Capacitance: Definition Revision Notes | A Level AQA Physics

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).

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]

1. $C = \frac{Q}{V}$ — [M1] Selecting the correct equation scores the method mark.
2. $Q = 6.0 \times 10^{-4}\ \text{C}$, $V = 12\ \text{V}$ — Both values are already in SI units — no conversion needed here.
3. $C = \frac{6.0 \times 10^{-4}}{12} = 5.0 \times 10^{-5}\ \text{F}$ — Dividing: $6.0 \times 10^{-4} \div 12 = 0.5 \times 10^{-4} = 5.0 \times 10^{-5}$. [A1] Answer with correct unit.
4. $C = 5.0 \times 10^{-5}\ \text{F} = 50\ \mu\text{F}$ — 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.

Answer: $C = 5.0 \times 10^{-5}\ \text{F}$ (or 50 μF)

Worked example: Calculating Charge Stored

A 220 μF capacitor is charged to a potential difference of 9.0 V. Calculate the charge stored on the capacitor. [2 marks]

1. $C = 220\ \mu\text{F} = 220 \times 10^{-6}\ \text{F} = 2.20 \times 10^{-4}\ \text{F}$ — Always convert prefixes before substituting. This is where marks are lost most often.
2. $Q = CV$ — [M1] Rearranging $C = Q/V$ correctly.
3. $Q = 2.20 \times 10^{-4} \times 9.0$ — Both values now in SI units: farads and volts.
4. $Q = 2.20 \times 10^{-4} \times 9.0 = 19.8 \times 10^{-4} = 1.98 \times 10^{-3}\ \text{C}$ — Arithmetic check: $2.20 \times 9.0 = 19.80$, then carry the $\times 10^{-4}$ factor. [A1] Answer with unit.

Answer: $Q = 1.98 \times 10^{-3}\ \text{C}$ (or 1.98 mC)

Worked example: Calculating Potential Difference

A capacitor of capacitance 470 nF stores a charge of $2.35 \times 10^{-5}$ C. Calculate the potential difference across the capacitor. [2 marks]

1. $C = 470\ \text{nF} = 470 \times 10^{-9}\ \text{F} = 4.70 \times 10^{-7}\ \text{F}$ — Nano means $10^{-9}$. Write out the conversion explicitly to avoid errors.
2. $V = \frac{Q}{C}$ — [M1] Correct rearrangement of $C = Q/V$.
3. $V = \frac{2.35 \times 10^{-5}}{4.70 \times 10^{-7}}$ — Both numerator and denominator are in SI units.
4. $V = \frac{2.35}{4.70} \times \frac{10^{-5}}{10^{-7}} = 0.5 \times 10^{2} = 50\ \text{V}$ — Dividing coefficients: $2.35 \div 4.70 = 0.5$. Dividing powers: $10^{-5} \div 10^{-7} = 10^{-5-(-7)} = 10^{2}$. So $V = 0.5 \times 10^{2} = 50\ \text{V}$. [A1]

Answer: $V = 50\ \text{V}$

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.