Uniform electric fields
Cambridge A-Level Physics (9702) · Unit 18: Electric fields · 7 flashcards
Uniform electric fields is topic 18.2 in the Cambridge A-Level Physics (9702) syllabus , positioned in Unit 18 — Electric fields , alongside Electric fields and field lines, Electric force between point charges and Electric field of a point charge. In one line: E = ΔV / d, where E is the electric field strength (V/m or N/C), ΔV is the potential difference (V), and d is the distance between the plates (m). This equation is specific to uniform electric fields.
Marked as A2 Level: examined at A Level in Paper 4 (A Level Structured Questions) and Paper 5 (Planning, Analysis and Evaluation). It is not tested on the AS-only papers (Papers 1, 2 and 3).
The deck below contains 7 flashcards — 1 definition, 4 key concepts and 2 calculations — covering the precise wording mark schemes reward. Use the definition card to lock down command-word answers (define, state), then move on to the concept and calculation cards to handle explain, describe, calculate and compare questions.
Equation that relates electric field strength (E), potential difference (ΔV), and distance (d) in a uniform electric field
E = ΔV / d, where E is the electric field strength (V/m or N/C), ΔV is the potential difference (V), and d is the distance between the plates (m). This equation is specific to uniform electric fields.
What the Cambridge 9702 syllabus says
Official 2025-2027 spec · A2 LevelThese are the exact learning outcomes Cambridge sets for this topic. The candidate is expected to be able to do each of these on the relevant paper.
- recall and use E = ∆V / ∆d to calculate the field strength of the uniform field between charged parallel plates
- describe the effect of a uniform electric field on the motion of charged particles
Cambridge syllabus keywords to use in your answers
These are the official Cambridge 9702 terms tagged to this section. Mark schemes credit responses that use the exact term — weave them into your answers verbatim rather than paraphrasing.
Tips to avoid common mistakes in Uniform electric fields
- › Divide the total charge by 1.6 x 10^-19 to express the charge value in terms of e.
- › Draw separate vector arrows for the force from each charge and then determine the resultant direction using the parallelogram of forces.
- › Always double-check the syllabus formulae for point charges: Force is (Q₁Q₂)/(4πε₀r²) whereas Field Strength is Q/(4πε₀r²).
- › Apply the rule that a negative charge experiences a force in the opposite direction to the electric field to balance gravitational weight acting downwards.
- › Draw and describe field lines for a point charge or isolated sphere as radial lines that appear to originate from the center.
State the equation that relates electric field strength (E), potential difference (ΔV), and distance (d) in a uniform electric field.
E = ΔV / d, where E is the electric field strength (V/m or N/C), ΔV is the potential difference (V), and d is the distance between the plates (m). This equation is specific to uniform electric fields.
Describe the motion of a positively charged particle placed in a uniform electric field.
A positively charged particle will experience a force in the direction of the electric field. It will accelerate in that direction with a constant acceleration, assuming the electric field is the only force acting on the particle.
A potential difference of 500V is applied across two parallel plates separated by 0.05m. Calculate the electric field strength between the plates.
Using E = ΔV / d, E = 500V / 0.05m = 10,000 V/m. The electric field strength is 10,000 V/m (or 10,000 N/C).
How does the electric field strength change between parallel plates if the separation distance doubles, while the potential difference remains constant?
Since E = ΔV / d, if d doubles and ΔV remains constant, the electric field strength E is halved.
Describe the effect of a uniform electric field on the vertical motion of an electron initially moving horizontally through the field (assume no gravity).
The electron will experience a constant upward force (opposite to the field direction). This results in a constant upward acceleration, causing the electron to follow a parabolic trajectory, similar to projectile motion.
Explain why the electric field between two parallel plates is considered uniform.
The electric field is uniform because the field lines are parallel and equally spaced between the plates (excluding edge effects). This means the electric field strength and direction are the same at all points between the plates.
If a proton is accelerated from rest through a potential difference of 1000V, what is its kinetic energy gain?
The kinetic energy gained by the proton is equal to the work done on it by the electric field, which is qΔV. Therefore, KE gain = (1.60 x 10⁻¹⁹ C)(1000 V) = 1.60 x 10⁻¹⁶ J.
Review the material
Read full revision notes on Uniform electric fields — definitions, equations, common mistakes, and exam tips.
Read NotesMore topics in Unit 18 — Electric fields
Uniform electric fields sits alongside these A-Level Physics decks in the same syllabus unit. Each uses the same spaced-repetition system, so progress in one informs the next.
Key terms covered in this Uniform electric fields deck
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