Year 12 Physics Module 6 ⏱ ~40 min 5 MC · 2 Short Answer Lesson 3 of 21

Work, Energy and Potential

Particle accelerators use electric potential differences to transfer energy to charged particles. The signed energy ledger is the key: define the start and end potentials, then use $\Delta U=q\Delta V$, $W_{\mathrm{field}}=-q\Delta V$ and $\Delta K=-q\Delta V$. For an accelerating potential magnitude, the positive gain is $|q\Delta V|$.

Today's hook: CERN's Large Hadron Collider (first collision run, 23 November 2009) accelerates protons to 0.999999991c using a 27 km ring. If you double the accelerating voltage in any stage, does the proton come out twice as fast, and why does this matter for reaching near-light-speed?

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Orient with signs

Set a start-to-finish potential convention before predicting how an electron or proton gains kinetic energy.

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

Use the available foundations worksheet for guided practice on this lesson.

Build Foundations & guided practice

Before you read, predict

An electron is released from rest near the negative plate of a parallel-plate capacitor. It accelerates across the gap and strikes the positive plate.

Before reading on, consider:

If the battery voltage is doubled (same plate separation), does the electron hit the plate with twice the speed, four times the speed, or something else?
If the plate separation is doubled but the battery stays the same, does the electron arrive with more, less, or the same kinetic energy?
A proton is released from rest near the positive plate of the same capacitor. Does it reach the negative plate with the same kinetic energy as the electron?

Warm-up: A charge $q$ moves through a potential difference $\Delta V$. The work done by the electric field on the charge is:

Learning Intentions

goals

Know, Energy Relationships

With $\Delta V=V_f-V_i$, work by the electric field is $W_{\mathrm{field}}=-q\Delta V$
Change in electric potential energy: $\Delta U = q\Delta V$
When only the electric field does work, $\Delta K=-q\Delta V$; the positive gain magnitude is $|q\Delta V|$

Understand, Independence from Path

Why KE gain depends only on the accelerating-potential magnitude $|\Delta V|$, not on plate separation
Why, for motion along the electric force in a uniform field, $|W_{\mathrm{field}}|=|q|Ed=|q\Delta V|$
The difference between eV and joules as energy units

Can Do, Calculate and Compare

Calculate final speed from a stated accelerating-potential magnitude
Compare KE and speed for particles given the same accelerating-potential magnitude
Identify when non-relativistic approximations break down

Scan these before reading

vocab

Work done by electric field$W_{\mathrm{field}}=-q\Delta V$ for $\Delta V=V_f-V_i$.

Electric potential energyEnergy stored due to a charge's position in an electric field.

Kinetic energy change$\Delta K=-q\Delta V$ when the electric field is the only force doing work.

Electronvolt (eV)Energy unit: $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$, the positive energy magnitude $e\times1\ \text{V}$.

EquipotentialA surface where every point is at the same electric potential.

Cross-lesson links: L02 described how an electric force changes a charged particle's trajectory. L03 now tracks the associated transfer between electric potential energy and kinetic energy. L04 introduces magnetic forces, which can redirect a charged particle without changing its kinetic energy.

Core Content

Build the energy ledger

Relate potential difference, system potential energy, field work and kinetic energy without losing the sign.

Work and Energy in Electric Fields

+5 XP

How potential difference converts to kinetic energy

Picture an electron sitting still at the negative plate of a parallel-plate capacitor. It moves to the positive plate, from lower to higher potential. The electric field does positive work on the electron, converting electric potential energy into kinetic energy. Write the start and finish potentials before using a magnitude: for example, $V_i=0\ \text{V}$ and $V_f=+5000\ \text{V}$ give $\Delta V=+5000\ \text{V}$ for that electron.

Work in a Uniform Field

$W_{\mathrm{field}}=\vec F\cdot\Delta\vec r=-q\Delta V$

For motion along the electric force in a uniform field, $|W_{\mathrm{field}}|=|q|Ed=|q\Delta V|$.

The crucial insight is that the work depends only on the potential difference $\Delta V$ and the charge $q$, not on the path taken or the plate separation. If you double the separation while keeping the same battery, the field magnitude $E$ is halved but $d$ is doubled, so $E d=|\Delta V|$ stays constant.

By the work-energy theorem, this work becomes kinetic energy:

Kinetic Energy Gain

$\Delta K=-q\Delta V$

For acceleration from rest, the positive gain is $\tfrac{1}{2}mv^2=|q\Delta V|$

$v=\sqrt{\dfrac{2|q\Delta V|}{m}}$, within the non-relativistic model

Figure 1, Energy conversion for an electron moving from $V_i=0\ \text{V}$ to $V_f>0$. Here $q=-e$, so $\Delta U=-e\Delta V$, $W_{\mathrm{field}}=\Delta K=+e\Delta V$, and total energy (PE + KE) is conserved.

Stop and check

An electron moves from rest through an accelerating-potential magnitude of 100 V, gaining kinetic energy $K$. A second electron has an accelerating-potential magnitude of 200 V. Does the second electron have twice the kinetic energy, twice the speed, or both? Explain.

Worked example, Electron gun at 5000 V

An electron starts from rest at $V_i=0\ \text{V}$ and finishes at $V_f=+5000\ \text{V}$. Calculate (a) $\Delta U$, $W_{\mathrm{field}}$ and its final kinetic energy in joules and eV, (b) its final speed.

Signed ledger. $q=-e$ and $\Delta V=+5000\ \text{V}$, so $\Delta U=q\Delta V=-5000\ \text{eV}$, $W_{\mathrm{field}}=-q\Delta V=+5000\ \text{eV}$ and $\Delta K=+5000\ \text{eV}=\mathbf{8.0\times10^{-16}\ \text{J}}$.
Part (b). $\tfrac12m_ev^2=\Delta K$, so $v=\sqrt{2\Delta K/m_e}$.
Substitution gives $v=\mathbf{4.19\times10^7\ \text{m/s}}$.
Relativistic check: $v/c \approx 0.14$, so this is a close non-relativistic estimate rather than an exact speed. Use relativistic energy-momentum reasoning when the approximation matters.

Define $\Delta V=V_f-V_i$. Then $\Delta U=q\Delta V$, $W_{\mathrm{field}}=-q\Delta V$, and, when electric work is the only work, $\Delta K=-q\Delta V$. For acceleration from rest use the positive gain $|q\Delta V|$ and $v=\sqrt{2|q\Delta V|/m}$.

Pause, copy the highlighted relationships and formula into your book before moving on.

An electron accelerates from rest. If the accelerating-potential magnitude is doubled, its final speed will be multiplied by:

Map potential and field

Read equipotentials and the potential gradient, then connect an electronvolt to a measurable energy change.

Equipotentials and the Electronvolt

+5 XP

Why the volt is both a potential unit and an energy unit

We just saw that the signed kinetic-energy change is $\Delta K=-q\Delta V$, while an acceleration problem often quotes the positive gain $|q\Delta V|$. That raises a question: what do equipotential surfaces look like between plates, and why do physicists use electronvolts instead of joules? This card answers it → equipotentials are parallel planes; 1 eV = $1.60 \times 10^{-19}$ J.

Between parallel plates, surfaces parallel to the plates are equipotentials every point on such a surface has the same electric potential. A charge can move along an equipotential without any change in electric potential energy.

Figure 2, Equipotential lines are parallel to the plates. Moving perpendicular to them does work; moving along them does not.

The electronvolt (eV) is an energy-unit magnitude: $1\ \text{eV}$ is the energy magnitude associated with one elementary charge and a potential difference magnitude of one volt.

Electronvolt definition

$1\ \text{eV} = e \times 1\ \text{V} = 1.60 \times 10^{-19}\ \text{J}$

This is incredibly convenient in atomic and particle physics because typical electron energies are measured in eV, keV, or MeV, much easier than writing powers of ten in joules.

HSC Tip

When a question asks for energy "in electronvolts," you do not need to multiply by $e$ and convert to joules. An electron moving through a 250 V potential rise gains 250 eV of kinetic energy; its potential energy decreases by 250 eV.

Non-relativistic limit

The equation $v=\sqrt{2|q\Delta V|/m}$ is only valid when $v\ll c$ (a useful check is $v<0.1c$). For electrons, the classical result reaches $0.1c$ at about 2.6 kV. Near light speed, use relativistic energy instead of $\tfrac12mv^2$.

Potential gradient

Electric field points toward decreasing potential. In one dimension, $E_x=-dV/dx$: the gradient of a $V$–$x$ graph is $-E_x$, while the area under an $E_x$–$x$ graph is $-\Delta V$. Potential is a property of the field location; potential energy $U=qV$ belongs to the charge–field system and therefore also depends on the signed charge.

Stop and check

An electron and a proton are each accelerated from rest through the same accelerating-potential magnitude of 1000 V. Which has more kinetic energy? Which is moving faster? By what factor?

Worked example, Comparing electron and proton speeds

A proton starts from rest at $V_i=+5000\ \text{V}$ and finishes at $V_f=0\ \text{V}$, alongside the electron above. Calculate the proton's speed and compare it with the electron's speed.

Signed ledger. $q=+e$ and $\Delta V=-5000\ \text{V}$, so $\Delta U=-5000\ \text{eV}$, $W_{\mathrm{field}}=\Delta K=+5000\ \text{eV}$.
$v_p=\sqrt{2e(5000\ \text{V})/m_p}=\mathbf{9.79\times10^5}\ \text{m/s}$.
Comparison: Both particles gain the same kinetic energy magnitude, 5000 eV, but because $m_p/m_e = 1836$, the proton is about $\sqrt{1836} \approx 43$ times slower than the electron. Thus $v \propto 1/\sqrt{m}$ for fixed $|q\Delta V|$.

Equipotential surface: constant electric potential, no work done moving along it. $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$; an electron moving through a potential rise of magnitude $V$ gains $V$ eV. Same $|q\Delta V|$ gives the same KE magnitude, but speed $v \propto 1/\sqrt{m}$: the proton is $\sqrt{1836} \approx 43$ times slower than the electron.

Pause, write the highlighted definition and ratio into your book before moving on.

An electron and a proton accelerated from rest through the same accelerating-potential magnitude gain the same kinetic-energy magnitude.

Doubling the plate separation (same voltage) doubles the final kinetic energy of the accelerated particle.

Moving a charge along an equipotential surface does zero work.

Calculate particle acceleration

Use an accelerating potential magnitude to compare kinetic energy and speed for particles with different charge-to-mass ratios.

Activity 1, Particle Acceleration Calculations

ApplyBand 4

Apply the work-energy theorem to charged particles

An electron is accelerated from rest through an accelerating potential magnitude of 400 V. Calculate its final speed.
A proton is accelerated from rest through the same 400 V magnitude. Calculate its final speed. By what factor is the electron faster than the proton?
The plate separation is increased from 2 cm to 4 cm while the accelerating potential magnitude stays at 400 V. Does the proton's final speed change? Explain using $|W_{\mathrm{field}}|=|q|Ed$.

An alpha particle ($q = 2e$, $m = 4 \times 1.67 \times 10^{-27}$ kg) is accelerated from rest through an accelerating-potential magnitude of 1000 V. Its kinetic energy is _____ eV.

Analyse an electron gun

Apply the signed ledger to a concrete device, then distinguish the field model from a real accelerating system.

Interactive Tool, Electric Fields Open fullscreen ↗

Use the parallel-plate potential/energy mode for this lesson. Set the start and finish potentials, predict the sign of $\Delta V$ and $\Delta U$, then test the result. The signed ledger and labelled figures above remain the complete accessible explanation, so the interactive is optional.

Activity 2, Electron Gun Analysis

UnderstandBand 5

Explain the operation of an electron gun using energy principles

In an electron gun, electrons start from rest at $V_i=0\ \text{V}$ and reach an electrode at $V_f=+2500\ \text{V}$, separated by 5 mm.

Calculate the kinetic energy of each electron in joules.
Calculate the speed of each electron.
Explain why the kinetic energy does not depend on the distance between the electrodes.
The electrons enter a region with no electric field. Describe their motion using Newton's first law.

Consolidate the model

Resolve the common distance and mass misconceptions, then carry a precise signed model into independent practice.

Wrap-up, Misconceptions and Summary

Misconceptions, final check

Wrong: "Doubling the plate separation gives the electron more time to accelerate, so it gains more kinetic energy."

Right: for the same accelerating potential magnitude, $\Delta K=|q\Delta V|$. The field magnitude $E$ halves when $d$ doubles, but the particle travels twice as far, so $|W_{\mathrm{field}}|=|q|Ed=|q|\Delta V$ is unchanged. The final kinetic energy and speed are unchanged.

Wrong: "The proton must gain more kinetic energy because it is heavier."

Right: with $\Delta V=V_f-V_i$, $\Delta K=-q\Delta V$. If the question gives only the magnitude of an accelerating potential difference, the positive gain is $|q\Delta V|$. Mass then affects the resulting speed.

Copy into your books

Key Definitions

Signed ledger: $\Delta U=q\Delta V$, $W_{\mathrm{field}}=-q\Delta V$
Signed change: $\Delta K=-q\Delta V$; accelerating gain magnitude: $|q\Delta V|$
$1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$

Speed Formula

$v = \sqrt{2|q\Delta V|/m}$ in the non-relativistic model
$v \propto 1/\sqrt{m}$ (same $|q\Delta V|$)
Valid only if $v \ll c$

Equipotentials

Equal potential, no work moving along them
Perpendicular to electric field lines
Parallel to plates in uniform field

Key Principles

KE magnitude depends on $|\Delta V|$, not on $d$
Same accelerating-energy magnitude $|q\Delta V|$ → same KE, different speeds
Work-energy theorem underpins all of this

Three of these statements about charged particle acceleration are correct. Pick the odd one out.

Independent practice

Show what you can do without prompts

Complete a shuffled question-bank set, then construct a signed solution for an accelerating charge.

Quick recall, work, energy and potential

+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, 7 marks

+5 XP

ApplyBand 4(3 marks) 1. An electron starts from rest at $V_i=0\ \text{V}$ and finishes at $V_f=+800\ \text{V}$. (a) State its kinetic-energy gain in eV and joules. (b) Calculate its final speed. (c) If the plate separation is then halved while the same 800 V potential rise is maintained, state what happens to the final speed and explain why.

1 mark: correct KE in both units · 1 mark: correct speed · 1 mark: identifies speed is unchanged with correct reasoning

AnalyseBand 5(4 marks) 2. A proton and an alpha particle ($q = 2e$, $m = 4 \times m_p$) are each accelerated from rest through the same accelerating-potential magnitude. (a) Compare their kinetic-energy gains. (b) Calculate $v_\alpha / v_p$. (c) Explain the physical significance of the charge-to-mass ratio $q/m$ in particle acceleration.

1 mark: alpha has twice the KE (because $q = 2e$) · 1 mark: correct speed ratio · 1 mark: links $q/m$ to speed · 1 mark: clear physical reasoning

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 (3 marks): (a) The stated 800 V is an accelerating-potential magnitude, so the positive kinetic-energy gain is 800 eV; in joules: $800 \times 1.60 \times 10^{-19} = 1.28 \times 10^{-16}\ \text{J}$ (1 mark). (b) $\tfrac{1}{2}m_e v^2 = |q\Delta V| \Rightarrow v = \sqrt{2(1.60 \times 10^{-19})(800)/(9.11 \times 10^{-31})} = 1.68 \times 10^7\ \text{m/s}$ (1 mark). (c) Speed is unchanged: halving $d$ doubles $E$, but the particle travels half the distance. The work magnitude $|W_{\mathrm{field}}|=|q|Ed=|q|\Delta V$ remains constant, so final kinetic energy and speed are unchanged (1 mark).

Q2 (4 marks): (a) For the stated accelerating potential magnitude, the positive gain is $|q\Delta V|$. The alpha particle has charge magnitude $2e$, so it gains twice the proton's kinetic energy (1 mark). (b) $v=\sqrt{2|q\Delta V|/m}$. Thus $v_\alpha/v_p=\sqrt{(2e/4m_p)/(e/m_p)}=1/\sqrt{2}\approx0.707$ (1 mark). (c) For the same accelerating potential magnitude, final speed depends on $|q|/m$. State the start and end potentials to recover the signed relation $\Delta K=-q\Delta V$ (2 marks).

Final step

Retrieve, reflect and finish

Retrieve the sign conventions, revisit the accelerator prediction and record the model condition you will check next time.

Check what actually stuck

How did your thinking change?

At the start you were asked about CERN's Large Hadron Collider, first proton-proton collisions 23 November 2009, and whether doubling the accelerating voltage doubles the proton's speed.

The answer is no. For an accelerating potential magnitude, $v=\sqrt{2|q\Delta V|/m}$, so speed scales as $\sqrt{|\Delta V|}$. Doubling that magnitude increases speed by $\sqrt{2}\approx1.41$, not by 2. The relation is non-relativistic and must be replaced by relativistic energy-momentum reasoning near light speed.

Plate separation makes no difference to the final kinetic-energy magnitude, only $|\Delta V|$ matters, although separation does change the field magnitude $E=|\Delta V|/d$ in the uniform parallel-plate model.

Extend: An alpha particle ($He^{2+}$, mass $6.64 \times 10^{-27}$ kg, charge $2e$) is accelerated from rest through an accelerating-potential magnitude of 1000 V. Calculate its final kinetic energy in eV and its final speed. How does its speed compare to a proton through the same accelerating-potential magnitude?

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