Physics • Year 12 • Module 6 • Lesson 3
Work, Energy and Potential
Lock in the core vocabulary, energy-conversion relationships, and the electronvolt unit before tackling harder questions.
1. Term–definition match
The definitions below are shuffled. In the right-hand column write the matching term from this list: work done by electric field, electric potential energy, kinetic energy gain, electronvolt (eV), equipotential, potential difference, charge-to-mass ratio, non-relativistic limit, conservation of energy, work-energy theorem. 10 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | The energy transferred to a charged particle when the electric field exerts a force over a displacement; equals $W = q\Delta V$ for any path between two potentials. | |
| 1.2 | Energy stored due to a charge’s position in an electric field; changes as the charge moves between potentials. | |
| 1.3 | The increase in kinetic energy of a particle starting from rest after being accelerated through a potential difference; $\Delta K = q\Delta V$. | |
| 1.4 | A unit of energy equal to $1.60 \times 10^{-19}$ J; the energy gained by one elementary charge accelerated through 1 V. | |
| 1.5 | A surface on which every point is at the same electric potential; no work is done moving a charge along it. | |
| 1.6 | The difference in electric potential between two points; measured in volts (V). | |
| 1.7 | The ratio $q/m$ that determines the final speed of a particle accelerated through a given potential difference; appears in $v = \sqrt{2(q/m)\Delta V}$. | |
| 1.8 | The condition under which $v = \sqrt{2q\Delta V/m}$ is valid; requires $v \ll c$ (approximately $v < 0.1c$). | |
| 1.9 | The principle that the total energy of an isolated system is constant; here, electric potential energy converts entirely to kinetic energy when no friction acts. | |
| 1.10 | The statement that the net work done on an object equals its change in kinetic energy: $W_{\text{net}} = \Delta K$. |
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 The work done on a charge moving between two plates depends only on the potential difference and the charge, not on the plate separation. T / F
2.2 If the plate separation is doubled while the battery voltage stays constant, the electron reaches the positive plate with twice the kinetic energy. T / F
2.3 An electron accelerated from rest through 500 V gains 500 eV of kinetic energy. T / F
2.4 A proton and an electron accelerated through the same potential difference from rest arrive with different kinetic energies because they have different masses. T / F
2.5 Moving a charge along an equipotential surface does zero work against the electric field. T / F
2.6 If voltage doubles (plate separation unchanged), the final speed of an electron accelerated from rest also doubles. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word or phrase is used once. 8 marks (1 per blank)
Word bank:
charge-to-mass ratio · electronvolt · equipotential · kinetic · path · potential difference · relativistic · work-energy theorem
When a charge accelerates between parallel plates, the electric field does work equal to $W = q\Delta V$, where $\Delta V$ is the ___________ between the plates. The crucial insight is that this work is independent of the ___________ taken between the two potentials. By the ___________, all of this work converts into ___________ energy: $\Delta K = q\Delta V$. The final speed therefore depends on the particle’s ___________, so a proton and an electron gain identical kinetic energy but reach very different speeds through the same voltage. In atomic physics, energy is commonly expressed in the ___________, where $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$. Surfaces of constant potential are called ___________ surfaces; moving along them requires no work. The formula $v = \sqrt{2q\Delta V/m}$ breaks down above approximately 2.5 kV for electrons, where ___________ corrections are needed.
4. Function recall
Answer each question in 1–2 sentences using precise terms from the lesson. 8 marks (2 each)
4.1 Explain why the final kinetic energy of a charge accelerated from rest between parallel plates does not depend on the distance between the plates.
4.2 State the formula for the final speed of a particle of charge $q$ and mass $m$ accelerated from rest through potential difference $\Delta V$. Identify what property determines whether the proton or the electron moves faster.
4.3 Define the electronvolt and explain why it is a convenient unit in atomic and particle physics.
4.4 Describe what an equipotential surface is and state the work done when a charge moves along one.
5. Formula review — show the steps
For each scenario, identify the formula, substitute values, and state the answer with correct units. Show all working. 12 marks (3 each)
5.1 An electron ($q = 1.60 \times 10^{-19}\ \text{C}$, $m = 9.11 \times 10^{-31}\ \text{kg}$) is accelerated from rest through 200 V. Calculate the kinetic energy gained in (i) joules and (ii) electronvolts.
5.2 Using the kinetic energy from 5.1, calculate the final speed of the electron. ($m_e = 9.11 \times 10^{-31}\ \text{kg}$)
5.3 A proton ($q = 1.60 \times 10^{-19}\ \text{C}$, $m = 1.67 \times 10^{-27}\ \text{kg}$) is accelerated from rest through the same 200 V. Calculate its final speed and compare it to the electron’s speed from 5.2.
5.4 The voltage in question 5.1 is now doubled to 400 V (same electron, from rest). Without fully recalculating, state how the kinetic energy and the final speed each change. Justify using the relevant formulas.
Q1 — Term–definition match
1.1 work done by electric field • 1.2 electric potential energy • 1.3 kinetic energy gain • 1.4 electronvolt (eV) • 1.5 equipotential • 1.6 potential difference • 1.7 charge-to-mass ratio • 1.8 non-relativistic limit • 1.9 conservation of energy • 1.10 work-energy theorem.
Q2 — True / false with correction
2.1 True. $W = q\Delta V$ contains only charge and potential difference — plate separation does not appear.
2.2 False. Doubling plate separation at constant voltage leaves the kinetic energy unchanged. The electric field halves but the particle travels twice as far; $W = qEd = q(V/d)(2d) = qV$ is the same. The electron arrives with the same kinetic energy (and same speed).
2.3 True. An electron through 500 V gains $\Delta K = eV = 500\ \text{eV}$. No unit conversion needed in electronvolts.
2.4 False. Both the proton and electron gain identical kinetic energy equal to $q\Delta V$ (same charge magnitude, same $\Delta V$). They have different masses, which means they arrive at different speeds, not different kinetic energies.
2.5 True. On an equipotential surface, $\Delta V = 0$, so $W = q\Delta V = 0$.
2.6 False. Doubling the voltage doubles the kinetic energy ($\Delta K = q\Delta V$), but since $\frac{1}{2}mv^2 = q\Delta V$, the speed scales as $v \propto \sqrt{\Delta V}$. Doubling $\Delta V$ multiplies speed by $\sqrt{2} \approx 1.41$, not 2.
Q3 — Cloze paragraph
In order: potential difference / path / work-energy theorem / kinetic / charge-to-mass ratio / electronvolt / equipotential / relativistic.
Q4.1 — Independence from plate separation
The work done by the electric field equals $W = q\Delta V$. When plate separation increases, the electric field $E = V/d$ decreases, but the particle travels a longer distance; the product $W = qEd = q(V/d)d = qV$ is unchanged. Therefore the kinetic energy gain depends only on the charge and the potential difference, not on the separation.
Q4.2 — Final speed formula
$v = \sqrt{2q\Delta V / m}$. The final speed is determined by the charge-to-mass ratio $q/m$: a higher $q/m$ gives a higher speed through the same $\Delta V$. The electron has a charge-to-mass ratio about 1836 times larger than the proton, so it reaches a much higher speed despite gaining the same kinetic energy.
Q4.3 — Electronvolt
One electronvolt is the kinetic energy gained by a particle carrying one elementary charge ($1.60 \times 10^{-19}\ \text{C}$) when accelerated through a potential difference of 1 V; $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$. It is convenient because atomic-scale energies are on the order of eV to MeV, avoiding the need to write very small powers of ten in joules.
Q4.4 — Equipotential surface
An equipotential surface is a surface on which every point has the same electric potential. Because $\Delta V = 0$ between any two points on the surface, the work done by the electric field on a charge moving along it is $W = q\Delta V = 0$.
Q5.1 — Kinetic energy at 200 V
(i) $\Delta K = q\Delta V = (1.60 \times 10^{-19})(200) = \mathbf{3.20 \times 10^{-17}\ \text{J}}$
(ii) Numerically, $\Delta K = 200\ \text{eV}$ (one electron through 200 V = 200 eV).
Q5.2 — Electron speed at 200 V
$\frac{1}{2}m_e v^2 = 3.20 \times 10^{-17}\ \text{J}$
$v = \sqrt{\frac{2 \times 3.20 \times 10^{-17}}{9.11 \times 10^{-31}}} = \sqrt{7.03 \times 10^{13}} = \mathbf{8.38 \times 10^6\ \text{m/s}}$
Check: $v/c \approx 0.028$ — safely non-relativistic.
Q5.3 — Proton speed at 200 V
$v_p = \sqrt{\frac{2q\Delta V}{m_p}} = \sqrt{\frac{2(1.60 \times 10^{-19})(200)}{1.67 \times 10^{-27}}} = \sqrt{3.83 \times 10^{10}} = \mathbf{1.96 \times 10^5\ \text{m/s}}$
The proton’s speed is about 43 times slower than the electron’s ($v \propto 1/\sqrt{m}$ for same $q\Delta V$; $\sqrt{m_p/m_e} \approx \sqrt{1836} \approx 42.8$). Both particles gain the same kinetic energy (200 eV).
Q5.4 — Effect of doubling voltage
Kinetic energy: $\Delta K = q\Delta V$ scales linearly with $\Delta V$, so doubling $V$ from 200 V to 400 V doubles the kinetic energy from $3.20 \times 10^{-17}\ \text{J}$ to $6.40 \times 10^{-17}\ \text{J}$. Speed: $v = \sqrt{2q\Delta V/m} \propto \sqrt{\Delta V}$, so doubling $\Delta V$ multiplies speed by $\sqrt{2} \approx 1.41$. The new speed is $8.38 \times 10^6 \times \sqrt{2} \approx 1.18 \times 10^7\ \text{m/s}$.