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 — proving cathode rays were particles (electrons) with mass 1/1836 that of a hydrogen atom. Every CRT television ever built, right up to 2007, used this identical parabolic deflection principle.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
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
Core Content
Why charged particles in uniform E-fields follow projectile-like motion
When a charged particle enters the uniform electric field between parallel plates with some initial velocity, it experiences a constant force perpendicular to the plates. If the initial velocity has a component parallel 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 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; an electron curves upward because the force on a negative charge opposes the downward E-field.
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: $v = \sqrt{v_x^2 + v_y^2}$
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 $a = qE/m$.
A charged particle entering a uniform electric field with horizontal velocity follows a parabola — identical to projectile motion with $a = qE/m$ 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?
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:
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? Compare the magnitude of the deflection $|y|$ to half the plate separation $d/2$. If $|y| > d/2$, the particle strikes the plate before exiting.
Figure 2 — Horizontal velocity remains constant; vertical velocity changes due to constant electric acceleration.
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: $t = L/v_x$. Vertical deflection: $y = \tfrac{1}{2}at^2$ (check $|y| < d/2$ to confirm 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.
A particle hits the plate if the vertical deflection $|y|$ exceeds half the plate separation $d/2$.
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.
An electron is fired horizontally at $2.0 \times 10^6$ m/s into the uniform electric field between two horizontal plates. The plates are 10 cm long and 2.0 cm apart, with a potential difference of 250 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.
$E = V/d = 250 / 0.020 = 1.25 \times 10^4$ V/m (downward)
For the electron ($q = -e$, $m_e = 9.11 \times 10^{-31}$ kg):
$a = qE/m = (-1.60 \times 10^{-19})(1.25 \times 10^4) / (9.11 \times 10^{-31})$
$a = 2.20 \times 10^{15}$ m/s$^2$ upward (force on negative charge opposes downward field)
Horizontal motion is uniform: $t = L/v_x$
$t = 0.10 / (2.0 \times 10^6) = 5.0 \times 10^{-8}$ s = 50 ns
$y = u_y t + \tfrac{1}{2} a t^2 = 0 + \tfrac{1}{2}(2.20 \times 10^{15})(5.0 \times 10^{-8})^2$
$y = 2.75 \times 10^{-3}$ m = 2.75 mm upward
The plate separation is 2.0 cm, so $d/2 = 10$ mm. Since 2.75 mm < 10 mm, the electron exits safely.
$v_x = 2.0 \times 10^6$ m/s (unchanged)
$v_y = u_y + at = 0 + (2.20 \times 10^{15})(5.0 \times 10^{-8}) = 1.10 \times 10^8$ m/s
$v = \sqrt{(2.0 \times 10^6)^2 + (1.10 \times 10^8)^2} \approx 1.10 \times 10^8$ m/s
$\theta = \tan^{-1}(v_y/v_x) = \tan^{-1}(55) \approx 89°$ above horizontal
Worked example result: $E = 1.25 \times 10^4$ V/m, $a = 2.20 \times 10^{15}$ m/s², $t = 50$ ns, $y = 2.75$ mm (safe, since $d/2 = 10$ mm), exit $\theta \approx 89°$. 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?
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 deflection if you double the voltage, lengthen the plates, or speed up the particle? This card answers it → the deflection formula $y = qEL^2/(2mv_x^2)$ reveals the proportionalities.
The vertical deflection formula $y = \tfrac{1}{2}at^2 = \tfrac{1}{2}\cdot\dfrac{qE}{m}\cdot\dfrac{L^2}{v_x^2}$ reveals exactly which variables matter.
- 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
$y = \dfrac{qEL^2}{2mv_x^2}$
Because $y \propto L^2$ and $y \propto 1/v_x^2$, doubling the plate length quadruples the deflection, while doubling the entry speed reduces deflection to one quarter. The fastest way to prevent a particle hitting the plate is to increase its horizontal speed.
Deflection formula (horizontal entry): $y = \dfrac{qEL^2}{2mv_x^2}$. Key proportionalities: $y \propto E$, $y \propto L^2$, $y \propto 1/v_x^2$. Proton deflects $\approx 1836$ times less than an electron 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).
Practise the step-by-step method for particle trajectory problems
- An electron is fired horizontally at $2.0 \times 10^6$ m/s into a field of 5000 V/m between plates 10 cm long and 2 cm apart. Calculate the deflection and determine if it hits the plate.
- Repeat (1) but for a proton entering at the same speed. Which quantity changes, which stays the same?
- An electron is fired at $4.0 \times 10^6$ m/s into the same field. Without recalculating, predict how the deflection compares to (1). Verify by calculating.
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 $t = L/v_x = 0.08 / (1.0 \times 10^6) = _____ \times 10^{-8}$ s.
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.
ApplyBand 4(2 marks) 1. An electron enters horizontally at $3.0 \times 10^6$ m/s between plates with $E = 8000$ V/m and length 8.0 cm. Calculate (a) the time of flight and (b) the vertical deflection.
1 mark: correct time of flight with working · 1 mark: correct deflection with working
ApplyBand 5(2 marks) 2. A proton is fired horizontally at $1.0 \times 10^6$ m/s between plates with $E = 2000$ V/m and length 8.0 cm. Calculate the vertical deflection and determine if it hits a plate 2.0 cm apart.
1 mark: correct deflection · 1 mark: correct plate-hit determination with reasoning
EvaluateBand 6(4 marks) 3. An electron is fired horizontally at $4.0 \times 10^6$ 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 strength. (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 $\tan^{-1}(v_y/v_x)$
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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) $t = L/v_x = 0.08 / (3.0 \times 10^6) = 2.67 \times 10^{-8}$ s. (b) $a = (1.60 \times 10^{-19} \times 8000) / (9.11 \times 10^{-31}) = 1.41 \times 10^{15}$ m/s$^2$; $y = \tfrac{1}{2}at^2 = \tfrac{1}{2}(1.41 \times 10^{15})(2.67 \times 10^{-8})^2 = 5.0 \times 10^{-4}$ m $= 0.50$ 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. (b) $a = (1.60 \times 10^{-19} \times 8000)/(9.11 \times 10^{-31}) = 1.41 \times 10^{15}$ m/s$^2$; $t = 0.05/(4.0 \times 10^6) = 1.25 \times 10^{-8}$ s; $y = \tfrac{1}{2}(1.41 \times 10^{15})(1.25 \times 10^{-8})^2 = 1.10 \times 10^{-3}$ m $= 1.10$ mm. Since 1.10 mm $< 7.5$ mm, the electron exits safely. (c) $v_y = at = (1.41 \times 10^{15})(1.25 \times 10^{-8}) = 1.76 \times 10^7$ m/s; $\theta = \tan^{-1}(1.76 \times 10^7 / 4.0 \times 10^6) = \tan^{-1}(4.4) \approx 77°$ above horizontal.
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 answer: acceleration $a = qE/m = (1.6 \times 10^{-19} \times 2.0 \times 10^4)/(9.11 \times 10^{-31}) \approx 3.5 \times 10^{15}$ m/s². Time in the plates: $t = L/v_x = 0.05/3.0 \times 10^7 \approx 1.67 \times 10^{-9}$ s. Vertical deflection: $y = \tfrac{1}{2}at^2 \approx 4.9 \times 10^{-3}$ m ≈ 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.