Work, Energy and Potential
When CERN's Large Hadron Collider accelerated its first proton beams in 2009, each proton in the 27 km ring had been boosted through a total potential difference equivalent to 6.5 × 10¹² eV (6.5 TeV). The 1,232 superconducting dipole magnets that bent the beam each cost €1 million — but the energy input itself came entirely from the simple principle that a charge accelerated through a potential difference gains kinetic energy equal to qΔV.
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 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:
Know — Energy Relationships
- Work done by electric field: $W = qEd = q\Delta V$
- Change in electric potential energy: $\Delta U = q\Delta V$
- Kinetic energy gain: $\Delta K = q\Delta V$ (from rest)
Understand — Independence from Path
- Why KE gain depends only on $\Delta V$, not on plate separation
- Why $E = V/d$ and $W = qEd$ give the same result for parallel plates
- The difference between eV and joules as energy units
Can Do — Calculate and Compare
- Calculate final speed of a charge accelerated through a known potential difference
- Compare KE and speed for different particles through the same $\Delta V$
- Identify when non-relativistic approximations break down
Core Content
How potential difference converts to kinetic energy
Picture an electron sitting still at the negative plate of a parallel-plate capacitor. When the switch is closed, it instantly accelerates across the gap and slams into the positive plate — arriving much faster than it started. The electric field between the plates did work on the electron, converting electric potential energy into kinetic energy. For any uniform field between parallel plates, the work done is:
$W = Fd = qEd$
Since $E = V/d$, this becomes $W = qV$
The crucial insight is that the work done depends only on the potential difference $V$ and the charge $q$, not on the path taken or the plate separation. If you double the separation while keeping the same battery, $E$ is halved but $d$ is doubled — the product $Ed = V$ stays constant.
By the work-energy theorem, this work becomes kinetic energy:
$\Delta K = q\Delta V$
If the particle starts from rest: $\tfrac{1}{2}mv^2 = q\Delta V$
$v = \sqrt{\dfrac{2q\Delta V}{m}}$ — final speed depends on $q/m$ (charge-to-mass ratio)
Figure 1 — Energy conversion as an electron accelerates from negative to positive plate. Total energy (PE + KE) is conserved throughout.
An electron accelerates from rest through 100 V, gaining kinetic energy $K$. A second electron accelerates from rest through 200 V. Does the second electron have twice the kinetic energy, twice the speed, or both? Explain.
An electron is accelerated from rest through a potential difference of 5000 V. Calculate (a) its final kinetic energy in joules and eV, (b) its final speed.
- Part (a). In eV: $\Delta K = e \times 5000\ \text{V} = \mathbf{5000\ \text{eV}}$. In joules: $\Delta K = (1.60 \times 10^{-19})(5000) = \mathbf{8.0 \times 10^{-16}\ \text{J}}$.
- Part (b). $\tfrac{1}{2}m_e v^2 = eV$ → $v = \sqrt{2eV/m_e} = \sqrt{2(1.60 \times 10^{-19})(5000)/(9.11 \times 10^{-31})} = \mathbf{4.19 \times 10^7}\ \text{m/s}$.
- Relativistic check: $v/c \approx 0.14$ — borderline relativistic. For HSC, the non-relativistic answer is accepted.
Work done by electric field: $W = q\Delta V$. KE gain from rest: $\Delta K = q\Delta V$; final speed $v = \sqrt{2q\Delta V/m}$. KE depends only on charge and potential difference — NOT on plate separation or path taken.
Pause — copy the highlighted relationships and formula into your book before moving on.
An electron accelerates from rest through a potential difference. If the voltage is doubled, the final speed of the electron will be multiplied by:
Why the volt is both a potential unit and an energy unit
We just saw that a charge gains kinetic energy $\Delta K = q\Delta V$ regardless of path or separation. 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 a unit of energy defined as the work done when one elementary charge is accelerated through a potential difference of one volt:
$1\ \text{eV} = e \times 1\ \text{V} = 1.60 \times 10^{-19}\ \text{J}$
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.
When a question asks for energy "in electronvolts," you do not need to multiply by $e$ and convert to joules. An electron through 250 V simply gains 250 eV of kinetic energy.
The equation $v = \sqrt{2qV/m}$ is only valid when $v \ll c$ (roughly $v < 0.1c$). For electrons, this breaks down above about 2.5 kV. In HSC Physics, you will generally stay in the non-relativistic regime unless told otherwise.
An electron and a proton are each accelerated from rest through 1000 V. Which has more kinetic energy? Which is moving faster? By what factor?
A proton is accelerated from rest through 5000 V alongside an electron. Calculate the proton's speed and compare to the electron's speed from earlier.
- $v_p = \sqrt{2eV/m_p} = \sqrt{2(1.60 \times 10^{-19})(5000)/(1.67 \times 10^{-27})} = \mathbf{9.79 \times 10^5}\ \text{m/s}$.
- Comparison: Both particles gain the same kinetic energy (5000 eV). But because $m_p/m_e = 1836$, the proton is about $\sqrt{1836} \approx 43$ times slower than the electron.
- Key insight: $v \propto 1/\sqrt{m}$ for fixed $qV$.
Equipotential surface: constant electric potential — no work done moving along it. $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$; an electron through $V$ volts gains $V$ eV. Same $q\Delta V$ → same KE, 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 potential difference gain the same kinetic energy.
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.
Apply the work-energy theorem to charged particles
- An electron is accelerated from rest through 400 V. Calculate its final speed.
- A proton is accelerated from rest through 400 V. 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 voltage stays at 400 V. Does the proton's final speed change? Explain using $W = qEd$.
An alpha particle ($q = 2e$, $m = 4 \times 1.67 \times 10^{-27}$ kg) is accelerated from rest through 1000 V. Its kinetic energy is _____ eV.
Explain the operation of an electron gun using energy principles
In an electron gun, electrons are accelerated from rest through a potential difference of 2500 V between two electrodes 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.
Misconceptions — final check
Copy into your books
Key Definitions
- Work: $W = qEd = q\Delta V$
- KE gain: $\Delta K = q\Delta V$ (from rest)
- $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$
Speed Formula
- $v = \sqrt{2q\Delta V/m}$
- $v \propto 1/\sqrt{m}$ (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 depends on $\Delta V$, not on $d$
- Same $q\Delta V$ → 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.
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(3 marks) 1. An electron is accelerated from rest through a potential difference of 800 V. (a) State the kinetic energy in eV and in joules. (b) Calculate the final speed of the electron. (c) If the plate separation is then halved but the voltage stays at 800 V, 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 potential difference $\Delta V$. (a) Compare their kinetic energies. (b) Calculate the ratio of their final speeds $v_\alpha / v_p$. (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) KE = 800 eV; in joules: $800 \times 1.60 \times 10^{-19} = 1.28 \times 10^{-16}\ \text{J}$ (1 mark). (b) $\tfrac{1}{2}m_e v^2 = eV \Rightarrow v = \sqrt{2(1.60 \times 10^{-19})(800)/(9.11 \times 10^{-31})} = 1.68 \times 10^7\ \text{m/s}$ (1 mark). (c) Speed is unchanged: halving $d$ doubles $E$, but the particle travels half the distance. Work $W = qEd = q(V/d)d = qV$ remains constant, so final KE and speed are unchanged (1 mark).
Q2 (4 marks): (a) $\Delta K = q\Delta V$. For the proton, $q = e$; for the alpha, $q = 2e$. So alpha gains twice the kinetic energy (1 mark). (b) $v = \sqrt{2q\Delta V/m}$. For proton: $v_p = \sqrt{2e\Delta V/m_p}$. For alpha: $v_\alpha = \sqrt{2(2e)\Delta V/(4m_p)} = \sqrt{e\Delta V/m_p}$. Ratio: $v_\alpha/v_p = \sqrt{e\Delta V/m_p} / \sqrt{2e\Delta V/m_p} = 1/\sqrt{2} \approx 0.707$ (1 mark). (c) Final speed depends on $q/m$. The alpha has $q/m = 2e/4m_p = e/(2m_p)$, while the proton has $q/m = e/m_p$. A higher $q/m$ means greater acceleration for the same field — electrons have enormous $q/m$ compared to nuclei, making them easy to accelerate to high speeds (1 mark + 1 mark for full reasoning).
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. Because $v = \sqrt{2q\Delta V/m}$, speed scales as $\sqrt{\Delta V}$. Doubling the voltage increases speed by $\sqrt{2} \approx 1.41$, not by 2. You need to quadruple the voltage to double the speed. The LHC gets around this by running protons through thousands of acceleration stages, each adding energy — the total corresponds to 6.5 TeV per beam.
Plate separation makes no difference to the final kinetic energy — only the potential difference $\Delta V$ matters (though separation does change the field strength $E = V/d$).
Extend: An alpha particle ($He^{2+}$, mass $6.64 \times 10^{-27}$ kg, charge $2e$) is accelerated from rest through 1000 V. Calculate its final kinetic energy in eV and its final speed. How does its speed compare to a proton through the same voltage?
Five timed questions on work, energy and potential. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
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