Physics 9702 · syllabus 18
Cambridge A Level Physics 9702 — Electric fields topical past papers
Electric fields and the forces they exert: uniform fields between plates, Coulomb's law, the field and potential of a point charge, and potential energy.
- Questions
- 305
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 35
Anyone can browse these pages. Opening the questions and their mark schemes takes a free Momentum account.
What Electric fields covers
An electric field is a region in which a charge feels a force, and its strength is the force per unit positive charge, E = F/q. Between two charged parallel plates the field is uniform, with strength E = ΔV/Δd, so a charged particle there feels a constant force. An electron entering at right angles to the field follows a parabola, exactly as a projectile does under gravity.
Around a point charge, or outside a charged sphere, the field follows an inverse-square law. Coulomb's law gives the force between two charges, F = Q₁Q₂/(4πε₀r²), and the field strength is E = Q/(4πε₀r²). Electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity, V = Q/(4πε₀r), and the field strength equals the negative of the potential gradient.
The structure mirrors gravitational fields, with one key difference: charges can repel as well as attract, so electric potential and potential energy can be positive or negative, while gravitational potential is always negative. This is A Level content, examined on Paper 4, and it underpins capacitance, the motion of charged particles in magnetic fields and the photoelectric effect.
In the syllabus
The headings this topic has in the Cambridge International 2025–2027 syllabus, as the board prints them.
- 18 Electric fields
- 18.1 Electric fields and field lines
- 18.2 Uniform electric fields
- 18.3 Electric force between point charges
- 18.4 Electric field of a point charge
- 18.5 Electric potential
What you should be able to do
- Define electric field strength, and draw field lines for point charges, pairs of charges and parallel plates
- Use E = ΔV/Δd and F = qE to analyse the motion of charged particles between plates
- Use Coulomb's law and E = Q/(4πε₀r²) for point charges and charged spheres
- Define electric potential and use V = Q/(4πε₀r) and Eₚ = Qq/(4πε₀r)
- Find field strength as the negative gradient of a potential–distance graph
- Compare electric and gravitational fields, stating similarities and differences
Key ideas
- E = F/q (N C⁻¹, equivalent to V m⁻¹); in a uniform field E = ΔV/Δd
- Coulomb's law F = Q₁Q₂/(4πε₀r²), with 1/(4πε₀) = 8.99 × 10⁹ m F⁻¹
- Point charge: E = Q/(4πε₀r²) and V = Q/(4πε₀r)
- Electric potential energy of two charges Eₚ = Qq/(4πε₀r); work done moving a charge q through a p.d. ΔV is qΔV
- Field strength = −(potential gradient), E = −ΔV/Δx
How Electric fields has been examined
From 2022 to 2026, Electric fields was on Paper 4 · A Level Structured Questions in 11 of 11 exam series, worth about 11% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 1 · Multiple Choice | 1 | 1% | 1 of 11 |
| Paper 4 · A Level Structured Questions | 34 | 11% | 11 of 11 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 10.
Filed in Momentum under: Electric Fields.
A worked example
An electron enters the space midway between two parallel plates at 2.0 × 10⁷ m s⁻¹, moving parallel to them. The plates are 4.0 cm long and 2.0 cm apart, with a p.d. of 500 V between them. Find the electron's deflection across the field as it leaves the plates.
- Field strength: E = V/d = 500 / 0.020 = 2.5 × 10⁴ V m⁻¹
- Acceleration towards the positive plate: a = eE/m = 1.60 × 10⁻¹⁹ × 2.5 × 10⁴ / 9.11 × 10⁻³¹ = 4.39 × 10¹⁵ m s⁻²
- Time between the plates: t = 0.040 / 2.0 × 10⁷ = 2.0 × 10⁻⁹ s
- Deflection: y = ½at² = ½ × 4.39 × 10¹⁵ × (2.0 × 10⁻⁹)² = 8.8 × 10⁻³ m
- This is less than the 1.0 cm to the plate, so the electron leaves the field without striking it
Answer: 8.8 × 10⁻³ m (8.8 mm) towards the positive plate
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Leaving the plate separation in centimetres in E = ΔV/Δd — 500 V across 2.0 cm is 2.5 × 10⁴ V m⁻¹, not 250 V m⁻¹
- Treating the path between plates as part of a circle — the force is constant in size and direction, so the path is a parabola, unlike the circle in a magnetic field
- Squaring r in the potential formula — potential goes as 1/r, field strength as 1/r²
- Forgetting that the velocity component parallel to the plates is unchanged — the field acts only along the field lines, so the time between the plates is plate length ÷ entry speed
- Adding field strengths from two charges as if they were scalars — field strength is a vector, whereas potentials add as scalars, with their signs
Exam technique
- Treat deflection between plates as a projectile: constant velocity along the plates, constant acceleration qE/m across them, and the time linking the two
- Draw field lines from positive to negative, meeting conducting surfaces at right angles, evenly spaced between parallel plates and curving at the edges
- When comparing with gravitational fields, write matched pairs: both inverse-square, both with potentials varying as 1/r, but gravity only attracts
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.