Physics • Year 12 • Module 6: Electromagnetism • Lesson 2

Trajectories in Electric Fields

Lock in the core vocabulary, the SUVAT equations for charged-particle motion, and the direction rules for parabolic trajectories before tackling calculations.

Build · Vocab & Recall

1. Term–definition match

The definitions below are shuffled. In the right-hand column write the matching term from this list: trajectory, parabolic path, time of flight, electric acceleration, deflection, independence of motion, uniform electric field, horizontal velocity, vertical velocity component, impact velocity. 10 marks (1 each)

#DefinitionMatching term
1.1The complete path followed by a particle through space as a function of time.
1.2The curved path traced by a particle under a constant perpendicular force; the same shape as a projectile under gravity.
1.3The duration a charged particle spends inside the region between two charged plates.
1.4The constant acceleration given by a = qE/m produced on a charged particle in a uniform electric field.
1.5The perpendicular displacement of a particle from its original straight-line path due to a transverse electric force.
1.6The principle that the horizontal and vertical motions of a particle in a field do not affect each other.
1.7An electric field that has the same magnitude and direction at every point, produced between two large parallel plates.
1.8The component of a particle's velocity parallel to the plates, which remains constant throughout the motion.
1.9The component of a particle's velocity perpendicular to the plates, which increases uniformly due to the electric force.
1.10The resultant speed and direction of a particle at the moment it exits the electric field region.
Stuck? Revisit the Key Terms panel and Cards 1 and 2 in the lesson.

2. True or false — with correction

Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)

2.1 A charged particle entering a uniform electric field perpendicular to the field lines follows a circular path.    T  /  F

2.2 The horizontal component of a charged particle's velocity remains constant as it moves between parallel plates (ignoring gravity).    T  /  F

2.3 An electron and a proton fired horizontally into the same electric field will both curve in the same direction.    T  /  F

2.4 The time of flight between plates is calculated using t = L / vx, where L is the plate length and vx is the horizontal speed.    T  /  F

2.5 The vertical deflection of a charged particle between plates is independent of the particle's mass.    T  /  F

2.6 Increasing the horizontal launch speed of a particle increases its time of flight between plates of fixed length.    T  /  F

Stuck? Revisit Cards 1 and 2, especially the independence-of-motion principle and the formula a = qE/m.

3. Fill-in-the-blank paragraph

Use the word bank to complete the passage. Each word is used once. 8 marks (1 per blank)

Word bank:

parabola  ·  horizontal  ·  constant  ·  qE/m  ·  perpendicular  ·  charge  ·  opposite  ·  increases

When a charged particle enters a uniform electric field at right angles to the field lines, the electric force acts ___________ to its initial velocity. Because this force is constant, the particle traces a ___________ — exactly like a projectile in gravity. The ___________ component of velocity is unchanged throughout the motion, while the vertical speed ___________ uniformly due to the constant electric acceleration. This acceleration is given by ___________, and its direction depends on the sign of the particle's ___________. An electron curves in the ___________ direction to a proton in the same field, because the two particles carry opposite signs of charge. The ___________ nature of horizontal motion means t = L/vx gives the time of flight.

Stuck? Revisit Card 1 (The Parabolic Path) and the independence of motion principle in the lesson.

4. Function recall

Answer each question in 1–2 sentences using precise terms from the lesson. 8 marks (2 each)

4.1 State the two SUVAT equations used to find the vertical deflection and vertical exit speed of a charged particle between plates.

4.2 Why does the trajectory of a charged particle in a uniform electric field have the same mathematical form as projectile motion under gravity?

4.3 How do you determine whether a charged particle will strike a plate before exiting the field?

4.4 Explain why increasing the horizontal launch speed of an electron reduces its deflection between plates of fixed length.

Stuck? Revisit Card 2 and the worked example in the lesson.

5. Formula identification & variable naming

For each formula below, name the quantity each symbol represents and state its SI unit. 12 marks (1 per cell)

FormulaWhat it calculatesKey variable and unit
t = L / vx
a = qE / m
y = ½at²
vy = at
v = √(vx² + vy²)
θ = tan−1(vy / vx)
Stuck? Revisit the formula panel in Cards 1 and 2 of the lesson.
Answers — Do not peek before attempting

Q1 — Term–definition match

1.1 trajectory • 1.2 parabolic path • 1.3 time of flight • 1.4 electric acceleration • 1.5 deflection • 1.6 independence of motion • 1.7 uniform electric field • 1.8 horizontal velocity • 1.9 vertical velocity component • 1.10 impact velocity.

Q2 — True / false with correction

2.1 False. A charged particle entering a uniform field with velocity perpendicular to the field follows a parabolic path, not a circular one. A circular path requires a force that always points toward the centre, which is not the case for a uniform field.

2.2 True. No horizontal force acts (ignoring gravity), so vx remains constant throughout.

2.3 False. An electron and a proton curve in opposite directions because they carry opposite signs of charge. In a downward E-field, the force on a positive proton is downward (F = qE) but the force on a negative electron is upward.

2.4 True. Horizontal motion is uniform, so t = L / vx applies directly.

2.5 False. The deflection y = ½at² = ½(qE/m)t² depends on the charge-to-mass ratio q/m. A more massive particle with the same charge has a smaller acceleration and therefore smaller deflection.

2.6 False. A higher horizontal speed means the particle crosses the same plate length L in less time (t = L/vx), so the time of flight decreases.

Q3 — Cloze paragraph

In order: perpendicular / parabola / horizontal / increases / qE/m / charge / opposite / constant.

Q4.1 — SUVAT equations for vertical motion

Vertical deflection: y = uyt + ½at² (for a horizontal entry, uy = 0, so y = ½at²). Vertical exit speed: vy = uy + at (simplifies to vy = at for horizontal entry). Here a = qE/m.

Q4.2 — Why the same form as projectile motion

In projectile motion, gravity provides a constant downward force on the mass (F = mg), giving a constant acceleration g. In a uniform electric field, the electric force F = qE is also constant in magnitude and direction, giving constant acceleration a = qE/m. Because both situations involve a constant transverse force with no force along the initial velocity direction, the mathematics (and therefore the shape of the trajectory) is identical.

Q4.3 — Condition for striking a plate

Calculate the deflection y = ½at² at time t = L/vx. If the magnitude |y| exceeds half the plate separation (d/2), the particle strikes the plate before exiting. If |y| ≤ d/2, the particle exits the field without hitting a plate.

Q4.4 — Higher horizontal speed reduces deflection

A higher horizontal speed vx means the particle spends less time between the plates: t = L/vx decreases. Since vertical deflection is y = ½at² and a is unchanged, a shorter time t gives a much smaller y (deflection scales as t²). The electric force has less time to act, so the particle is deflected less.

Q5 — Formula identification

t = L/vx: Time of flight between the plates (s); L = plate length (m), vx = horizontal speed (m/s).

a = qE/m: Electric acceleration of the particle (m/s²); q = charge (C), E = field strength (V/m), m = mass (kg).

y = ½at²: Vertical deflection from the entry path (m); a = electric acceleration (m/s²), t = time of flight (s).

vy = at: Vertical speed at exit (m/s); a = electric acceleration (m/s²), t = time of flight (s).

v = √(vx² + vy²): Resultant (impact) speed at exit (m/s); vx and vy are horizontal and vertical components respectively.

θ = tan−1(vy/vx): Angle of exit velocity above/below the horizontal (°); vy and vx are the two velocity components at exit.