The big idea: An inkjet printer steers tiny charged ink drops between two charged plates — and every drop feels the same steady push, wherever it sits in the gap. That even push is a uniform electric field: put a voltage across two parallel plates and you make one.
Uniform means the same strength everywhere — the field lines are evenly spaced and parallel, running straight from the + plate to the − plate.
A uniform field between parallel plates: the lines are evenly spaced and parallel, pointing from the + plate to the − plate. Evenly spaced means the field is the same strength everywhere between the plates.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
Spot it: Uniform field = evenly-spaced, parallel field lines.
The lines point from + to − (the direction a positive charge would be pushed). The field is the same strength all the way across the gap.
Free preview
This is the free notes preview
You're reading the free notes. Aimnova Pro unlocks the full study experience — and you can try it with your first topic free to keep:
- FlashcardsLock in vocabulary and key terms with spaced repetition.
- Practice questionsAnswer exam-style questions and get instant AI marking.
- Mock exams & past-paper vaultSit full mocks and see exactly how examiners award marks.
- Personalised study planA daily plan built around your exam date and weak areas.
The strength of the uniform field depends on the voltage across the plates and the gap between them. The bigger the voltage, or the smaller the gap, the stronger the field:
- electric field strength between the plates (V m⁻¹, or N C⁻¹)
- potential difference (voltage) between the plates (V)
- separation (gap) between the plates (m)
V = E d. Cover the one you want: E and d side by side → multiply (V = E d); V above d → divide (E = V ÷ d); V above E → divide (d = V ÷ E).
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
The force on a charge: A charge q placed in the field feels a force given by the field-strength definition (also in the booklet):
F = qE — the bigger the charge or the stronger the field, the bigger the push.
- electric force on the charge (N)
- the charge in the field (C, coulombs)
- electric field strength (N C⁻¹, or V m⁻¹)
Two parallel plates are 0.020 m apart with a potential difference of 600 V between them. Find the electric field strength between them.
Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
Know your predicted grade
Take timed mock exams and get detailed feedback on every answer. See exactly where you're losing marks.
How this is tested — parallel plates are usually a two-part Paper 2 calculation in the Fields theme:
Paper 2 — the field
- Calculate the field E = V/d from the plate voltage and separation, and draw the evenly-spaced field lines.
Paper 2 — the energy
- Find the energy gained by a charge crossing the plates with W = qV, then state it in electronvolts (eV).
The classic trap: Forgetting to convert the gap to metres, or mixing up the gap d with the plate length.
Work done moving a charge: W = qV: Moving a charge q through a potential difference V does work on it:
W = qV (energy in joules). For a charge released from rest, this work becomes its kinetic energy.
This one is not in the data booklet — remember it. It is the very meaning of voltage: energy per unit charge.
- work done moving the charge (J, joules) — its energy gain
- the charge moved (C, coulombs)
- potential difference moved through (V, volts)
An alpha particle of charge +3.2 × 10⁻¹⁹ C is released from the positive plate. The plates are 0.050 m apart with a potential difference of 250 V. (a) Find the uniform electric field strength between the plates. (b) The alpha particle crosses the full 250 V; find its kinetic energy at the negative plate, then state it in electronvolts. (1 eV = 1.6 × 10⁻¹⁹ J.)
Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.