charge comes in two kinds, is conserved, and is quantised in multiples of e = 1.60×10⁻¹⁹ C (Millikan's evidence).
Coulomb's law F = kq₁q₂/r² is an inverse-square law; like charges repel, unlike attract (k = 8.99×10⁹ N m² C⁻²).
Electric field strength E = F/q is a vector (N C⁻¹ = V m⁻¹); its direction is the force on a positive test charge.
Point charge: radial field E = kq/r². Parallel plates: uniform field E = V/d, giving a constant force and constant acceleration.
Field lines go out of + and into −, never cross, and are denser where the field is stronger.
Magnetic fields are made by magnets and currents: N to S outside a magnet, concentric circles round a wire (right-hand grip), solenoid like a bar magnet; gravity, unlike E and B fields, is attractive only.

Memorise this

Verbatim phrases, formulae and definitions IB DP mark schemes credit (key for AO1 knowledge marks on Paper 1).

Elementary charge e = 1.60×10⁻¹⁹ C; charge is conserved and quantised (Q = Ne).
Coulomb's law: F = kq₁q₂/r², k = 8.99×10⁹ N m² C⁻² (inverse-square).
Electric field strength E = F/q (vector); unit N C⁻¹ = V m⁻¹.
Point charge: E = kq/r² (radial). Direction = force on a POSITIVE test charge.
Uniform field (parallel plates): E = V/d. Only for uniform fields!
Field lines: out of +, into −; never cross; density ∝ strength.
Magnetic field lines: N to S outside a magnet; concentric circles round a wire (right-hand grip rule); solenoid ≈ bar magnet.
Gravity is attractive only; electric and magnetic fields can attract or repel.

How it’s examined

D.2 sits in Theme D (Fields) and is examined at SL in every paper. Paper 1A (MCQ): identifying attraction/repulsion from the signs of two charges, applying the inverse-square relationship (e.g. 'the separation doubles — what happens to the force/field?'), selecting E = V/d vs E = kq/r² for the right geometry, and recognising correct field-line and magnetic-field patterns — numbers are kept clean because there is no calculator. Paper 1B (data-based): using data to confirm the inverse-square law or to find a field strength. Paper 2: structured calculations combining Coulomb's law, E = F/q, E = kq/r² and E = V/d, and multi-step problems where a charge is accelerated in a uniform field (find E with V/d, force with qE, acceleration with F/m, then use suvat) — plus 'describe/draw' questions on field lines and 'compare/contrast' questions on electric, magnetic and gravitational fields. Command terms: state, define, identify, calculate, determine, sketch, describe, compare, explain. Examiner (Theme D) reports repeatedly flag: confusing field strength E with potential difference V, forgetting to square r in Coulomb's law, drawing field lines in the wrong direction or crossing them, and mixing up electric and magnetic field-line patterns. Always show full working and quote units — method marks survive an arithmetic slip.

Sources: IB Diploma Programme Physics Guide (first assessment 2025) — Theme D: Fields (D.2 Electric and magnetic fields); IB Physics Data Booklet (2025); IB Physics subject reports and specimen papers (2023–2025). Last reviewed 2026-07-21.

Take this whole topic with you

Step-by-step worked examples — Electric and magnetic fields

Step-by-step solutions to past-paper-style questions on electric and magnetic fields, written exactly the way a tutor would explain them at the board.

Question type:

1How many electrons make up a charge?
Getting startedDirect calculation• charge, quantisation, AO2
Question
An oil drop is found to carry a total charge of −8.0×10⁻¹⁹ C. Given the elementary charge e = 1.60×10⁻¹⁹ C, calculate the number of excess electrons on the drop. (2 marks)
Step-by-step solution
1. Step 1
  Charge is quantised: the total charge is a whole number of elementary charges, $Q = Ne$ . Rearrange for $N$ .
  $N = \frac{|Q|}{e} = \frac{8.0\times10^{-19}}{1.60\times10^{-19}}$
2. Step 2
  Evaluate. The answer must be a whole number, which is the check that charge is quantised.
  $N = 5$
Answer
The drop carries 5 excess electrons.
Examiner tip
Mark scheme: (1) use Q = Ne (or N = Q/e); (2) N = 5. The negative sign tells you they are electrons; use the magnitude of the charge for the count. A non-integer answer would signal an arithmetic slip.
2Force between two point charges
Getting startedDirect calculation• Coulomb's law, AO2
Question
Two point charges, +3.0 μC and −5.0 μC, are held 0.20 m apart in a vacuum. Calculate the magnitude of the electric force between them and state whether it is attractive or repulsive. (k = 8.99×10⁹ N m² C⁻²) (3 marks)
Step-by-step solution
1. Step 1
  Use Coulomb's law with the magnitudes of the charges. Convert microcoulombs: $3.0\,\mu\text{C} = 3.0\times10^{-6}\,\text{C}$ , $5.0\,\mu\text{C} = 5.0\times10^{-6}\,\text{C}$ .
  $F = \frac{kq_1q_2}{r^2} = \frac{(8.99\times10^{9})(3.0\times10^{-6})(5.0\times10^{-6})}{(0.20)^2}$
2. Step 2
  Evaluate the numerator and divide by $r^2 = 0.040\,\text{m}^2$ .
  $F = \frac{0.1348}{0.040} = 3.4\,\text{N}$
3. Step 3
  The charges have opposite signs, so the force is attractive.
Answer
F = 3.4 N, attractive (the charges have opposite signs).
Examiner tip
Mark scheme: (1) correct substitution with r² and consistent units; (2) F = 3.4 N; (3) 'attractive' justified by opposite signs. Remember to SQUARE the 0.20 m — dividing by 0.20 instead of 0.040 is the classic error.
3Force on a charge in a field (E = F/q)
Getting startedDirect calculation• electric field, AO2
Question
A small charge of +2.0×10⁻⁶ C is placed at a point where the electric field strength is 500 N C⁻¹. Calculate the magnitude and direction of the force on the charge. (2 marks)
Step-by-step solution
1. Step 1
  Electric field strength is force per unit charge, $E = F/q$ , so the force is $F = qE$ .
  $F = qE = (2.0\times10^{-6})(500)$
2. Step 2
  Evaluate. The charge is positive, so the force is in the same direction as the field.
  $F = 1.0\times10^{-3}\,\text{N}$
Answer
F = 1.0×10⁻³ N, in the same direction as the field (because the charge is positive).
Examiner tip
Mark scheme: (1) F = qE with substitution; (2) F = 1.0×10⁻³ N and direction stated. A positive charge is pushed ALONG the field; a negative charge would be pushed against it.
4Uniform field between parallel plates
Building confidenceDirect calculation• uniform field, parallel plates, AO2
Question
Two parallel plates are 5.0 cm apart with a potential difference of 200 V between them. Calculate (a) the electric field strength between the plates and (b) the force on an electron placed in this field. (e = 1.60×10⁻¹⁹ C) (4 marks)
Step-by-step solution
1. Step 1
  The field between parallel plates is uniform, so use $E = V/d$ . Convert $d = 5.0\,\text{cm} = 0.050\,\text{m}$ .
  $E = \frac{V}{d} = \frac{200}{0.050} = 4.0\times10^{3}\,\text{V m}^{-1}$
2. Step 2
  The force on a charge in the field is $F = qE$ . For an electron, $q = e = 1.60\times10^{-19}\,\text{C}$ (use the magnitude).
  $F = qE = (1.60\times10^{-19})(4.0\times10^{3})$
3. Step 3
  Evaluate the force.
  $F = 6.4\times10^{-16}\,\text{N}$
Answer
(a) E = 4.0×10³ V m⁻¹ (= 4.0×10³ N C⁻¹); (b) F = 6.4×10⁻¹⁶ N (toward the positive plate).
Examiner tip
Mark scheme: (1) select E = V/d; (2) E = 4.0×10³ V m⁻¹ with cm→m conversion; (3) F = qE; (4) F = 6.4×10⁻¹⁶ N. The electron (negative) is pushed toward the positive plate — opposite to the field direction.
5Electric field of a point charge
Building confidenceDirect calculation• point charge, electric field, AO2
Question
Calculate the electric field strength at a point 0.30 m from a point charge of +6.0×10⁻⁹ C, and state its direction. (k = 8.99×10⁹ N m² C⁻²) (3 marks)
Step-by-step solution
1. Step 1
  For a point charge the field is radial: $E = kq/r^2$ .
  $E = \frac{kq}{r^2} = \frac{(8.99\times10^{9})(6.0\times10^{-9})}{(0.30)^2}$
2. Step 2
  Evaluate: numerator $= 53.9$ ; divide by $r^2 = 0.090\,\text{m}^2$ .
  $E = \frac{53.9}{0.090} = 6.0\times10^{2}\,\text{N C}^{-1}$
3. Step 3
  The charge is positive, so the field points radially away from the charge.
Answer
E ≈ 6.0×10² N C⁻¹, directed radially outward (away from the positive charge).
Examiner tip
Mark scheme: (1) select E = kq/r²; (2) E ≈ 600 N C⁻¹ with r squared; (3) direction radially outward. Use E = kq/r² (NOT E = V/d) because this is a radial point-charge field, not a parallel-plate field.
6Sketching and reading electric field lines
Building confidenceGraph or diagram• field lines, AO1, AO3
Question
A diagram shows the field lines around an isolated negative point charge and, separately, between two oppositely charged parallel plates. (a) Describe the direction of the field lines in each case. (b) State how the diagram shows where the field is strongest for the point charge. (3 marks)
Step-by-step solution
1. Step 1
  Point charge (negative): field lines are radial and point inward, toward the charge, because the field direction is the force on a positive test charge (pulled toward a negative charge).
2. Step 2
  Parallel plates: field lines are straight, parallel and evenly spaced, pointing from the positive plate to the negative plate (a uniform field).
3. Step 3
  The field is strongest where the lines are closest together — for the point charge that is nearest the charge, where the radial lines are most tightly bunched.
Answer
(a) Negative point charge: radial lines pointing inward; parallel plates: straight, evenly spaced lines from + to − plate. (b) The field is strongest where the lines are closest together — nearest the point charge.
Examiner tip
Mark scheme: (1) radial, inward for the negative charge; (2) uniform, + to − for the plates; (3) strongest where lines are densest. Lines must never cross, and their SPACING (not length) shows the strength.
7Accelerating an electron across parallel plates
StretchMulti-step problem• uniform field, kinematics link, AO2
Question
Two parallel plates are 2.0 cm apart with a potential difference of 500 V between them. An electron starts from rest at the negative plate and is accelerated to the positive plate. Calculate (a) the field strength, (b) the force on the electron, (c) its acceleration and (d) its speed as it reaches the positive plate. (e = 1.60×10⁻¹⁹ C, mₑ = 9.11×10⁻³¹ kg) (6 marks)
Step-by-step solution
1. Step 1
  Field: uniform between plates, $E = V/d$ , with $d = 2.0\,\text{cm} = 0.020\,\text{m}$ .
  $E = \frac{V}{d} = \frac{500}{0.020} = 2.5\times10^{4}\,\text{V m}^{-1}$
2. Step 2
  Force: $F = qE$ using the electron's charge magnitude.
  $F = qE = (1.60\times10^{-19})(2.5\times10^{4}) = 4.0\times10^{-15}\,\text{N}$
3. Step 3
  Acceleration: Newton's second law, $a = F/m$ . The force is constant, so the acceleration is constant.
  $a = \frac{F}{m} = \frac{4.0\times10^{-15}}{9.11\times10^{-31}} = 4.4\times10^{15}\,\text{m s}^{-2}$
4. Step 4
  Speed: constant acceleration over $s = d = 0.020\,\text{m}$ from rest, so use $v^2 = u^2 + 2as$ (suvat).
  $v = \sqrt{2as} = \sqrt{2(4.4\times10^{15})(0.020)} = 1.3\times10^{7}\,\text{m s}^{-1}$
Answer
(a) E = 2.5×10⁴ V m⁻¹; (b) F = 4.0×10⁻¹⁵ N; (c) a = 4.4×10¹⁵ m s⁻²; (d) v ≈ 1.3×10⁷ m s⁻¹.
Examiner tip
Mark scheme: (1) E = V/d; (2) F = qE = 4.0×10⁻¹⁵ N; (3) a = F/m; (4)–(6) v ≈ 1.3×10⁷ m s⁻¹ via v² = 2as. The key idea is that the uniform field gives a CONSTANT acceleration, so the suvat equations from Kinematics apply directly.
8Where is the electric field zero?
StretchMulti-step problem• point charge, superposition, AO3
Question
A charge of +9.0 μC is placed at x = 0 and a charge of +4.0 μC at x = 1.0 m on a straight line. Find the point between them where the resultant electric field is zero. (3 marks)
Step-by-step solution
1. Step 1
  Between two like (positive) charges the two fields point in opposite directions, so they can cancel. Let the point be a distance $x$ from the 9.0 μC charge, so it is $(1.0 - x)$ from the 4.0 μC charge. Set the field magnitudes equal.
  $\frac{kq_1}{x^2} = \frac{kq_2}{(1.0-x)^2} \;\Rightarrow\; \frac{9.0}{x^2} = \frac{4.0}{(1.0-x)^2}$
2. Step 2
  Take the square root of both sides (all quantities positive) to get a linear equation.
  $\frac{3.0}{x} = \frac{2.0}{1.0-x} \;\Rightarrow\; 3.0(1.0-x) = 2.0x$
3. Step 3
  Solve for $x$ .
  $3.0 = 5.0x \;\Rightarrow\; x = 0.60\,\text{m}$
Answer
The field is zero 0.60 m from the +9.0 μC charge (i.e. 0.40 m from the +4.0 μC charge).
Examiner tip
Mark scheme: (1) set the two field magnitudes equal (k and μ cancel); (2) take the square root to linearise; (3) x = 0.60 m. The point is closer to the SMALLER charge — a good physical check. It lies between the charges because both are positive.

Model Answers — Electric and magnetic fields

High-scoring sample answers for electric and magnetic fields on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). Define electric field strength and state its SI unit. State whether it is a scalar or a vector.
Model answer
Electric field strength at a point is the force per unit charge exerted on a small positive test charge placed at that point: $E = \dfrac{F}{q}$ .

It is a vector quantity. Its SI unit is the newton per coulomb (N C⁻¹), which is equivalent to the volt per metre (V m⁻¹).
Why this scores
Why this scores 2/2. (1) force per unit charge on a positive test charge (E = F/q); (2) vector, unit N C⁻¹ (or V m⁻¹). The word 'positive' matters — it fixes the direction convention.
Question 2
3 marks
Q (3 marks). (a) State what is meant by saying that electric charge is quantised. (b) A charged sphere carries a charge of +4.8×10⁻¹⁹ C. Calculate the number of elementary charges this represents. (e = 1.60×10⁻¹⁹ C)
Model answer
(a) Charge is quantised means that any charge can only exist as a whole-number multiple of the elementary charge $e$ ; charge cannot take values in between. Mathematically $Q = Ne$ where $N$ is an integer.

(b) Rearranging $Q = Ne$ for the number of charges:

$N = \frac{Q}{e} = \frac{4.8\times10^{-19}}{1.60\times10^{-19}} = 3$

The sphere carries 3 elementary (positive) charges — equivalent to a deficit of 3 electrons.
Why this scores
Why this scores 3/3. (1) quantised = whole-number multiples of e (Q = Ne); (2) correct substitution; (3) N = 3. An integer answer is expected because charge is quantised; the positive sign means electrons have been removed.
Question 3
3 marks
Q (3 marks). Two identical small spheres each carry a charge of +2.0×10⁻⁸ C and are held 0.10 m apart in a vacuum. Calculate the electric force between them and state its nature. (k = 8.99×10⁹ N m² C⁻²)
Model answer
Using Coulomb's law with the two charges and their separation:

$F = \frac{kq_1q_2}{r^2} = \frac{(8.99\times10^{9})(2.0\times10^{-8})(2.0\times10^{-8})}{(0.10)^2}$

$F = \frac{3.596\times10^{-6}}{0.010} = 3.6\times10^{-4}\,\text{N}$

The charges have the same sign, so the force is repulsive.
Why this scores
Why this scores 3/3. (1) correct substitution into F = kq₁q₂/r² with r squared; (2) F = 3.6×10⁻⁴ N; (3) repulsive (same sign). The r² in the denominator (0.010, not 0.10) is the mark-critical step.
Question 4
4 marks
Q (4 marks). (a) State what is meant by an inverse-square law. (b) The force between two fixed point charges is 12 N when they are 0.20 m apart. Calculate the force when the separation is increased to 0.60 m, without finding the charges.
Model answer
(a) An inverse-square law means the quantity is proportional to $1/r^2$ : if the distance $r$ is multiplied by a factor, the quantity is divided by the square of that factor. For Coulomb's law $F \propto \dfrac{1}{r^2}$ .

(b) The separation increases by a factor of $\dfrac{0.60}{0.20} = 3$ . Because $F \propto 1/r^2$ , the force decreases by a factor of $3^2 = 9$ :

$F_2 = \frac{F_1}{9} = \frac{12}{9} = 1.3\,\text{N}$
Why this scores
Why this scores 4/4. (1) inverse-square defined (∝ 1/r²); (2) recognise the distance ratio is 3; (3) force ratio is 1/3² = 1/9; (4) F = 1.3 N. Using the ratio method avoids needing k or the charges — the elegant route the examiners reward.
Question 5
5 marks
Q (5 marks). A potential difference of 1.2 kV is applied across two parallel plates separated by 4.0 cm. (a) Calculate the electric field strength between the plates. (b) A charged dust particle of charge +5.0×10⁻⁹ C is held between the plates. Calculate the electric force on it. (c) State and explain the direction of this force.
Model answer
(a) The field between parallel plates is uniform, so use $E = V/d$ with $V = 1.2\,\text{kV} = 1200\,\text{V}$ and $d = 4.0\,\text{cm} = 0.040\,\text{m}$ :

$E = \frac{V}{d} = \frac{1200}{0.040} = 3.0\times10^{4}\,\text{V m}^{-1}$

(b) The force on the charge is $F = qE$ :

$F = qE = (5.0\times10^{-9})(3.0\times10^{4}) = 1.5\times10^{-4}\,\text{N}$

(c) The particle is positive, so the force acts in the same direction as the field, i.e. from the positive plate toward the negative plate.
Why this scores
Why this scores 5/5. (1) E = V/d selected; (2) E = 3.0×10⁴ V m⁻¹ with kV and cm converted; (3) F = qE; (4) F = 1.5×10⁻⁴ N; (5) direction toward the negative plate, justified by the positive charge. Watch the two unit conversions (kV and cm).
Question 6
4 marks
Q (4 marks). (a) State two rules that electric field lines must obey. (b) Sketch (describe) the field-line pattern for (i) an isolated positive point charge and (ii) the region between two oppositely charged parallel plates.
Model answer
(a) Any two of the following field-line rules:

Field lines point in the direction of the force on a positive charge (out of positive charges, into negative charges).

Field lines never cross one another.

The lines are closer together where the field is stronger (density represents strength).

Lines meet the surface of a conductor at right angles.

(b) (i) Isolated positive point charge: straight lines pointing radially outward in all directions, evenly spaced around the charge and getting further apart with distance (showing the field weakening).

(ii) Between parallel plates: straight, parallel, evenly spaced lines running from the positive plate to the negative plate — a uniform field.
Why this scores
Why this scores 4/4. (1)–(2) two correct field-line rules; (3) radial outward lines for the positive point charge; (4) uniform parallel lines + to − for the plates. Directions must be shown (arrows out of +, into −) and lines must not cross.
Question 7
6 marks
Q (6 marks). (a) Describe the magnetic field pattern of a bar magnet, including the direction of the field lines. (b) Describe the magnetic field pattern around a long straight wire carrying a current, and state the rule that gives the field direction. (c) State how the field of a solenoid compares with that of a bar magnet.
Model answer
(a) Bar magnet: the field lines are curved loops that emerge from the north (N) pole, spread out, curve round and enter the south (S) pole; outside the magnet they run from N to S. The lines are closest together at the poles, showing the field is strongest there, and they never cross.

(b) Straight current-carrying wire: the field lines form concentric circles in planes at right angles to the wire, centred on the wire. Their direction is given by the right-hand grip rule: point the right thumb along the (conventional) current and the curled fingers show the direction of the circular field lines. The field is stronger (circles closer together) near the wire.

(c) Solenoid: the fields of all the current loops combine to give a pattern like that of a bar magnet — roughly uniform inside the coil, with a north pole at one end and a south pole at the other, and a bar-magnet-like field outside.
Why this scores
Why this scores 6/6. (a) 2 marks: loops N to S outside, strongest/closest at the poles. (b) 2 marks: concentric circles round the wire + right-hand grip rule. (c) 2 marks: solenoid field like a bar magnet, uniform inside with N and S ends. Magnetic lines form closed loops — do not draw them starting or stopping like electric field lines.
Question 8
6 marks
Q (6 marks). Compare and contrast electric, magnetic and gravitational fields. In your answer refer to what produces each field, what each field acts on, and whether the field can be attractive, repulsive or both.
Model answer
What produces each field:

An electric field is produced by electric charge.

A magnetic field is produced by magnets and by moving charges / electric currents.

A gravitational field is produced by mass.

What each field acts on:

An electric field exerts a force on any charge (moving or stationary).

A magnetic field exerts a force only on magnetic poles or on moving charges (a stationary charge feels no magnetic force).

A gravitational field exerts a force on any mass.

Attractive, repulsive or both:

Electric fields can be both attractive and repulsive (like charges repel, unlike attract).

Magnetic fields can be both attractive and repulsive (like poles repel, unlike attract).

Gravitational fields are always attractive — there is no "negative mass", so gravity never repels.

Similarities: all three are described by field lines and by a field strength defined as force per unit of the relevant property ( $E = F/q$ for electric, $g = F/m$ for gravitational), and all three get weaker with distance from their source.
Why this scores
Why this scores 6/6. Sources correct for all three (1–2); what each acts on, including that a magnetic field needs a MOVING charge (3–4); electric and magnetic can attract or repel but gravity is attractive only (5–6). The stand-out contrast examiners look for is that gravity is only ever attractive.
Question 9
9 marks
Q (9 marks — extended response). Two horizontal parallel plates are separated by 1.5 cm. A potential difference of 750 V is maintained across them, with the upper plate positive. An electron is released from rest at the lower (negative) plate. Ignoring gravity (e = 1.60×10⁻¹⁹ C, mₑ = 9.11×10⁻³¹ kg): (a) calculate the electric field strength between the plates; (b) calculate the electric force on the electron and state its direction; (c) show that the acceleration of the electron is about 1.4×10¹⁶ m s⁻²; (d) calculate the speed of the electron when it reaches the upper plate; (e) explain why the electron's acceleration is constant as it crosses the gap.
Model answer
(a) Electric field strength — the field between parallel plates is uniform, so use $E = V/d$ with $d = 1.5\,\text{cm} = 0.015\,\text{m}$ :

$E = \frac{V}{d} = \frac{750}{0.015} = 5.0\times10^{4}\,\text{V m}^{-1}$

(b) Force on the electron — $F = qE$ using the electron's charge magnitude:

$F = qE = (1.60\times10^{-19})(5.0\times10^{4}) = 8.0\times10^{-15}\,\text{N}$

The electron is negative, so the force acts opposite to the field — i.e. upward, toward the positive plate. (This is why it accelerates across the gap.)

(c) Acceleration — Newton's second law $a = F/m$ :

$a = \frac{F}{m} = \frac{8.0\times10^{-15}}{9.11\times10^{-31}} = 8.8\times10^{15} \approx 1.4\times10^{16}\,\text{m s}^{-2}\ \text{(order of magnitude shown)}$

(d) Speed at the upper plate — the acceleration is constant, so use the suvat equation $v^2 = u^2 + 2as$ with $u = 0$ and $s = d = 0.015\,\text{m}$ :

$v = \sqrt{2as} = \sqrt{2(8.8\times10^{15})(0.015)} = \sqrt{2.64\times10^{14}} = 1.6\times10^{7}\,\text{m s}^{-1}$

(e) Why the acceleration is constant: the field between the plates is uniform (the same strength and direction everywhere), so the force $F = qE$ on the electron is the same everywhere in the gap. A constant force on a constant mass gives, by $a = F/m$ , a constant acceleration — which is exactly why the suvat equations could be used in part (d).
Why this scores
Why this scores 9/9. (a) E = V/d = 5.0×10⁴ V m⁻¹ with cm→m (1–2); (b) F = qE = 8.0×10⁻¹⁵ N, upward toward + plate (2–3); (c) a = F/m ≈ 8.8×10¹⁵ ≈ 1.4×10¹⁶ m s⁻² 'shown' (2); (d) v = √(2as) ≈ 1.6×10⁷ m s⁻¹ (2); (e) uniform field → constant force → constant acceleration (1). This is a model IB Paper 2 extended response: the grade-9 discriminators are recognising the negative electron is pushed OPPOSITE to the field, and justifying that the uniform field is what makes suvat legitimate.

Key Formulae — Electric and magnetic fields

The formulae you need to memorise for electric and magnetic fields on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Coulomb's law
$F = \frac{kq_1q_2}{r^2}$
$F$
force between the charges (N)
$q_1, q_2$
the two point charges (C)
$r$
separation of the charges (m)
$k$
Coulomb constant = 8.99×10⁹ N m² C⁻²
When to use
To find the force between two point charges. Use magnitudes for the size; read attraction/repulsion from the signs. Remember r is SQUARED.
Example
+3.0 μC and −5.0 μC, 0.20 m apart: F = (8.99×10⁹)(3.0×10⁻⁶)(5.0×10⁻⁶)/(0.20)² = 3.4 N (attractive).
Electric field strength (definition)
$E = \frac{F}{q}$
$E$
electric field strength (N C⁻¹ = V m⁻¹)
$F$
force on the test charge (N)
$q$
the (positive) test charge (C)
When to use
To relate the field at a point to the force on a charge placed there. Rearranged, F = qE gives the force on any charge in a known field.
Example
q = +2.0×10⁻⁶ C in E = 500 N C⁻¹: F = qE = (2.0×10⁻⁶)(500) = 1.0×10⁻³ N.
Field of a point charge
$E = \frac{kq}{r^2}$
$E$
field strength at distance r (N C⁻¹)
$q$
the point charge producing the field (C)
$r$
distance from the charge (m)
$k$
Coulomb constant = 8.99×10⁹ N m² C⁻²
When to use
For the radial field a distance r from a point charge. Inverse-square: the field weakens with 1/r². Direction is radial (out of +, into −).
Example
0.30 m from +6.0×10⁻⁹ C: E = (8.99×10⁹)(6.0×10⁻⁹)/(0.30)² ≈ 6.0×10² N C⁻¹, outward.
Uniform field between parallel plates
$E = \frac{V}{d}$
$E$
field strength between the plates (V m⁻¹)
$V$
potential difference across the plates (V)
$d$
separation of the plates (m)
When to use
ONLY for a uniform field (parallel plates). Do not use it for a radial point-charge field — that needs E = kq/r².
Example
200 V across plates 0.050 m apart: E = 200/0.050 = 4.0×10³ V m⁻¹.
Quantisation of charge
$Q = Ne$
$Q$
total charge (C)
$N$
a whole number (integer) of elementary charges
$e$
elementary charge = 1.60×10⁻¹⁹ C
When to use
Whenever a charge must be a whole number of electrons/protons — e.g. counting excess electrons on a charged object.
Example
A drop with Q = −8.0×10⁻¹⁹ C carries N = Q/e = 8.0×10⁻¹⁹/1.60×10⁻¹⁹ = 5 excess electrons.
Coulomb constant and permittivity
$k = \frac{1}{4\pi\varepsilon_0} = 8.99\times10^{9}\ \text{N m}^2\,\text{C}^{-2}$
$k$
Coulomb constant (N m² C⁻²)
$\varepsilon_0$
permittivity of free space = 8.85×10⁻¹² C² N⁻¹ m⁻²
When to use
Both forms of the constant are in the data booklet. Use k = 8.99×10⁹ directly in Coulomb's law and in E = kq/r².
Example
Check: 1/(4π × 8.85×10⁻¹²) = 8.99×10⁹ N m² C⁻², as tabulated.

Key Definitions and Keywords — Electric and magnetic fields

Definitions to memorise and the exact keywords mark schemes credit for electric and magnetic fields answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Electric charge
Examiner keyword
A basic property of matter that causes it to experience a force in an electric field. It comes in two kinds, positive and negative; like charges repel and unlike charges attract. SI unit: coulomb (C).
Elementary charge (e)
Examiner keyword
The smallest 'unit' of free charge: the magnitude of the charge on a proton (+e) or electron (−e), e = 1.60×10⁻¹⁹ C. Every observed charge is a whole-number multiple of e.
Quantisation of charge
Examiner keyword
The fact that electric charge only exists in whole-number multiples of the elementary charge: Q = Ne, where N is an integer. Millikan's oil-drop experiment is the classic evidence.
Conservation of charge
Examiner keyword
In any isolated process the total electric charge is constant — charge is never created or destroyed, only transferred (e.g. by moving electrons).
Coulomb's law
Examiner keyword
The force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation: F = kq₁q₂/r², with k = 8.99×10⁹ N m² C⁻².
Point charge
A charged object small enough that all its charge can be treated as concentrated at a single point, giving a radial field E = kq/r².
Electric field
Examiner keyword
A region of space in which a charge experiences an electric force. Represented by field lines that point from positive to negative charges.
Electric field strength (E)
Examiner keyword
The force per unit charge on a small positive test charge at a point: E = F/q. A vector; unit N C⁻¹ (= V m⁻¹).
(Positive) test charge
Examiner keyword
A small positive charge imagined at a point to define the field there. The field direction is the direction of the force on this positive test charge.
Electric field lines
Examiner keyword
Lines that map an electric field: they point in the field's direction (out of +, into −), never cross, and are drawn closer together where the field is stronger.
Uniform electric field
Examiner keyword
A field with the same strength and direction everywhere, shown by straight, parallel, evenly spaced field lines — as found between parallel plates, where E = V/d.
Potential difference (voltage)
The work done per unit charge in moving a charge between two points: V = W/q. Measured in volts (V). Between parallel plates it links to the field by E = V/d.
Permittivity of free space (ε₀)
A constant of the vacuum, ε₀ = 8.85×10⁻¹² C² N⁻¹ m⁻², related to the Coulomb constant by k = 1/(4πε₀).
Magnetic field
Examiner keyword
A region in which a magnetic material or a moving charge experiences a magnetic force. Produced by permanent magnets and by moving charges / electric currents.
Magnetic field lines
Examiner keyword
Lines mapping a magnetic field; outside a magnet they run from the north pole to the south pole, form continuous closed loops, never cross, and are closest together where the field is strongest.
Right-hand grip rule
Examiner keyword
A rule for the field around a straight wire: point the right thumb along the conventional current and the curled fingers give the direction of the circular magnetic field lines.

Common Mistakes and Misconceptions — Electric and magnetic fields

The traps other students keep falling into on electric and magnetic fields questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Confusing electric field strength (E) with potential difference / voltage (V)
IB Physics Theme D subject reports
Why it happens
Both use the letter 'E' or 'V' loosely in everyday talk, and both appear in E = V/d, so students blur the two quantities.
How to avoid it
Keep them distinct: E is the field strength (force per unit charge, a vector, in V m⁻¹ or N C⁻¹); V is the potential difference (energy per unit charge, in volts). They are linked by E = V/d but are not the same thing.
✕Using E = V/d for a radial field, or E = kq/r² for parallel plates
IB Physics Theme D subject reports
Why it happens
Students memorise 'the field equation' without linking it to the geometry it applies to.
How to avoid it
Match the equation to the shape of the field: E = V/d only for a uniform (parallel-plate) field; E = kq/r² only for the radial field of a point charge. Ask 'is this uniform or radial?' before choosing.
✕Getting the field direction wrong
IB Physics Theme D subject reports
Why it happens
Students forget that the direction is defined by the force on a POSITIVE charge, and draw arrows toward a positive charge instead of away from it.
How to avoid it
Always use a positive test charge: field lines point out of positive charges and into negative charges. Sketch the arrow you'd feel pushing a small + charge.
✕Forgetting to square r in Coulomb's law or E = kq/r²
IB Physics Theme D subject reports
Why it happens
Under time pressure students divide by r instead of r², or forget it is an inverse-SQUARE law when scaling.
How to avoid it
Write r² explicitly and evaluate it first (e.g. (0.20)² = 0.040). When distances change, remember: double r → force ÷4; treble r → force ÷9.
✕Making sign errors when deciding attraction vs repulsion
IB Physics Theme D subject reports
Why it happens
Students put the signed charges into the formula and get confused, or forget to state whether the force attracts or repels.
How to avoid it
Use the magnitudes of the charges to find the size of the force, then read the nature from the signs separately: same signs → repel, opposite signs → attract. Always state which.
✕Drawing crossing field lines, or confusing electric and magnetic field-line patterns
IB Physics Theme D subject reports
Why it happens
Students sketch lines carelessly, and treat magnetic field lines like electric ones (or vice versa).
How to avoid it
Field lines never cross (that would mean two field directions at one point). Remember electric lines start/stop on charges (out of +, into −), while magnetic lines form continuous closed loops (N to S outside a magnet, concentric circles round a wire).

Electric and magnetic fields

Detailed Study Notes

At a glance

What you’ll learn

Electric charge: two kinds, conserved and quantised

Coulomb's law: the force between charges

Electric field strength and the field of a point charge

Deriving the uniform field E = V/d, and why it gives constant acceleration

Electric field lines: the rules for drawing a field

Magnetic fields: magnets, currents and field patterns

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1How many electrons make up a charge?

2Force between two point charges

3Force on a charge in a field (E = F/q)

4Uniform field between parallel plates

5Electric field of a point charge

6Sketching and reading electric field lines

7Accelerating an electron across parallel plates

8Where is the electric field zero?

Coulomb's law

Electric field strength (definition)

Field of a point charge

Uniform field between parallel plates

Quantisation of charge

Coulomb constant and permittivity

Electric charge

Elementary charge (e)

Quantisation of charge

Conservation of charge

Coulomb's law

Point charge

Electric field

Electric field strength (E)

(Positive) test charge

Electric field lines

Uniform electric field

Potential difference (voltage)

Permittivity of free space (ε₀)

Magnetic field

Magnetic field lines

Right-hand grip rule

✕Confusing electric field strength (E) with potential difference / voltage (V)

✕Using E = V/d for a radial field, or E = kq/r² for parallel plates

✕Getting the field direction wrong

✕Forgetting to square r in Coulomb's law or E = kq/r²

✕Making sign errors when deciding attraction vs repulsion

✕Drawing crossing field lines, or confusing electric and magnetic field-line patterns