Year 12 Physics Module 6 · Electromagnetism ⏱ ~45 min 5 MC · 3 Short Answer Lesson 2 of 21

Trajectories in Electric Fields

In 1897, J.J. Thomson at the Cavendish Laboratory, Cambridge, fired cathode rays sideways through perpendicular electric and magnetic fields inside a glass tube. By balancing the two deflections he calculated e/m = 1.76 × 10¹¹ C/kg, showing cathode rays were particles far lighter than any atom. That ratio alone does not give the electron mass: an independent measurement of the charge, which Millikan supplied in 1909, was needed before the 1/1836 mass ratio could be worked out. The same crossed-field deflection principle is used in cathode-ray oscilloscopes; television picture tubes steered their beams with magnetic coils instead.

Today's hook: In Thomson's 1897 cathode ray tube experiment at the Cavendish Laboratory, an electron entered a 5 cm long deflection region with horizontal velocity 3.0 × 10⁷ m/s. The electric field between the plates was 2.0 × 10⁴ V/m. Predict: does the electron exit the plates still moving horizontally, or does it leave at an angle, and why does the path have the same shape as a thrown ball?

0/5TASKS

Your guided path0 of 8 done · keep going

You’re here

Orient and predict

Set the plate geometry and predict the direction and shape of an electron’s path.

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response Master Mastery tasks Exam Exam-style questions

Before you read, predict

An electron is fired horizontally at high speed into the gap between two parallel plates. The top plate is positive, the bottom plate is negative.

Before reading on, answer:

Sketch the path you think the electron will follow: straight line, parabola, circle, or something else?
A proton is fired horizontally at the same speed into the same gap. Will its path curve the same way, the opposite way, or stay straight?
If the plates were longer, would the particle hit the plate or exit the gap? What factors determine this?

Warm-up, when a charged particle enters a uniform electric field perpendicular to its velocity, what type of path does it follow?

Learning Intentions

goals

Know, Trajectory Equations

A charged particle in a uniform E-field follows a parabolic trajectory
Horizontal motion: constant velocity ($a_x = 0$)
Vertical motion: constant acceleration ($a_y = qE/m$)

Understand, Parallels to Projectiles

Why the trajectory is identical in form to projectile motion
Why electric acceleration replaces gravitational acceleration
How charge sign determines the direction of curvature

Can Do, Calculate & Predict

Calculate time of flight, range, and vertical deflection
Calculate impact velocity (magnitude and direction)
Predict whether a particle will strike a plate or exit the field

Scan these before reading

vocab

TrajectoryThe path followed by a particle through space as a function of time.

Time of flightThe time a particle spends between the plates before exiting.

Electric acceleration$a = qE/m$, the constant acceleration produced by a uniform electric field.

DeflectionThe perpendicular displacement of a particle from its original path due to a transverse force.

Independence of motionHorizontal and vertical motions are independent; one does not affect the other.

Cross-lesson links: L01 derived the electric field between plates. L02 applies it to a moving charge. When the entry velocity is perpendicular to a uniform electric field, the components follow the same mathematical model as a projectile, with electric force replacing gravity. Thomson's CRT is a practical example of this controlled deflection.

Core Content

Set the frame and model

Resolve motion into components and use the signed electric acceleration to explain curvature.

The Parabolic Path

+5 XP

Why charged particles in uniform E-fields follow projectile-like motion

For the standard plate model, the particle enters with velocity perpendicular to the uniform electric field, parallel to the plates. It then experiences a constant electric force perpendicular to the plates. The resulting motion is two-dimensional with:

Horizontal ($x$): No force, so $a_x = 0$ and $v_x$ is constant.
Vertical ($y$): Constant force $F = qE$, so $a_y = qE/m$ is constant.

This perpendicular-entry case is mathematically identical to projectile motion under gravity, except that:

Gravity ($g = 9.8$ m/s$^2$) is replaced by electric acceleration ($a = qE/m$)
The direction of $a$ depends on the sign of the charge
For an electron in a downward E-field, $a$ points upward
For a proton in the same field, $a$ points downward

Figure 1, A projectile curves downward due to gravity. Between large parallel plates, an electron entering to the right curves upward because its force is opposite the downward E-field. The axes and arrows define the velocity and force components.

SUVAT in Electric Fields

Horizontal: $x = v_x t$ ($v_x$ constant)

Vertical: $y = u_y t + \tfrac{1}{2} a t^2$ ($a = qE/m$)

Vertical velocity: $v_y = u_y + at$

Resultant: $v = \sqrt{v_x^2 + v_y^2}$

Stop & Check

An electron and a proton are fired horizontally at the same speed into the same uniform electric field (pointing downward). Which particle has the larger vertical acceleration? Which one hits the top plate first? Explain using $a = qE/m$.

For entry perpendicular to a uniform E-field, a charged particle follows a parabola, identical to projectile motion with signed $a = qE/m$ replacing $g$. Horizontal: $x = v_x t$ (constant). Vertical: $y = u_y t + \tfrac{1}{2}at^2$; resultant exit speed $v = \sqrt{v_x^2 + v_y^2}$.

Pause, copy the highlighted equations and analogy into your book before moving on.

A proton and an electron are fired horizontally into the same uniform electric field. Which statement is correct?

Resolve deflection and impact

Find the time in the field, vertical displacement and the condition for hitting a plate.

Range, Deflection, and Impact Velocity

+5 XP

Calculating where the particle ends up and how fast it is moving

We just saw that a charged particle in a uniform E-field follows a parabola with constant horizontal speed and constant vertical acceleration $a = qE/m$. That raises a question: how do you calculate the exact deflection, flight time, and impact velocity? This card answers it → systematic use of $t = L/v_x$, $y = \tfrac{1}{2}at^2$, and the Pythagoras result for exit speed.

To solve trajectory problems, treat the horizontal and vertical motions as completely independent:

Key Equations

Time between plates: $t = \dfrac{L}{v_x}$ $L$ = plate length, $v_x$ = horizontal speed (constant)

Vertical deflection: $y = u_y t + \tfrac{1}{2} a t^2$ $a = qE/m$ (watch the sign!)

Impact vertical velocity: $v_y = u_y + at$ $v_y$ at exit determines the subsequent trajectory

Impact speed: $v = \sqrt{v_x^2 + v_y^2}$ Direction found from $\tan^{-1}(v_y/v_x)$

Will it hit the plate? If the particle enters on the gap centreline, compare the magnitude of its deflection $|y|$ to half the plate separation $d/2$. If $|y| > d/2$, it strikes a plate before exiting. For any other entry position, compare the signed position with the actual plate boundaries.

Figure 2, Horizontal velocity remains constant; vertical velocity changes due to constant electric acceleration.

Stop & Check

An electron enters horizontally between two plates. During its flight, does its kinetic energy increase, decrease, or stay the same? What about a proton entering the same field? Explain using work done by the electric force.

Time of flight: $t = L/v_x$. For centreline entry, vertical deflection is $y = \tfrac{1}{2}at^2$ and $|y| < d/2$ confirms no plate strike. Exit speed: $v = \sqrt{v_x^2 + v_y^2}$; exit angle $\theta = \tan^{-1}(v_y/v_x)$.

Add the highlighted equations to your notes before the check below.

The horizontal speed of a charged particle between parallel plates remains constant throughout its flight (no horizontal force).

Increasing the horizontal launch speed of an electron between fixed plates increases its vertical deflection.

For a particle entering on the centreline, it hits a plate if $|y|$ exceeds half the plate separation $d/2$.

Work a deflection example

Use a signed acceleration, a plate-boundary test and velocity components in a full calculation.

Worked Example, Electron Deflection Between Plates

+5 XP

Calculate deflection, flight time, and impact velocity step by step

We just saw the key equations for trajectory problems. That raises a question: how do you apply them in sequence on an exam question? This card answers it → a four-step worked example: find $E$, find $a$, find $t$, find $y$ and exit angle.

Problem

An electron is fired horizontally at $2.0 \times 10^7$ m/s into the uniform electric field between two horizontal plates. The plates are 4.0 cm long and 2.0 cm apart, with a potential difference of 20 V (top plate positive).

(a) Calculate the electric field strength and the electron's acceleration.
(b) Calculate the time the electron spends between the plates.
(c) Calculate the vertical deflection of the electron as it exits the plates.
(d) Calculate the electron's velocity (magnitude and direction) as it exits.

Step 1, E-field and acceleration

$E = V/d = 20 / 0.020 = 1.0 \times 10^3$ V/m (downward)

For the electron ($q = -e$, $m_e = 9.11 \times 10^{-31}$ kg):

$a = qE/m = (-1.60 \times 10^{-19})(1.0 \times 10^3) / (9.11 \times 10^{-31})$

$a = 1.76 \times 10^{14}$ m/s$^2$ upward (force on negative charge opposes downward field)

Step 2, Time of flight

Horizontal motion is uniform: $t = L/v_x$

$t = 0.040 / (2.0 \times 10^7) = 2.0 \times 10^{-9}$ s = 2.0 ns

Step 3, Vertical deflection

$y = u_y t + \tfrac{1}{2} a t^2 = 0 + \tfrac{1}{2}(1.76 \times 10^{14})(2.0 \times 10^{-9})^2$

$y = 3.52 \times 10^{-4}$ m = 0.352 mm upward

The plate separation is 2.0 cm, so $d/2 = 10$ mm. Since 0.352 mm < 10 mm, the electron exits safely.

Step 4, Exit velocity

$v_x = 2.0 \times 10^7$ m/s (unchanged)

$v_y = u_y + at = 0 + (1.76 \times 10^{14})(2.0 \times 10^{-9}) = 3.52 \times 10^5$ m/s

$v = \sqrt{(2.0 \times 10^7)^2 + (3.52 \times 10^5)^2} \approx 2.0 \times 10^7$ m/s

$\theta = \tan^{-1}(v_y/v_x) = \tan^{-1}(0.0176) \approx 1.0°$ above horizontal

Worked example result: $E = 1.0 \times 10^3$ V/m, $a = 1.76 \times 10^{14}$ m/s² upward, $t = 2.0$ ns, $y = 0.352$ mm (safe, since $d/2 = 10$ mm), exit $\theta \approx 1.0°$. Step order: $E \to a \to t \to y \to$ compare to $d/2$.

Pause, write the highlighted step order into your book before moving on.

An electron enters horizontally at $3.0 \times 10^6$ m/s between plates 6.0 cm long. How long does it spend between the plates?

Reason with variables and limits

Use proportional reasoning, state the model conditions and connect motion graphs to the components.

Factors Affecting Deflection

+5 XP

Understanding which variables control how much a particle bends

We just saw how to calculate deflection step by step. That raises a question: what happens to the magnitude of deflection if you double the voltage, lengthen the plates, or speed up the particle? This card answers it → the deflection formula $|y| = |q|EL^2/(2mv_x^2)$ reveals the proportionalities.

For horizontal entry, the magnitude of vertical deflection is $|y| = \tfrac{1}{2}|a|t^2 = \tfrac{1}{2}\cdot\dfrac{|q|E}{m}\cdot\dfrac{L^2}{v_x^2}$. The sign and therefore the plate approached still come from $a_y=qE/m$ after you define the axes.

Increasing $E$ (stronger field): larger $a$, greater deflection, proportional to $E$
Increasing $L$ (longer plates): longer time in field, much greater deflection, proportional to $L^2$
Increasing $v_x$ (faster launch): shorter time in field, much smaller deflection, inversely proportional to $v_x^2$
Changing $d$ (plate separation) at fixed voltage: changes $E = V/d$, so deflection changes inversely with $d$
Changing particle mass: heavier particles (proton vs electron) have smaller $a = qE/m$, so much less deflection for the same conditions

Deflection formula (horizontal entry, $u_y = 0$)

$|y| = \dfrac{|q|EL^2}{2mv_x^2}$

Key insight

Because $|y| \propto L^2$ and $|y| \propto 1/v_x^2$, doubling the plate length quadruples the deflection magnitude, while doubling the entry speed reduces it to one quarter. Within this ideal plate model, increasing the horizontal speed is one way to reduce the chance of a plate strike.

Deflection magnitude (horizontal entry): $|y| = \dfrac{|q|EL^2}{2mv_x^2}$. Key proportionalities: $|y| \propto E$, $|y| \propto L^2$, $|y| \propto 1/v_x^2$. A proton's deflection magnitude is about 1836 times smaller than an electron's in the same field.

Add the highlighted formula and proportionalities to your notes before the check below.

Three of these changes increase the vertical deflection of an electron between fixed plates. Pick the odd one out (the one that decreases deflection).

Model, frame and graph check

Choose the laboratory frame and define $+x$ along the entry velocity. Choose a $+y$ direction, then give both $E_y$ and $a_y=qE_y/m$ their signs consistently. The familiar perpendicular-entry model requires a uniform field, negligible gravity and non-relativistic speed. With a different entry angle, resolve both initial velocity components and use the general component equations instead of the horizontal-launch shortcut.

On an $x$–$t$ graph the gradient is the constant $v_x$. On a $v_y$–$t$ graph the signed gradient is $a_y=qE/m$, and the area under that graph is the vertical displacement. For a centreline entry, test $|y|<d/2$ before reporting an exit velocity; otherwise compare the signed position with the actual plate boundary.

Apply and consolidate

Test a plate-boundary prediction, explain a CRT deflection and summarise the assumptions behind the model.

Activity 1, Trajectory Calculations

ApplyBand 4

Practise the step-by-step method for particle trajectory problems

An electron enters on the centreline, horizontally at $2.0 \times 10^7$ m/s, into a downward field of 500 V/m between plates 1.0 cm long and 2 cm apart. Calculate the deflection and determine if it hits the top plate.
Repeat (1) but for a proton entering on the centreline at the same speed. Which quantity changes, which stays the same?
An electron enters on the centreline at $4.0 \times 10^7$ m/s into the same field. Without recalculating, predict how the deflection magnitude compares to (1). Verify by calculating.

Activity 2, Concept Check

UnderstandBand 4

Explain the physics behind the parabolic trajectory

Cathode-ray tubes in old televisions used charged plates to deflect electron beams onto a phosphor screen. Explain, in terms of the physics of this lesson, how adjusting the plate voltage changes where the beam hits the screen. Include reference to $E$, $a$, $y$, and the direction of electron deflection.

Fill the gap. For a proton fired at $1.0 \times 10^6$ m/s into a field $E = 2000$ V/m between plates 8.0 cm long, the time of flight is $t = L/v_x = 0.08 / (1.0 \times 10^6) = \square \times 10^{-8}$ s.

Independent practice

Show what you can do without prompts

Complete a fresh question-bank set, then calculate and explain trajectories in HSC-style responses.

Quick recall, trajectories in electric fields

+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 8 marks

+5 XP

ApplyBand 4(2 marks) 1. An electron enters horizontally at $3.0 \times 10^7$ m/s between plates with an electric-field magnitude $E = 8000$ V/m and length 8.0 cm. Calculate (a) the time of flight and (b) the magnitude of the vertical deflection.

1 mark: correct time of flight with working · 1 mark: correct deflection with working

ApplyBand 5(2 marks) 2. A proton enters on the centreline, horizontally at $1.0 \times 10^6$ m/s, between plates with an electric-field magnitude $E = 2000$ V/m and length 8.0 cm. Calculate the deflection magnitude and determine whether it hits either plate, which are 2.0 cm apart.

1 mark: correct deflection · 1 mark: correct plate-hit determination with reasoning

EvaluateBand 6(4 marks) 3. An electron enters on the centreline, horizontally at $4.0 \times 10^7$ m/s, between two horizontal plates 5.0 cm long and 1.5 cm apart. The top plate is at +120 V and the bottom plate is at 0 V. (a) Calculate the electric-field magnitude and direction. (b) Show that the electron does not hit either plate before exiting. (c) Calculate the angle of the electron's velocity relative to the horizontal as it exits.

1 mark: $E = V/d$ correct · 1 mark: $y$ calculated and compared to $d/2$ · 1 mark: $v_y$ calculated · 1 mark: angle using $\tan^{-1}(v_y/v_x)$

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1 (2 marks): (a) $t = L/v_x = 0.08 / (3.0 \times 10^7) = 2.67 \times 10^{-9}$ s. (b) $a = (1.60 \times 10^{-19} \times 8000) / (9.11 \times 10^{-31}) = 1.41 \times 10^{15}$ m/s$^2$; $y = \tfrac{1}{2}at^2 = \tfrac{1}{2}(1.41 \times 10^{15})(2.67 \times 10^{-9})^2 = 5.0 \times 10^{-3}$ m $= 5.0$ mm.

Q2 (2 marks): $a = (1.60 \times 10^{-19} \times 2000) / (1.67 \times 10^{-27}) = 1.92 \times 10^{11}$ m/s$^2$; $t = 8.0 \times 10^{-8}$ s; $y = \tfrac{1}{2}(1.92 \times 10^{11})(8.0 \times 10^{-8})^2 = 6.1 \times 10^{-4}$ m $= 0.61$ mm. Since $0.61$ mm $< 10$ mm ($d/2$), the proton does not hit the plate.

Q3 (4 marks): (a) $E = V/d = 120/0.015 = 8000$ V/m, downward from the positive top plate to the lower-potential bottom plate. (b) The electron's acceleration is upward with magnitude $a = (1.60 \times 10^{-19} \times 8000)/(9.11 \times 10^{-31}) = 1.41 \times 10^{15}$ m/s$^2$; $t = 0.05/(4.0 \times 10^7) = 1.25 \times 10^{-9}$ s; $y = \tfrac{1}{2}(1.41 \times 10^{15})(1.25 \times 10^{-9})^2 = 1.10 \times 10^{-3}$ m $= 1.10$ mm upward. Since 1.10 mm $< 7.5$ mm, the electron exits safely. (c) $v_y = at = (1.41 \times 10^{15})(1.25 \times 10^{-9}) = 1.76 \times 10^6$ m/s upward; $\theta = \tan^{-1}(1.76 \times 10^6 / 4.0 \times 10^7) = \tan^{-1}(0.044) \approx 2.5°$ above horizontal.

Final step

Retrieve, reflect and finish

Retrieve the trajectory model, revisit your prediction and record the condition you will check next time.

Check what actually stuck

How did your thinking change?

At the start you were asked about Thomson's 1897 cathode ray tube experiment at the Cavendish Laboratory: an electron entering a 5 cm deflection region at 3.0 × 10⁷ m/s in a 2.0 × 10⁴ V/m field.

The acceleration magnitude is $|a| = |q|E/m = (1.6 \times 10^{-19} \times 2.0 \times 10^4)/(9.11 \times 10^{-31}) \approx 3.5 \times 10^{15}$ m/s², opposite the field for an electron. Time in the plates: $t = L/v_x = 0.05/(3.0 \times 10^7) \approx 1.67 \times 10^{-9}$ s. The deflection magnitude is $|y| = \tfrac{1}{2}|a|t^2 \approx 4.9 \times 10^{-3}$ m, or 4.9 mm. The electron exits at an angle, not horizontally. This is the parabolic trajectory Thomson measured to determine e/m = 1.76 × 10¹¹ C/kg.

Both electrons and protons follow parabolas in Thomson's apparatus, but they curve in opposite directions because the electric force on a negative charge opposes the field direction.

I have completed this lesson

← Previous: Uniform Electric Fields Next: Work, Energy and Potential →

Module overview · Physics · Checkpoint 1 · Module quiz