# Charged particle motion in E-fields

> CIE A-Level Physics · 9702 A2
> Source: https://www.owlsprep.com/study/cie-9702-u21-charged-particle-motion-in-e/

This sub-topic explains how charged particles move in uniform electric fields, covering acceleration parallel to fields and projectile-like deflection when entering perpendicular to fields, core concepts for common particle physics exam problems.

**Prerequisites:** [Uniform electric fields](https://www.owlsprep.com/study/cie-9702-u21-uniform-electric-fields/); [Projectile motion](https://www.owlsprep.com/study/cie-9702-u05-projectile-motion/)

## Learning objectives

- Calculate acceleration of a charged particle in a uniform electric field
- Analyze projectile-like motion of charged particles entering E-fields perpendicular to field lines
- Calculate transverse deflection and final velocity after crossing a uniform E-field
- Apply work-energy principles to find speed of accelerated charged particles

## Force and Acceleration in Uniform E-fields

**Electric force on a charged particle** — 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.

*Notation:* $F = qE$

*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 $1.67 \times 10^{-27} \text{ kg}$, charge $+1.60 \times 10^{-19} \text{ C}$) enters a uniform electric field of strength $250 \text{ N C}^{-1}$ 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.

*Calculator:* allowed

## Motion Parallel to the Electric Field

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:

1. $v = u + at$
2. $s = ut + \frac{1}{2} a t^2$
3. $v^2 = u^2 + 2 a s$

> **info**
>
> For a particle accelerated from rest through potential difference $V$, use the work-energy shortcut: work done by the field equals kinetic energy gained: $qV = \frac{1}{2}mv^2$, so $v = \sqrt{\frac{2qV}{m}}$. This avoids needing time or distance.

**Worked example:** An electron is accelerated from rest through 1200 V. Find its final speed ($m_e = 9.11 \times 10^{-31} \text{ kg}$, $e = 1.60 \times 10^{-19} \text{ C}$).

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:

   $$v \approx 2.1 \times 10^7 \text{ m s}^{-1}$$

*Calculator:* allowed

## Motion Perpendicular to the Electric Field

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.

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

*Notation:* $y$

**Worked example:** An electron travels horizontally at $2.0 \times 10^7 \text{ m s}^{-1}$ between parallel deflection plates of length 0.04 m. The vertical E-field strength is $1200 \text{ N C}^{-1}$. Calculate the vertical deflection while between the plates ($m_e = 9.11 \times 10^{-31} \text{ kg}$, $e = 1.60 \times 10^{-19} \text{ C}$).

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.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting that negative charge reverses the direction of acceleration/force
  - Why it fails: Magnitude will be correct, but direction is wrong, leading to incorrect deflection signs in coordinate problems
  - Correct: Assign a coordinate system first, then check the sign of charge when writing force/acceleration terms
- **Wrong:** Including gravitational force when calculating acceleration of an electron in an E-field
  - Why it fails: Gravity is many orders of magnitude weaker than electric force for small charged particles, so it has no measurable effect
  - Correct: Neglect gravity for electrons and other subatomic particles, only include it if explicitly requested
- **Wrong:** Treating acceleration as variable in a uniform E-field
  - Why it fails: Uniform E-field produces constant force, hence constant acceleration
  - Correct: Use standard constant-acceleration kinematic equations, not variable-force relationships
- **Wrong:** Adding a non-zero initial transverse velocity term for deflection
  - Why it fails: Particles enter perpendicular to the field with zero initial velocity in the transverse direction
  - Correct: Set $u_y = 0$, so $y = \frac{1}{2} a t^2$ is the correct equation for deflection

## Cheatsheet

| Quantity | Formula |
| --- | --- |
| Acceleration of charged particle | $a = \frac{qE}{m}$ |
| Speed from potential difference $V$ | $v = \sqrt{\frac{2qV}{m}}$ |
| Time crossing plates of length $L$ | $t = \frac{L}{u_x}$ |
| Transverse deflection (perpendicular entry) | $y = \frac{1}{2} \frac{qE}{m} \left(\frac{L}{u_x}\right)^2$ |
| Final transverse velocity | $v_y = \frac{qE L}{m u_x}$ |

## 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.

- [Electric potential](https://www.owlsprep.com/study/cie-9702-u21-electric-potential/)
- [Capacitance](https://www.owlsprep.com/study/cie-9702-u22-overview/)
- [Capacitance concepts](https://www.owlsprep.com/study/cie-9702-u22-capacitance-concepts/)

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