# Capacitance concepts

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u22-capacitance-concepts/

This module introduces core capacitance concepts for CIE A-Level Physics, including the definition of capacitance, parallel plate capacitor behaviour, and the effect of dielectrics. You will learn core formulas and problem-solving skills for common exam questions.

**Prerequisites:** [Electric potential difference and uniform electric fields](https://www.owlsprep.com/study/cie-9702-u18-electric-fields-potential/)

## Learning objectives

- Define capacitance and state its units and core definition formula
- Derive and use the formula for capacitance of a parallel plate capacitor
- Explain the effect of inserting a dielectric on capacitance
- Identify factors that determine the capacitance of a capacitor

## Definition of Capacitance

**Capacitance** — The ratio of the magnitude of charge stored on one plate of a capacitor to the potential difference across the capacitor, given by $C = \frac{Q}{V}$. The SI unit is the farad (F), where $1 \text{ F} = 1 \text{ C V}^{-1}$.

*Notation:* C

*Example:* A $10 \ \mu\text{F}$ capacitor stores $20 \ \mu\text{C}$ of charge when $2 \text{ V}$ is applied across its plates.

Capacitors are passive electronic components that store energy in the electric field between two separated conducting plates. When connected to a voltage source, equal and opposite charge accumulates on the two plates, creating a uniform electric field between them.

**Worked example:** A capacitor stores $6.0 \times 10^{-4} \text{ C}$ of charge when connected to a $12 \text{ V}$ battery. Calculate its capacitance.

1. Start with the definition of capacitance:
2. $$C = \frac{Q}{V}$$
3. Substitute the given values for charge and potential difference:
4. $$C = \frac{6.0 \times 10^{-4} \text{ C}}{12 \text{ V}} = 5.0 \times 10^{-5} \text{ F}$$
5. Convert to the commonly used microfarad unit for convenience:
6. $$C = 50 \ \mu\text{F}$$

## Parallel Plate Capacitor Capacitance

A parallel plate capacitor is the simplest and most common capacitor structure, made of two parallel conducting plates separated by a fixed distance. We can derive its capacitance from the properties of uniform electric fields.

**Derivation:** Derive capacitance for an air-filled parallel plate capacitor

*Starting from:* Uniform electric field between plates: $V = Ed$; Electric field strength: $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A \varepsilon_0}$

1. Substitute the expression for $E$ into the potential difference formula:
2. $$V = Ed = \frac{Q d}{A \varepsilon_0}$$
3. Rearrange to get $\frac{Q}{V}$, which equals capacitance by definition:
4. $$C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d}$$

*Conclusion:* For an air-filled parallel plate capacitor, capacitance depends only on plate area $A$, plate separation $d$, and the constant $\varepsilon_0 = 8.85 \times 10^{-12} \text{ F m}^{-1}$.

**Worked example:** A parallel plate air capacitor has plates of area $0.020 \text{ m}^2$ separated by $1.0 \text{ mm}$ of air. Calculate its capacitance.

1. Convert plate separation to SI units:
2. $$d = 1.0 \text{ mm} = 1.0 \times 10^{-3} \text{ m}$$
3. Substitute into the parallel plate capacitance formula:
4. $$C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 0.020}{1.0 \times 10^{-3}}$$
5. Calculate the final value:
6. $$C \approx 1.8 \times 10^{-10} \text{ F} = 180 \text{ pF}$$

## Effect of Dielectrics

**Dielectric** — An insulating material inserted between the plates of a capacitor to increase its capacitance. Common dielectrics include glass, paper, ceramic, and plastic.

When a dielectric is inserted, polar molecules in the material align with the existing electric field, reducing the net electric field strength between the plates. For a given charge, this reduces the potential difference $V$, so from $C = \frac{Q}{V}$, capacitance increases.

The new capacitance is given by $C = \frac{\varepsilon_r \varepsilon_0 A}{d}$, where $\varepsilon_r$ (relative permittivity) is always greater than 1 for insulating materials.

**Worked example:** The 180 pF air capacitor from the previous example has its air gap replaced with glass of relative permittivity $\varepsilon_r = 5.0$. Calculate the new capacitance.

1. Capacitance scales linearly with relative permittivity:
2. $$C_{\text{new}} = \varepsilon_r C_{\text{air}}$$
3. Substitute values:
4. $$C_{\text{new}} = 5.0 \times 180 \text{ pF} = 900 \text{ pF}$$

**Check your understanding**

Test your understanding of dielectric behaviour:

1. A capacitor is connected to a constant voltage battery, and a dielectric is inserted between the plates. What happens to the charge stored?

   - Charge increases
   - Charge decreases
   - Charge stays the same
   - Charge becomes zero

   *Why:* Voltage $V$ is constant, capacitance $C$ increases, so from $Q = CV$ charge $Q$ also increases.

## Common pitfalls

- **Wrong:** Forgetting to convert plate separation/area to SI units before calculation
  - Why it fails: This leads to answers wrong by multiple orders of magnitude, a very common exam error
  - Correct: Always convert all quantities to SI units (metres for length, m² for area) before substituting into capacitance formulas
- **Wrong:** Assuming capacitance increases when plate separation increases
  - Why it fails: Capacitance is inversely proportional to separation, so the relationship is the opposite
  - Correct: Remember $C \propto \frac{1}{d}$: increasing plate separation decreases capacitance for a parallel plate capacitor
- **Wrong:** Confusing permittivity of free space $\varepsilon_0$ and relative permittivity $\varepsilon_r$
  - Why it fails: Missing or swapping these values leads to answers wrong by 10+ orders of magnitude
  - Correct: $\varepsilon_0 = 8.85 \times 10^{-12} \text{ F m}^{-1}$ is a universal constant, $\varepsilon_r$ is dimensionless and specific to the dielectric material
- **Wrong:** Claiming capacitance depends on the charge stored or applied potential difference
  - Why it fails: $C = \frac{Q}{V}$ is a definition, not a dependency: capacitance is a fixed property of the capacitor itself
  - Correct: Capacitance depends only on the geometry of the capacitor and the dielectric between its plates, not $Q$ or $V$

## Cheatsheet

| Quantity | Symbol | Formula | Unit |
| --- | --- | --- | --- |
| Capacitance | C | $C = \frac{Q}{V}$ | farad (F) |
| Air-filled parallel plate C | C | $C = \frac{\varepsilon_0 A}{d}$ | farad (F) |
| Dielectric-filled parallel plate C | C | $C = \frac{\varepsilon_r \varepsilon_0 A}{d}$ | farad (F) |
| Permittivity of free space | $\varepsilon_0$ | $8.85 \times 10^{-12}$ | F m⁻¹ |
| Relative permittivity | $\varepsilon_r$ | $\geq 1$, dimensionless | - |

## What's next

Core capacitance concepts are the foundation for all other capacitor topics in CIE A-Level Physics. Mastering these basics makes it much easier to solve problems involving combinations of capacitors in series and parallel, calculate energy stored in capacitors, and analyze exponential charging and discharge of capacitors in circuits. These topics are regularly tested in both multiple-choice and structured written questions, so a solid understanding of capacitance concepts is critical for exam success.

- [Parallel plate capacitor](https://www.owlsprep.com/study/cie-9702-u22-parallel-plate-capacitor/)
- [Energy stored in a capacitor](https://www.owlsprep.com/study/cie-9702-u22-energy-stored-in-capacitor/)
- [Capacitors in series and parallel](https://www.owlsprep.com/study/cie-9702-u22-capacitors-in-series-and-parallel/)

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