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Acceleration Due to an Electric Field

Key idea: Use F = qE and a = qE/m for charged particles in uniform fields, including kinematics and energy methods (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Analyse charged-particle motion in uniform electric fields.

1. Definitions (Must Know)

A. Electric force on a charge

A charge q in an electric field of strength E experiences an electric force:

F = qE

Direction:

  • if q > 0, force is in the same direction as vector E,
  • if q < 0, force is opposite to vector E.

B. Acceleration in a uniform electric field

If the only force is the electric force (no gravity, no drag), then:

F = ma ⇒ a = qE/m

In a uniform field, E is constant, so a is constant.

C. Work done by a potential difference

When a charge moves through a potential difference Δ V, the work done by an external force is:

Wₑₓₜ = qΔ V

If the charge accelerates freely under the field, then Δ K = -Δ U and the kinetic energy gain is:

Δ K = -qΔ V

2. Key Ideas (What Earns Marks)

  • In a uniform field:
    • F = qE
    • a = qE/m
  • Between parallel plates:
    • E = Δ V/d (magnitude)
  • Two common solution routes:
    • Kinematics for motion over a known distance/time (constant a)
    • Energy for speed after a potential difference: Δ K = -qΔ V
Direction reminder

Draw vector E from the positive plate to the negative plate. Then decide the acceleration direction using the sign of q.

3. Detailed Explanations

A. Kinematics method (constant acceleration)

If you know E and the particle’s q and m, then a = qE/m is constant.

Use SUVAT with a stated axis (e.g. “upward is +”):

v = u + at, s = ut + (1/2)at², v² = u² + 2as

B. Energy method (accelerating through a p.d.)

If a charge is accelerated through a potential difference Δ V and other forces are negligible:

Δ K = -qΔ V

If it starts from rest:

(1/2)mv² = |q||Δ V|

This is especially useful for electron/proton “accelerating voltage” questions.

C. Motion between parallel plates (projectile-like path)

If a particle enters a uniform field with horizontal speed u:

  • horizontal motion: constant speed u (no horizontal force)
  • vertical motion: constant acceleration a = qE/m

Time in field of length L:

t = L/u

Vertical deflection (initial vertical speed 0):

y = (1/2)at²

4. Common Mistakes

  • Forgetting that electrons have q < 0 (so acceleration is opposite to vector E).
  • Using E = Δ V/d but leaving d in cm or mm.
  • Mixing up potential V and potential difference Δ V in energy steps.
  • Using qΔ V = (1/2)mv² when the particle did not start from rest (should be Δ K).

5. Exam Tips

  • Choose axes first; then let signs handle direction.
  • If the question asks only for a speed, use energy (faster).
  • If the question asks for deflection/angle while in the field, use kinematics with t = L/u.

6. Worked Examples

Modelled example 1

Electron accelerated through a p.d.

Core

Problem

An electron accelerates from rest through 2.0 kV. Find its speed using e = 1.60 × 10⁻¹⁹ C and mₑ = 9.11 × 10⁻³¹ kg.
Study the worked solution

  1. Use energy conservation

    Method

    Set kinetic-energy gain equal to e|Δ V|.

    Reason

    Only final speed is required and the electron starts from rest.

    Working

    (1/2)mₑv² = eΔ V

  2. Rearrange and evaluate

    Method

    Obtain 2.7 × 10⁷ m s⁻¹.

    Reason

    Convert kilovolts to volts before substitution.

    Working

    v = square root of ((2(1.60 × 10⁻¹⁹)(2.0 × 10³))/(9.11 × 10⁻³¹)) = 2.7 × 10⁷ m s⁻¹

Guided practice 2

Deflection between parallel plates

About 6 min

Problem

A proton enters horizontally at 2.0 × 10⁶ m s⁻¹ through plates of length 0.050 m where E = 1.0 × 10⁴ V m⁻¹. Find vertical deflection.

Try this before viewing the solution

Hints

Hint 1: separate axes
Use a_y = qE/m and t = L/uₓ.
Hint 2: vertical motion starts from rest
Then apply y = (1/2)a_yt².
View solution step by step

  1. Find vertical acceleration

    Method

    Use a = qE/m.

    Reason

    The uniform field supplies constant vertical force.

    Working

    a = 9.6 × 10¹¹ m s⁻².

  2. Find exposure time

    Method

    Use horizontal motion.

    Reason

    Horizontal velocity remains constant.

    Working

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

  3. Find deflection

    Method

    Use constant-acceleration kinematics.

    Reason

    Initial vertical velocity is zero.

    Working

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

Common misconception 3

Proton accelerated through a p.d.

Find and correct the mistake

Learner claim

A proton accelerates from rest through 500 V. A learner reuses an electron-speed result because both particles have charge magnitude e. Diagnose the claim and find the proton speed.

Try this before viewing the solution

Quantity causing a different speed

View solution step by step

  1. Use the proton mass

    Method

    Apply (1/2)mₚv² = eΔ V.

    Reason

    The energy gain is set by charge magnitude, but speed follows from the particle’s mass.

    Working

    v = square root of ((2eΔ V)/mₚ)

  2. Evaluate

    Method

    Obtain 3.1 × 10⁵ m s⁻¹.

    Reason

    The proton is much more massive than an electron.

    Working

    v = square root of ((2(1.60 × 10⁻¹⁹)(500))/(1.67 × 10⁻²⁷)) = 3.1 × 10⁵ m s⁻¹

Examiner practice 4

Find E from plate p.d. and separation

3 marks

Examination question

Parallel plates have p.d. 900 V and separation 15 mm. Find field magnitude. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Convert separation

    1 mark

    Method

    Use 15 × 10⁻³ m.

    Reason

    Field unit is per metre.

    Working

    d = 0.015 m.

  2. Use the uniform-field relation

    1 mark

    Method

    Apply E = Δ V/d.

    Reason

    Potential gradient is constant between the plates.

    Working

    E = 900/(15 × 10⁻³)

  3. Evaluate

    1 mark

    Method

    Obtain 6.0 × 10⁴ V m⁻¹.

    Reason

    Use field magnitude.

    Working

    E = 6.0 × 10⁴ V m⁻¹.

Challenge 5

Find acceleration from E

Minimal support

Independent transfer

An electron enters a uniform field E = 2.5 × 10⁴ V m⁻¹. Find acceleration magnitude and state its direction relative to vector E.

Try this before viewing the solution

Hints

Hint 1: separate magnitude and direction
Use |a| = eE/mₑ; then apply the negative charge sign.
View solution step by step

  1. Calculate magnitude

    Method

    Divide electric-force magnitude by electron mass.

    Reason

    |a| = |q|E/m.

    Working

    |a| = ((1.60 × 10⁻¹⁹)(2.5 × 10⁴))/(9.11 × 10⁻³¹) = 4.39 × 10¹⁵ m s⁻²

  2. State direction

    Method

    Point acceleration opposite vector E.

    Reason

    The electron charge is negative.

    Working

    vector a∥- vector E.

7. Mind Stretchers

Mind stretcher 1: Kinematics vs energyExtension

Show that using kinematics gives the same energy result: Δ K = -qΔ V for a charge moving a distance d along a uniform field.

Show Answer

In a uniform field, F = qE and E = Δ V/d, so F = qΔ V/d.

In a uniform field, Δ V = -Ed along the field direction, so Fd = qEd = -qΔ V.

By work–energy theorem, W_field = Δ K, so:

Δ K = -qΔ V

Since Δ U = qΔ V, we also have Δ K = -Δ U.

Mind stretcher 2: Why is the path a parabola?Extension

An electron enters a uniform field region with horizontal speed u. Explain why its path is a parabola (ignore relativity).

Show Answer

Horizontally there is no force, so horizontal velocity is constant: x = ut.

Vertically the force is constant (F = qE), so vertical acceleration is constant and y = (1/2)at².

Eliminate t using t = x/u to get y ∝ x², which is a parabola.

Mind stretcher 3: Optional (Enrichment)Extension

A. Relativistic speeds (beyond syllabus)

For large accelerating voltages, v can approach a significant fraction of c, and (1/2)mv² = eΔ V becomes inaccurate. In A Level questions, the non-relativistic model is normally assumed unless stated otherwise.

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