Year 11 Physics Module 4 ⏱ ~35 min 5 MC · 3 Short Answer Lesson 3 of 17

Electric Potential Energy & Work Done

A lightning bolt (Bureau of Meteorology Sydney, January 2018) transferred 5 C of charge across a 100 MV potential difference. Work done: W = QV = 5 × 10⁸ = 500 MJ, enough to power an average Sydney home for 46 years. However, the bolt lasted only 0.0002 s, delivering this energy at a peak power of 2,500 GW, 2.5× the entire Australian national grid capacity.

Today's hook: The Bureau of Meteorology recorded a lightning bolt over Sydney in January 2018 that transferred 5 C of charge across 100 MV, work done W = QV = 5 × 10⁸ = 500 MJ at a peak power of 2,500 GW, lasting only 0.0002 s. Yet a 100 kV Van de Graaff spark carries only ~0.1 J. Why does voltage alone not determine energy? The answer, W = QV, is why the bolt is lethal and the Van de Graaff is merely startling.

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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.

Before you read, predict

A Van de Graaff generator produces a potential difference of 100 kV. A single proton placed at the dome would experience an enormous force. But the generator's spark can only deliver a brief tingle, far less than a mains electric shock at 240 V.

Predict 1: If voltage is "push," why doesn't a 100 kV Van de Graaff deliver more energy than a 240 V power socket?

Predict 2: A proton and an electron are each accelerated from rest through 100 kV. Which gains more kinetic energy?

Warm-up, the unit of electric potential difference (voltage) is:

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Learning Intentions

goals

Know

Potential difference: $V = W/q$ (volts = joules per coulomb)
Electric potential energy: $U = qV$
Work in a uniform field: $W = qEd$
Equipotential lines are perpendicular to field lines

Understand

The difference between electric potential ($V$) and electric potential energy ($U$)
Why moving a charge along an equipotential requires no work
How a positive charge gains potential energy when moved against the field

Can Do

Solve problems involving work, charge, and potential difference
Calculate work done using $W = qEd$ for uniform fields
Sketch equipotential lines for point charges and parallel plates

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Scan these before reading

vocab

Potential difference ($V$)Work done per unit charge moving between two points: $V = W/q$. Unit: volt (V = J C⁻¹).

Electric potential energy ($U$)Energy stored by a charge in a field: $U = qV$. Property of the charge-field system. Unit: joule (J).

Equipotential lineA line connecting all points at the same electric potential. Always perpendicular to field lines. No work done moving along an equipotential.

Work done ($W$)$W = qV$ (universal) or $W = qEd$ (uniform field, displacement parallel to field).

Cross-lesson links: L02 showed electric fields as force maps. L03 introduces electric potential energy (U = qV) and the work done by electric fields. The work-energy theorem connects L03 back to M2 dynamics: W = ΔKE = ½mv². This connection is tested when charged particles are accelerated through a potential difference, in CRT tubes, particle accelerators, and electron guns.

Misconceptions to fix

✗ Wrong: High voltage always means high energy.

✓ Right: Energy depends on both voltage and charge: $W = qV$. A 100 kV Van de Graaff with microcoulombs stores less energy than a 12 V car battery with ampere-hours of charge.

✗ Wrong: Electric potential and electric potential energy are the same thing.

✓ Right: $V$ (potential) is a property of the field at a point, in volts. $U$ (potential energy) is a property of the charge-field system, in joules. $U = qV$ links them but they are not interchangeable.

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Electric Potential Difference

Core Content

Electric Potential Difference

+5 XP

Lift a book from the floor to a shelf, you do work against gravity and the book stores gravitational potential energy. Now carry a positive charge from a negative plate toward a positive plate inside a capacitor, the charge resists, just like the book resisting you, and again you have to do work. When you let go, the charge accelerates back, converting stored energy to kinetic energy, exactly as the book would fall.

Potential difference (voltage)

$$V = \frac{W}{q}$$

$1\ \text{V} = 1\ \text{J C}^{-1}$, a 12 V battery does 12 J of work per coulomb of charge.

Potential vs Potential Energy

Electric potential ($V$): Property of the field. Measured in volts. Analogy: the height of a shelf.
Electric potential energy ($U$): Property of the charge-field system. $U = qV$. Analogy: the gravitational PE of a specific mass on that shelf.

Real-World Anchor

A Van de Graaff at 100 kV stores only a few microcoulombs of charge. Energy in a spark: $W = qV \approx (10^{-6})(10^5) = 0.1\ \text{J}$, same as dropping a pen. A 240 V mains socket delivers many coulombs per second continuously. Voltage alone does not determine danger; the available charge does.

Electric potential difference $V = W/q$ (1 V = 1 J C⁻¹) measures work done per unit charge between two points. Electric potential energy $U = qV$ is a property of the charge-field system, not the field alone.

Pause, copy the highlighted definition into your book before moving on.

High voltage always means high energy stored.

Electric potential ($V$) and electric potential energy ($U$) are the same quantity.

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Work Done in Electric Fields

+5 XP

We just saw that potential difference $V$ measures energy per unit charge. That raises a question: how much work does the field do on a moving charge, and when does it gain or lose energy? This card answers it → $W = qV$ quantifies the work.

Rearranging the definition of potential difference gives the fundamental equation for work done on a charge.

Work done (universal)

$W = qV$ (valid for any electric field)

Work done in uniform field only

$W = qEd$ (displacement $d$ parallel to field; substitute $V = Ed$)

Gaining and Losing Potential Energy

A positive charge moved with the field: field does positive work on it → charge loses PE, gains KE.
A positive charge moved against the field: external work required → charge gains PE, loses KE.
A negative charge behaves in the opposite way.

Worked example, Proton in a particle accelerator

A proton accelerated from rest through $V = 5.0 \times 10^4\ \text{V}$. Find (a) work done, (b) final speed.

$W = qV = (1.60 \times 10^{-19})(5.0 \times 10^4) = 8.0 \times 10^{-15}\ \text{J}$
$W = \tfrac{1}{2}m_p v^2 \Rightarrow v = \sqrt{2W/m_p} = \sqrt{2(8.0 \times 10^{-15})/(1.67 \times 10^{-27})} = 3.1 \times 10^6\ \text{m s}^{-1}$ (~1% of $c$)

Work done by an electric field: $W = qV$ (universal) or $W = qEd$ (uniform field, displacement parallel to field). A positive charge moving with the field loses PE and gains KE; moving against the field gains PE. Energy conservation: $W = \Delta KE = \tfrac{1}{2}mv^2$ (starting from rest).

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

An electron is accelerated from rest through 200 V. The kinetic energy gained is:

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Equipotential Lines and Surfaces

+5 XP

We just saw that $W = qV$ gives the work done when a charge moves through a potential difference. That raises a question: what are the surfaces on which no work is done at all? This card answers it → equipotential lines connect all points at the same potential.

An equipotential line connects points that all have the same electric potential. No work is required to move a charge along an equipotential, because $W = q\Delta V$ and $\Delta V = 0$.

Key Properties

Perpendicular to field lines: Always cross field lines at right angles.
Closer spacing = stronger field: $E = -\Delta V / \Delta d$, rapid change in potential means strong field.
Conductors are equipotentials: In electrostatic equilibrium, the surface of any conductor is an equipotential.

A uniform electric field crosses vertical equipotential lines at right angles. A positive charge constrained to move along the 200 volt line has zero potential change and zero field work even though its electric force points across the line; motion with the field to 100 volts lowers potential energy. — Zero work along an equipotential does not mean zero force. For a positive charge, $\vec F=q\vec E$ still points across the equipotential; a constraint is needed to make the charge follow the equal-potential path.

Configuration	Equipotential Pattern
Single point charge	Concentric circles (2D) or spheres (3D) centred on charge
Parallel plates	Parallel straight lines perpendicular to field lines
Dipole	Oval curves enclosing each charge, perpendicular bisector at zero potential

An equipotential line connects points of equal electric potential; no work is done moving a charge along it ($W = q\Delta V = 0$). Equipotentials are always perpendicular to field lines; closer spacing means a stronger field. A conductor surface is an equipotential in electrostatic equilibrium.

Pause, write the highlighted definition into your book.

Three of these are correct about equipotential lines. Pick the odd one out.

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Activity 1, Calculate and Interpret: Work and Energy

ApplyBand 4

Parallel plates connected to 500 V, separated 2.5 cm. An alpha particle ($q = +3.2 \times 10^{-19}\ \text{C}$, $m = 6.6 \times 10^{-27}\ \text{kg}$) is released from rest at the positive plate.

Calculate the work done on the alpha particle as it moves to the negative plate.
Calculate $E = V/d$ and verify that $W = qEd$ gives the same result as $W = qV$.
Calculate the alpha particle's speed just before it reaches the negative plate.

An electron accelerated from rest through 500 V gains a kinetic energy of _____ J (give your answer in scientific notation to 2 sig. figs.).

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Activity 2, Calculate and Interpret: Equipotential Analysis

AnalyseBand 5

Equipotential lines around two point charges are labelled: +20 V, +10 V, 0 V, −10 V, −20 V.

Identify the sign of each charge by inspecting the equipotential values.
A +1.0 nC charge is moved from the +10 V equipotential to the 0 V equipotential. Calculate the work done by the field.
How much work would be required to move the same charge from 0 V back to +10 V? Explain.

A positive charge is moved from a region of low potential to high potential. The work done by the electric field is:

Wrap-up, Key Definitions

Potential difference

$V = W/q$. One volt = one joule per coulomb. Voltage is work per unit charge. Not the same as energy.

Work done on a charge

$W = qV$ (universal). $W = qEd$ (uniform fields only). Positive charges gain PE when moved against the field.

$V$ vs $U$

$V$ = field property (V). $U = qV$ = system property (J). Analogy: height vs gravitational PE of a specific mass at that height.

Equipotentials

Lines of constant $V$. Always ⊥ to field lines. No work to move along them. Closer spacing = stronger field.

✓

Show what you have learned

Quick recall, potential energy & work done

+5 XP

Short Answer, 10 marks

+5 XP

UnderstandBand 3(3 marks) 3. Distinguish between electric potential and electric potential energy, using an analogy to help explain your answer.

ApplyBand 4(3 marks) 4. Two parallel plates separated by 5.0 cm are connected to a 1.2 kV supply. Calculate: (a) electric field strength; (b) work done by the field when an electron moves from negative to positive plate; (c) change in electric potential energy of the electron.

EvaluateBand 6(4 marks) 5. A student claims a 100 kV Van de Graaff generator is more dangerous than a 12 V car battery. Evaluate this claim using potential difference, charge, and energy, and explain why the car battery can deliver far more total energy despite its lower voltage.

Show all answers

Activity 1, Model Answers

$W = qV = (3.2 \times 10^{-19})(500) = 1.6 \times 10^{-16}\ \text{J}$
$E = V/d = 500/(2.5 \times 10^{-2}) = 2.0 \times 10^4\ \text{V m}^{-1}$. Check: $W = qEd = (3.2 \times 10^{-19})(2.0 \times 10^4)(2.5 \times 10^{-2}) = 1.6 \times 10^{-16}\ \text{J}$ ✓
$v = \sqrt{2W/m} = \sqrt{2(1.6 \times 10^{-16})/(6.6 \times 10^{-27})} = 2.2 \times 10^5\ \text{m s}^{-1}$

Activity 2, Model Answers

Left charge is positive (equipotentials increase toward it: +20 V, +10 V). Right charge is negative (equipotentials decrease: −10 V, −20 V).
$W = q\Delta V = (1.0 \times 10^{-9})(0 - 10) = -1.0 \times 10^{-8}\ \text{J}$ (field does negative work, charge moves to lower potential).
$+1.0 \times 10^{-8}\ \text{J}$ of external work required. Moving against the field increases PE. Magnitude is equal, sign opposite.

Short Answer, Model Answers

Q3 (3 marks): Electric potential ($V$) is a property of the field at a point, in volts, the "electrical push" per unit charge. Electric potential energy ($U = qV$) is a property of the charge-field system, in joules. Analogy: $V$ is like the height of a diving board; $U$ is the gravitational PE of a specific diver on that board. Changing the diver's mass changes $U$ but not the height.

Q4 (3 marks): (a) $E = V/d = 1200/(5.0 \times 10^{-2}) = 2.4 \times 10^4\ \text{V m}^{-1}$. (b) $W = qV = (-1.60 \times 10^{-19})(1200) = -1.92 \times 10^{-16}\ \text{J}$ (magnitude $1.92 \times 10^{-16}\ \text{J}$). (c) $\Delta U = -W = +1.92 \times 10^{-16}\ \text{J}$ if field does negative work; electron gains PE and loses KE moving from − to + plate.

Q5 (4 marks): The claim confuses voltage with energy. Energy = $W = qV$. The Van de Graaff at 100 kV stores only ~microcoulombs: $W \approx (10^{-6})(10^5) = 0.1\ \text{J}$. A 12 V car battery stores ~50 Ah = $1.8 \times 10^5\ \text{C}$: $W = (1.8 \times 10^5)(12) = 2.16 \times 10^6\ \text{J}$, over 20 million times more energy. Danger depends on voltage, charge capacity, and body resistance, not voltage alone.

✓

Retrieve, reflect and finish

Check what actually stuck

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quiz

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How did your thinking change?

The Bureau of Meteorology Sydney lightning bolt from January 2018 in the hero delivered 500 MJ, W = QV = 5 C × 100 MV, at a peak power of 2,500 GW. That enormous energy came from both the high voltage AND the 5 C of charge transferred. Now look back at your Think First predictions about the Van de Graaff vs mains: the Van de Graaff only stores a few microcoulombs, so despite its 100 kV, W = qV ≈ 0.1 J, far less than the mains socket.

Both a proton and electron accelerated through the same voltage gain the same kinetic energy ($W = qV$, same $|q|$, same $V$), the same W = QV relationship that made the January 2018 lightning bolt so energetic.

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