What is a "volt", really? It is not just a number on a battery, it is a measure of energy per charge. A 9-volt battery gives 9 joules of energy to every coulomb of charge that passes through it. That single idea, captured by $W = qV$, connects the abstract field equations to the everyday language of circuits you are about to study.
A battery is labelled "9 V". You have seen volts on batteries, chargers and power points your whole life.
Before reading: in your own words, what do you think the "9 volts" is actually measuring? Write your best guess.
A volt tells you how many joules each coulomb carries
Voltage is simply how much energy is given to each unit of charge, a bigger voltage means more energy per coulomb.
The voltage (or potential difference) between two points is defined as the work done per unit charge in moving a charge between them:
$$V = \frac{W}{q} \qquad\Longleftrightarrow\qquad W = qV$$
This means:
Voltage is the work done per unit charge between two points: $V = \dfrac{W}{q}$, so $W = qV$. One volt is one joule per coulomb. Voltage is always a potential difference between two positions.
Pause, copy the highlighted definition into your book before moving on.
Did you get this? True or false: one volt is defined as one joule of energy per coulomb of charge.
Energy from charge and voltage
We just saw that $W = qV$. That raises a question: how do we use it to find the energy gained by a moving charge? This card answers it → substitute the charge and the voltage directly into $W = qV$.
Once you know how much charge moves and the voltage it moves through, the energy is a single multiplication.
Use charge in coulombs and voltage in volts; the work comes out in joules. The energy transferred equals the change in the charge's electric potential energy.
An electron ($q = 1.6 \times 10^{-19}\ \text{C}$) is accelerated through a potential difference of $250\ \text{V}$. Find the work done on the electron (the kinetic energy it gains).
The work done moving a charge through a potential difference is $W = qV$ (in joules). For a charge accelerated from rest, this work equals the kinetic energy it gains.
Add the highlighted equation and worked method to your notes before the check below.
Quick check: A charge of $2.0\ \text{C}$ moves through a potential difference of $6.0\ \text{V}$. How much work is done?
Pattern A, Calculate then connect
A charge of $+5.0 \times 10^{-6}\ \text{C}$ moves through a potential difference of $80\ \text{V}$. In your book: (1) calculate the work done with $W = qV$; (2) explain in one sentence why this work equals the change in the charge's electric potential energy; (3) state what the answer would be if the charge were doubled, and why.
Odd one out: three of these correctly express the WORK done moving a charge through a potential difference. Which one does not?
Dot point: Solve problems involving the work done when moving a charge in an electric field using the relationships between W, q and V
Voltage and the field are two views of the same energy
We just learned to find work from voltage. That raises a question: how does this fit with last lesson's $W = qEd$? This card answers it → they are the same energy, linked by $V = Ed$.
The two work equations are not rivals, they describe the identical energy, just written in terms of voltage or in terms of the field.
Last lesson the work done moving a charge through a uniform field was $W = qEd$. This lesson it is $W = qV$. Setting them equal for the same movement gives a neat link between voltage and field:
$$qV = qEd \quad\Longrightarrow\quad V = Ed \quad\Longleftrightarrow\quad E = \frac{V}{d}$$
This is exactly the parallel-plate relationship $E = \dfrac{V}{d}$ you met in L07, proof that the voltage view and the field view are perfectly consistent. Use $W = qV$ when you are given a voltage; use $W = qEd$ when you are given a field and a distance. Both give the same potential-energy change.
$W = qV$ and $W = qEd$ describe the same energy, linked by $V = Ed$ (so $E = \dfrac{V}{d}$). Use $W = qV$ when given a voltage and $W = qEd$ when given a field and distance.
Pause, copy the highlighted link into your book before the check below.
Last check: Decide if the statement is true or false.
For a charge moving through a uniform field, $W = qV$ and $W = qEd$ give the same work because $V = Ed$.
Match the pairs: Match each quantity to what it measures and its unit.
Dot point: Analyse the relationships between voltage, potential energy difference and work when changing the position of a charge in a uniform electric field using W = qV
A fresh set drawn from this lesson's question 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 5(4 marks) 1. An electron ($q = 1.6 \times 10^{-19}\ \text{C}$) is accelerated from rest through a potential difference of $500\ \text{V}$. Calculate the work done on the electron, and explain how the work done relates to the kinetic energy it gains and to its change in electric potential energy.
1 mark: uses $W = qV$ · 1 mark: $W = 8.0 \times 10^{-17}\ \text{J}$ · 1 mark: this work equals the kinetic energy gained · 1 mark: equals the loss in electric potential energy (conservation of energy)
ApplyBand 4(3 marks) 2. Moving a charge of $5.0\ \text{mC}$ between two points in an electric field requires $0.30\ \text{J}$ of work. Calculate the potential difference between the two points, then determine the work needed to move a charge of $2.0\ \text{mC}$ between the same two points.
1 mark: $V = W/q = 60\ \text{V}$ · 1 mark: uses the same $V$ for the second charge · 1 mark: $W = qV = 0.12\ \text{J}$
Dot point: Solve problems involving the work done when moving a charge in an electric field using the relationships between W, q and V
AnalyseBand 5(3 marks) 3. Show that $W = qV$ and $W = qEd$ are consistent with one another for a uniform field, and state what this tells you about the relationship between $V$, $E$ and $d$.
1 mark: states $E = V/d$ for a uniform field · 1 mark: substitutes to show $qEd = q(V/d)d = qV$ · 1 mark: concludes $V = Ed$, so voltage is field strength multiplied by separation
Dot point: Analyse the relationships between voltage, potential energy difference and work when changing the position of a charge in a uniform electric field using W = qV
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): Using $W = qV = (1.6 \times 10^{-19})(500) = 8.0 \times 10^{-17}\ \text{J}$. Because the electron starts from rest and the only force doing work is the electric force, by conservation of energy this work equals the kinetic energy the electron gains. It also equals the magnitude of the electron's loss in electric potential energy as it moves through the potential difference, the electric potential energy it gives up is converted entirely into kinetic energy.
Q2 (3 marks): Potential difference is the work done per unit charge, so $V = W/q = 0.30/(5.0 \times 10^{-3}) = 60\ \text{V}$. The potential difference is a property of the two points and of the field, not of the charge that happens to be moved between them, so it is still $60\ \text{V}$ for any other charge. For the smaller charge, $W = qV = (2.0 \times 10^{-3})(60) = 0.12\ \text{J}$. The work scales in direct proportion to the charge moved, so moving two fifths of the charge requires two fifths of the work.
Q3 (3 marks): For a uniform field the field strength is $E = V/d$, where $V$ is the potential difference across a separation $d$. Substituting this into the second expression gives $W = qEd = q(V/d)d$. The two factors of $d$ cancel, leaving $W = qV$, which is exactly the first expression. The two equations are therefore the same statement written in different variables, and they must agree. Rearranging $E = V/d$ gives $V = Ed$, which tells us that the potential difference between two points in a uniform field is the field strength multiplied by the separation measured along the field. A stronger field, or a greater separation, means a larger potential difference between the two points.
You can now define voltage as energy per charge, calculate work with $W = qV$, and link it to the field through $V = Ed$. This wraps up electrostatics, next lesson begins electric circuits with the idea of current, $I = \dfrac{q}{t}$.
↩ Back to focus-area hubAt the start you guessed what "9 volts" measures. It measures energy per charge, 9 joules for every coulomb that passes through. Voltage is not energy by itself; the energy delivered is $W = qV$, which also equals the change in the charge's electric potential energy.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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