Charged particle motion in E-fields

CIE A-Level Physics· Unit 21: Electric fields, Subtopic 5· 15 min read

1. Force and Acceleration in Uniform E-fields★★☆☆☆⏱ 5 min

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📘 Definition

Electric force on a charged particle

$F = q E$

Magnitude of force on a point charge in a uniform electric field equals the product of charge and field strength. Direction is parallel to $E$ for positive charges, antiparallel for negative charges.

Example:

An electron (charge $- e$ ) experiences force opposite to the direction of the electric field.

From Newton's second law, constant electric force produces constant acceleration. Rearranging $F = ma$ gives the standard expression for acceleration:

a = \frac{F}{m} = \frac{qE}{m}

📐 Worked Example

A proton (mass $1.67 \times 1 0^{- 27} kg$ , charge $+ 1.60 \times 1 0^{- 19} C$ ) enters a uniform electric field of strength $250 N C^{- 1}$ parallel to its motion. Calculate its acceleration.

1
Recall the formula for acceleration of a charged particle:
$a = \frac{qE}{m}$
2
Substitute the given values:
$a = \frac{(1.60 \times 10^{-19})(250)}{1.67 \times 10^{-27}}$
3
Calculate the final result:
$a \approx 2.4 \times 10^{10} \text{ m s}^{-2}$

Exam tip:

Always check the sign of charge when finding acceleration direction; negative charge reverses the direction relative to the electric field.

2. Motion Parallel to the Electric Field★★☆☆☆⏱ 5 min

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When a charged particle moves parallel to the electric field, it undergoes constant acceleration linear motion, identical to a mass falling vertically in a uniform gravitational field. We use standard constant-acceleration kinematic equations:

$v = u + a t$
$s = u t + \frac{1}{2} a t^{2}$
$v^{2} = u^{2} + 2 a s$

📐 Worked Example

An electron is accelerated from rest through 1200 V. Find its final speed ( $m_{e} = 9.11 \times 1 0^{- 31} kg$ , $e = 1.60 \times 1 0^{- 19} C$ ).

1
Apply work-energy principle:
$qV = \frac{1}{2} m_e v^2$
2
Rearrange for $v$ :
$v = \sqrt{\frac{2qV}{m_e}}$
3
Substitute values ( $q = e$ ):
$v = \sqrt{\frac{2(1.60 \times 10^{-19})(1200)}{9.11 \times 10^{-31}}} \approx \sqrt{4.21 \times 10^{14}}$
4
Final speed:
$v \approx 2.1 \times 10^7 \text{ m s}^{-1}$

3. Motion Perpendicular to the Electric Field★★★☆☆⏱ 6 min

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When a charged particle enters a uniform E-field with initial velocity perpendicular to the field, its motion is identical to projectile motion under gravity: we separate motion into two independent components: constant velocity along the original direction, and constant acceleration perpendicular to the original direction.

📘 Definition

Transverse deflection

$y$

Total perpendicular displacement of the charged particle from its original straight-line path after crossing the electric field region.

📐 Worked Example

An electron travels horizontally at $2.0 \times 1 0^{7} m s^{- 1}$ between parallel deflection plates of length 0.04 m. The vertical E-field strength is $1200 N C^{- 1}$ . Calculate the vertical deflection while between the plates ( $m_{e} = 9.11 \times 1 0^{- 31} kg$ , $e = 1.60 \times 1 0^{- 19} C$ ).

1
Calculate time between plates (constant horizontal velocity):
$t = \frac{L}{u_x} = \frac{0.04}{2.0 \times 10^7} = 2.0 \times 10^{-9} \text{ s}$
2
Find vertical acceleration:
$a = \frac{eE}{m_e} = \frac{(1.60 \times 10^{-19})(1200)}{9.11 \times 10^{-31}} \approx 2.1 \times 10^{14} \text{ m s}^{-2}$
3
Use kinematic equation (initial $u_{y} = 0$ ):
$y = \frac{1}{2} a t^2$
4
Calculate deflection:
$y = 0.5 (2.1 \times 10^{14}) (2.0 \times 10^{-9})^2 = 4.2 \times 10^{-4} \text{ m} = 0.42 \text{ mm}$

Exam tip:

Velocity along the original direction of motion never changes, because there is no force in that direction for a uniform perpendicular E-field.

4. Common Pitfalls

Wrong move:

Forgetting that negative charge reverses the direction of acceleration/force

Why:

Magnitude will be correct, but direction is wrong, leading to incorrect deflection signs in coordinate problems

Correct move:

Assign a coordinate system first, then check the sign of charge when writing force/acceleration terms

Wrong move:

Including gravitational force when calculating acceleration of an electron in an E-field

Why:

Gravity is many orders of magnitude weaker than electric force for small charged particles, so it has no measurable effect

Correct move:

Neglect gravity for electrons and other subatomic particles, only include it if explicitly requested

Wrong move:

Treating acceleration as variable in a uniform E-field

Why:

Uniform E-field produces constant force, hence constant acceleration

Correct move:

Use standard constant-acceleration kinematic equations, not variable-force relationships

Wrong move:

Adding a non-zero initial transverse velocity term for deflection

Why:

Particles enter perpendicular to the field with zero initial velocity in the transverse direction

Correct move:

Set $u_{y} = 0$ , so $y = \frac{1}{2} a t^{2}$ is the correct equation for deflection

5. Quick Reference Cheatsheet

Quantity	Formula
Acceleration of charged particle	$a = \frac{q E}{m}$
Speed from potential difference $V$	$v = \frac{2 q V}{m}$
Time crossing plates of length $L$	$t = \frac{L}{u _{x}}$
Transverse deflection (perpendicular entry)	$y = \frac{1}{2} \frac{q E}{m} (\frac{L}{u _{x}})^{2}$
Final transverse velocity	$v_{y} = \frac{q E L}{m u _{x}}$

6. Frequently Asked

Is acceleration constant in a uniform electric field?

Yes, for uniform $E$ , the electric force is constant, so acceleration $a = q E / m$ is constant, matching gravitational acceleration near Earth's surface.

Why do we ignore gravity for electrons in E-fields?

Electrons have extremely small mass, so gravitational force is ~ $1 0^{14}$ times smaller than typical electric force, so it can be safely neglected for all standard problems.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

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Going deeper

syllabusOfficial CIE 9702 SyllabusFor official syllabus updates

What's Next

Understanding charged particle motion in uniform electric fields is fundamental for explaining particle deflection in cathode ray tubes, particle accelerators, and mass spectrometry, all common CIE exam question contexts. This subtopic builds on your AS-level knowledge of projectile motion and uniform electric fields, and directly connects to motion of charged particles in magnetic fields, where combined crossed E and B fields are used to select particle velocities. Mastery of kinematic separation of motion components here will help you solve more complex combined field problems in later topics, and reinforces Newtonian mechanics applied to electromagnetism.