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Uniform Electric Fields (Parallel Plates)

Key idea: Use E = V/d and F = qE to analyse uniform electric fields and the motion of charged particles between parallel plates (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Calculate field strength and force in uniform electric fields.
  • Analyse charged-particle motion in uniform electric fields.

1. Definitions (Must Know)

A. Uniform electric field

A uniform electric field is a field where electric field strength E has constant magnitude and direction (approximately true between large, parallel plates).

B. Electric field strength, E

Electric field strength, E, is force per unit positive charge:

E = F/q

Unit: N C⁻¹ (equivalently V m⁻¹).

C. Potential difference, Δ V

Potential difference, Δ V, is work done per unit charge by an external force:

Δ V = W/q

Unit: V, where 1 V = 1 J C⁻¹.

D. Potential gradient (1D)

Along a chosen axis x, the electric field is the negative potential gradient:

E = -dV/dx

2. Key Ideas (What Earns Marks)

  • Between parallel plates separated by distance d with potential difference Δ V:
    • E = (Δ V)/d (magnitude)
    • direction: from the positive plate to the negative plate
  • Field and potential are linked by:
    • E = -dV/dx (1D)
    • for a uniform field, E ≈ -(Δ V)/(Δ x)
  • Force on a charge in a uniform field:
    • F = qE
  • Motion is kinematics with constant acceleration:
    • a = qE/m
    • negative charge accelerates opposite to the field direction.
Direction and sign

vector E points from higher potential to lower potential (for a positive test charge). A negative charge accelerates opposite to vector E.

Exam pitfall: using E = V/d outside uniform regions

E = ΔV/d is valid where field lines are approximately parallel and evenly spaced (central plate region). Near edges, fringing means the field is no longer uniform.

3. Detailed Explanations

A. Why |E| = |Δ V|/d between parallel plates

Choose the x-axis along the field. Electric field is the negative potential gradient:

Eₓ = -dV/dx

The gradient is constant in a uniform field. Across plate separation d,

Δ V = -Eₓd

Therefore the magnitude relationship used for plate calculations is

|E| = (|Δ V|)/d

The minus sign carries the direction: the field points from higher potential to lower potential.

B. Motion of a charged particle in a uniform field

Because F = qE is constant, the acceleration is constant:

a = F/m = qE/m

So you can use kinematics:

  • If the initial velocity is parallel to the field: 1D constant acceleration.
  • If the initial velocity is perpendicular to the field: uniform velocity in one direction + uniform acceleration in the other (a “projectile-like” path).

Projectile-like path in a uniform electric field (example)

A parabolic y–x trajectory for a charged particle entering a uniform field with an initial horizontal velocity.

Scroll across the graph to read all labels.

A parabolic y–x trajectory for a charged particle entering a uniform field with an initial horizontal velocity.A parabolic y–x trajectory for a charged particle entering a uniform field with an initial horizontal velocity.
Horizontal motion is uniform (x = ut), vertical motion has constant acceleration (y = ½at²), so eliminating t gives y ∝ x² (a parabola).
Open full-size graph
View figure data
Values for Projectile-like path in a uniform electric field (example)
Horizontal distance, x (cm)Example trajectory (proton, values from worked example)
00
10.012
20.048
30.108
40.192
50.3

4. Common Mistakes

  • Using E = V/d without converting d to metres.
  • Forgetting q can be negative (direction of acceleration flips).
  • Mixing up units: 1 V/m = 1 N/C.
  • Using potential difference when the question needs displacement along the field direction.

5. Exam Tips

  • Always define axes and the sign convention (which direction is +).
  • Use magnitudes first, then state the direction separately if signs are messy.
  • Sanity check: a positive charge accelerates towards lower potential (towards the negative plate).

6. Worked Examples

Modelled example 1

Find E and the force on an electron

Core

Problem

Parallel plates have 500 V across 0.020 m. Find E, then the force on an electron including direction.
Study the worked solution

  1. Find the field

    Method

    Use E = Δ V/d.

    Reason

    Potential changes linearly in the central uniform region.

    Working

    E = 500/0.020 = 2.5 × 10⁴ V m⁻¹

  2. Find force magnitude

    Method

    Use |F| = |q|E.

    Reason

    The electron-charge magnitude sets force magnitude.

    Working

    |F| = (1.60 × 10⁻¹⁹)(2.5 × 10⁴) = 4.0 × 10⁻¹⁵ N

  3. State direction

    Method

    Point towards the positive plate.

    Reason

    Electron force is opposite the field.

    Working

    vector Fₑ is opposite vector E.

Guided practice 2

Acceleration of a proton in a uniform field

About 4 min

Problem

A proton is in a uniform 1.0 × 10⁴ V m⁻¹ field. Using q = 1.60 × 10⁻¹⁹ C and m = 1.67 × 10⁻²⁷ kg, find its acceleration.

Try this before viewing the solution

Unit: m s⁻²

Hints

Hint 1: combine force and motion
Use a = qE/m.
View solution step by step

  1. Substitute

    Method

    Use a = qE/m.

    Reason

    The field gives a constant electric force.

    Working

    a = ((1.60 × 10⁻¹⁹)(1.0 × 10⁴))/(1.67 × 10⁻²⁷)

  2. Evaluate

    Method

    Obtain 9.6 × 10¹¹ m s⁻² along the field.

    Reason

    The proton is positively charged.

    Working

    a ≈ 9.6 × 10¹¹ m s⁻².

Common misconception 3

Deflection while passing between plates

Find and correct the mistake

Learner claim

A proton enters horizontally at 2.0 × 10⁶ m s⁻¹ through a 0.050 m region where E = 1.0 × 10⁴ V m⁻¹. A learner uses vertical deflection to set the exposure time. Diagnose the setup and find y.

Try this before viewing the solution

Time in field

View solution step by step

  1. Find transit time

    Method

    Use horizontal length and speed.

    Reason

    No horizontal electric force changes u.

    Working

    t = 0.050/(2.0 × 10⁶) = 2.5 × 10⁻⁸ s

  2. Find vertical acceleration

    Method

    Use a = qE/m.

    Reason

    The vertical field gives constant force.

    Working

    a = 9.6 × 10¹¹ m s⁻².

  3. Find deflection

    Method

    Use y = (1/2)at².

    Reason

    Initial vertical speed is zero.

    Working

    y = (1/2)(9.6 × 10¹¹)(2.5 × 10⁻⁸)² = 3.0 × 10⁻⁴ m

Examiner practice 4

Find potential difference from E and d

3 marks

Examination question

Parallel plates are 12 mm apart with E = 3.0 × 10⁴ V m⁻¹. Find the p.d. magnitude. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Convert separation

    1 mark

    Method

    Use 12 × 10⁻³ m.

    Reason

    Field is given per metre.

    Working

    d = 0.012 m.

  2. Use the relation

    1 mark

    Method

    Apply Δ V = Ed.

    Reason

    The field is uniform.

    Working

    Δ V = (3.0 × 10⁴)(12 × 10⁻³)

  3. Evaluate

    1 mark

    Method

    Obtain 360 V.

    Reason

    Volt per metre times metre gives volt.

    Working

    Δ V = 3.6 × 10² V.

Challenge 5

Find separation from a required field

Minimal support

Independent transfer

Design a plate separation for E = 2.0 × 10⁵ V m⁻¹ using Δ V = 1.0 kV. State the uniform-field model boundary.

Try this before viewing the solution

Hints

Hint 1: convert kilovolts
Use d = Δ V/E with 1.0 kV = 1.0 × 10³ V.
View solution step by step

  1. Calculate separation

    Method

    Divide p.d. by field.

    Reason

    E = Δ V/d in the uniform region.

    Working

    d = (1.0 × 10³)/(2.0 × 10⁵) = 5.0 × 10⁻³ m = 5.0 mm

  2. State the boundary

    Method

    Use the central region of large parallel plates and neglect fringing.

    Reason

    The relation assumes nearly constant field.

    Working

    Plate size ≫ separation.

7. Mind Stretchers

Mind stretcher 1: Show that 1 V/m = 1 N/CExtension

Show Answer

1 V/m = (1 J/C)/(1 m) = (1 N m)/(C m) = 1 N/C

Mind stretcher 2: Why is the field approximately uniform between plates?Extension

Explain why the electric field is approximately uniform in the central region between two large, parallel plates.

Show Answer

In the central region, the field lines are roughly straight, parallel, and equally spaced, so the field has nearly constant direction and magnitude.

Non-uniformity occurs mainly near the edges due to fringing, but this is small if the plate size is much larger than the separation.

Mind stretcher 3: Simulation Bridge: Electric Field ExplorerExtension

Concept Explorer: Electric Field Explorer

Switch between parallel-plate and point-charge models, move the probe, and test E-V-r relationships with guided prompts.

BetaA LevelFieldsBest for: A Level electric fields revision
  • E = V/d
  • V = kQ/r
  • E = kQ/r²
  • Field Direction

Open the full interactive simulation on its own page

Use the standalone simulation page for the live controls, SVG scene, run modes, and scoring flow.

Open Interactive Simulation

The lesson stays lightweight and links out to the dedicated simulation page.

Explore uniform fields in the Electric Field Explorer.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027