Capacitance: Definition Revision Notes | A Level AQA Physics
Capacitance is the charge stored per unit potential difference across a capacitor: C = Q/V. Free AQA notes with 3 worked examples and 9 questions answered.
Updated 12 Sept 2026 · Revise Right Team
Capacitance: Definition is part of Capacitance (A-level only) in A Level AQA Physics.
5 min
Reading time
31
Lessons in this topic
A Level AQA
Pathway
Reviewed by Revise Right
Content review
What this lesson covers
Capacitance is the charge stored per unit potential difference across a capacitor: C = Q/V. Free AQA notes with 3 worked examples and 9 questions answered.
Define capacitance using the equation $C = \frac{Q}{V}$ and state its unit
Explain what capacitance represents physically in terms of charge stored per unit potential difference
Calculate capacitance, charge, or potential difference given two of the three quantities
Apply the definition of capacitance to interpret and compare capacitors in exam-style contexts
Key ideas to keep in view
Use this lesson to stay inside Capacitance (A-level only) while connecting the detail back to the wider Fields and their consequences (A-level only) topic.
Capacitance: The charge stored per unit potential difference across a capacitor. Measured in farads (F). Defined by $C = Q/V$.
Farad (F): The SI unit of capacitance. One farad means one coulomb of charge is stored for every one volt of potential difference across the capacitor.
Potential difference (V): The work done per unit charge between two points in a circuit. Measured in volts (V). In a capacitor context, this is the voltage across the two plates.
Charge stored (Q): The magnitude of charge on one plate of a capacitor. Measured in coulombs (C). Equal and opposite charges sit on the two plates.
Revision notes
Read the core explanations from Capacitance: Definition before testing yourself from memory.
1. Why Capacitance Matters
Every time a camera flash fires, a defibrillator delivers a shock, or a computer's memory holds data for a fraction of a second, a capacitor is doing the work. These devices store electrical charge — and the quantity that tells you how well they do that is called capacitance.
At A Level, capacitance is not a vague idea — it has a precise, calculable definition that sits at the heart of every question on this topic. Get this definition locked in now, and the rest of the capacitance unit follows naturally.
2. Building the Definition from First Principles
Picture a simple capacitor: two parallel conducting plates separated by an insulating gap. When you connect a battery across them, charge builds up — positive charge on one plate, negative on the other. The battery's voltage drives this charge onto the plates, and the plates hold it there even after the battery is removed.
Here is the key physical insight: a larger voltage pushes more charge onto the plates. For a given capacitor, the charge $Q$ stored is directly proportional to the potential difference $V$ applied across it. That proportionality constant is the capacitance $C$.
Notice what the equation $C = Q/V$ is really saying: capacitance is a ratio. It tells you how many coulombs of charge you get for every volt you apply. A capacitor with $C = 200\ \mu\text{F}$ stores twice as much charge as one with $C = 100\ \mu\text{F}$ when both are connected to the same voltage source.
This proportionality is the reason the graph of $Q$ against $V$ for a capacitor is a straight line through the origin — the gradient of that line equals $C$. You will meet this graph again when studying energy stored in a capacitor, so keep it in mind.
3. The Farad and Real-World Scale
The farad is an enormous unit in practice. A capacitor with $C = 1\ \text{F}$ would store one coulomb of charge for every volt applied — that is a staggering amount of charge for a small component. Almost every capacitor you encounter in electronics has a capacitance measured in microfarads ($\mu\text{F}$, $10^{-6}$ F), nanofarads (nF, $10^{-9}$ F), or picofarads (pF, $10^{-12}$ F).
Prefix
Symbol
Value in farads
Typical use
microfarad
$\mu$F
$10^{-6}$ F
Power supply smoothing, audio circuits
nanofarad
nF
$10^{-9}$ F
Radio-frequency tuning circuits
picofarad
pF
$10^{-12}$ F
High-frequency electronics, sensors
Converting these prefixes correctly is a common source of dropped marks. Always convert to SI base units (farads and coulombs) before substituting into $C = Q/V$.
4. The Formula and Its Rearrangements
AQA exam questions on capacitance definition will ask you to calculate any one of the three quantities — $C$, $Q$, or $V$ — given the other two. The starting equation is always $C = Q/V$, and rearranging it is straightforward.
Each rearrangement follows directly from the definition — there is no new physics here, only algebra. The most important habit is to check units at every step: charge must be in coulombs, voltage in volts, and capacitance in farads before you substitute.
5. Worked Calculations
Theory is one thing — exam questions are another. The following walkthroughs cover all three rearrangements of $C = Q/V$. Work through each step yourself before revealing the next one.
Worked example: Calculating Capacitance from Q and V
A capacitor stores a charge of $6.0 \times 10^{-4}$ C when connected to a 12 V supply. Calculate the capacitance of the capacitor. [2 marks]
$C = \frac{Q}{V}$ — [M1] Selecting the correct equation scores the method mark.
$Q = 6.0 \times 10^{-4}\ \text{C}$, $V = 12\ \text{V}$ — Both values are already in SI units — no conversion needed here.
$C = 5.0 \times 10^{-5}\ \text{F} = 50\ \mu\text{F}$ — Converting to microfarads: $5.0 \times 10^{-5}\ \text{F} = 50 \times 10^{-6}\ \text{F} = 50\ \mu\text{F}$. Either form is acceptable in the mark scheme.
A 220 μF capacitor is charged to a potential difference of 9.0 V. Calculate the charge stored on the capacitor. [2 marks]
$C = 220\ \mu\text{F} = 220 \times 10^{-6}\ \text{F} = 2.20 \times 10^{-4}\ \text{F}$ — Always convert prefixes before substituting. This is where marks are lost most often.
$Q = CV$ — [M1] Rearranging $C = Q/V$ correctly.
$Q = 2.20 \times 10^{-4} \times 9.0$ — Both values now in SI units: farads and volts.
$Q = 2.20 \times 10^{-4} \times 9.0 = 19.8 \times 10^{-4} = 1.98 \times 10^{-3}\ \text{C}$ — Arithmetic check: $2.20 \times 9.0 = 19.80$, then carry the $\times 10^{-4}$ factor. [A1] Answer with unit.
6. Physical Interpretation and Common Misconceptions
Capacitance is a property of the capacitor itself — determined by its physical construction (plate area, plate separation, and the material between the plates). It does not change when you charge or discharge the capacitor.
Think of it this way: the capacitance is like the size of a bucket. A bigger bucket (larger $C$) holds more water (charge) for the same depth of water (voltage). Changing how much water is in the bucket does not change the bucket's size.
This proportional relationship — $Q \propto V$ for a fixed $C$ — means that if you double the voltage across a capacitor, you double the charge stored. The ratio $Q/V$ stays constant. That constant ratio is, by definition, the capacitance.
7. Exam Technique and Mark-Winning Strategies
AQA questions on capacitance definition tend to fall into two categories: direct calculation questions (2–3 marks) and conceptual 'explain' or 'describe' questions (1–2 marks). Each type rewards a slightly different approach.
For calculation questions, the mark scheme awards one mark for a correct equation or method and one mark for the correct answer with units. Writing $C = Q/V$ explicitly at the start of your working secures the method mark even if you make an arithmetic error later.
For 'explain' questions — for example, 'Explain what is meant by a capacitance of 100 μF' — the examiner wants you to link the numerical value to the physical meaning. A full-mark answer would state: a capacitance of 100 μF means the capacitor stores $100 \times 10^{-6}$ C of charge for every 1 V of potential difference applied across it.
The core spec requirement for this lesson is the definition $C = Q/V$ — and you have now seen it from every angle: its physical meaning, its unit, all three rearrangements, and the exam language needed to score full marks when explaining it.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-12.
Capacitance is the charge stored per unit potential difference across a capacitor: C = Q/V. Its unit is the farad, where 1 F = 1 C V⁻¹. Capacitance is a property of the capacitor itself, set by its plate area, plate separation and the material between the plates, so it does not change as the capacitor charges or discharges.
What are capacitors used for?
Capacitors store electrical charge in devices such as camera flashes, defibrillators and computer memory that holds data for a fraction of a second. They are also used for power supply smoothing, audio circuits, radio-frequency tuning and high-frequency electronics.
What is a farad?
A farad is the unit of capacitance: 1 F = 1 C V⁻¹, so a 1 F capacitor stores one coulomb of charge for every volt applied. That is enormous in practice, so most capacitors are measured in microfarads (10⁻⁶ F), nanofarads (10⁻⁹ F) or picofarads (10⁻¹² F).
How do you calculate capacitance?
Use C = Q/V with charge in coulombs and pd in volts. A capacitor storing 6.0 × 10⁻⁴ C on a 12 V supply has C = 6.0 × 10⁻⁴ ÷ 12 = 5.0 × 10⁻⁵ F, or 50 µF. A Q-V graph for a capacitor is a straight line through the origin whose gradient equals C.
How do you calculate the charge stored on a capacitor?
Convert the capacitance to farads, then use Q = CV. A 220 µF capacitor charged to 9.0 V stores Q = 2.20 × 10⁻⁴ × 9.0 = 1.98 × 10⁻³ C, or 1.98 mC. Converting prefixes before substituting is where marks are most often lost.
What does a capacitance of 100 µF mean?
A capacitance of 100 µF means the capacitor stores 100 × 10⁻⁶ C of charge for every 1 V of potential difference applied across it. Charged to 12 V, it stores Q = CV = 100 × 10⁻⁶ × 12 = 1.2 × 10⁻³ C, and doubling the voltage doubles the charge stored.
Does capacitance change with voltage?
Capacitance does not change when a capacitor is charged or discharged: it is set by its physical construction, the plate area, plate separation and material between the plates. Doubling the voltage doubles the charge stored, so the ratio Q/V, the capacitance, stays constant.
How do you calculate the voltage across a capacitor?
The voltage across a capacitor is V = Q/C. A 470 nF capacitor storing 2.35 × 10⁻⁵ C has V = 2.35 × 10⁻⁵ ÷ 4.70 × 10⁻⁷ = 0.5 × 10² = 50 V, after first converting 470 nF to 4.70 × 10⁻⁷ F.
What are microfarads, nanofarads and picofarads?
Capacitance is usually measured in microfarads, µF, equal to 10⁻⁶ F, nanofarads, nF, equal to 10⁻⁹ F, or picofarads, pF, equal to 10⁻¹² F, because the farad is an enormous unit. Microfarads suit power supply smoothing, nanofarads radio-frequency tuning, and picofarads high-frequency electronics.