# Parallel plate capacitor

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u22-parallel-plate-capacitor/

This module covers the structure, core capacitance formula, and behaviour of parallel plate capacitors, the most common configuration in A-level Physics. We derive the formula, explain dielectric effects, and work through common exam problems.

**Prerequisites:** [Basic definition of capacitance](https://www.owlsprep.com/study/cie-9702-u22-capacitance-basics/); [Uniform electric fields between charged plates](https://www.owlsprep.com/study/cie-9702-u18-uniform-electric-fields/)

## Learning objectives

- Derive the capacitance formula for a parallel plate capacitor
- Explain how dielectrics affect parallel plate capacitance
- Calculate changes to capacitance, charge, energy for varying parameters
- Solve common exam problems involving parallel plate configurations

## Structure and Derivation of Capacitance

**Parallel Plate Capacitor** — A capacitor consisting of two parallel conducting plates separated by a small insulating gap, that stores equal and opposite charge on the plates when connected to a potential difference.

*Notation:* C

*Example:* Most practical capacitors use rolled parallel plate geometry to fit large area into a small volume.

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

*Starting from:* Uniform electric field between plates and definition of capacitance

1. For charge $Q$ on plates of area $A$, the uniform electric field between plates is:
2. $$E = \frac{Q}{\varepsilon_0 A}$$
3. Potential difference $V$ across plates separated by distance $d$:
4. $$V = Ed = \frac{Q d}{\varepsilon_0 A}$$
5. Substitute into definition $C = Q/V$:
6. $$C = \frac{Q}{Q d / \varepsilon_0 A} = \frac{\varepsilon_0 A}{d}$$

*Conclusion:* Air has relative permittivity ≈ 1, so this formula holds for air-filled parallel plate capacitors.

**Worked example:** A square parallel plate capacitor has side length 10 cm, plate separation 1 mm. Calculate its capacitance. ($\varepsilon_0 = 8.85 \times 10^{-12}$ F m⁻¹)

1. Convert all quantities to SI units:
2. $$A = (0.10 \text{ m})^2 = 0.01 \text{ m}^2, \quad d = 1 \times 10^{-3} \text{ m}$$
3. Substitute into $C = \frac{\varepsilon_0 A}{d}$:
4. $$C = \frac{8.85 \times 10^{-12} \times 0.01}{1 \times 10^{-3}} = 8.85 \times 10^{-11} \text{ F} = 88.5 \text{ pF}$$
5. Final answer: 88.5 pF

> **Exam tip:** Always convert lengths to meters before calculating area and capacitance. Unit errors are the most common cause of lost marks here.

## Effect of Dielectric Materials

Filling the gap between plates with an insulating dielectric increases capacitance. The dielectric polarises in the electric field, reducing the net electric field between plates for the same stored charge.

**Relative Permittivity (Dielectric Constant)** — A dimensionless constant equal to the ratio of capacitance with a full dielectric filling to capacitance of the same air-filled capacitor.

*Notation:* $\varepsilon_r$

*Example:* $\varepsilon_r \approx 1$ for air/vacuum, $\varepsilon_r \approx 2.5$ for paper, $\varepsilon_r \approx 80$ for water.

The general formula for a dielectric-filled parallel plate capacitor is:

$$C = \frac{\varepsilon_0 \varepsilon_r A}{d} = \varepsilon_r C_0$$

**Worked example:** An air-filled parallel plate capacitor has capacitance 40 pF. It is filled with a dielectric of $\varepsilon_r = 3.2$, what is the new capacitance?

1. Use the relationship $C = \varepsilon_r C_0$, where $C_0 = 40$ pF:
2. $$C = 3.2 \times 40 \text{ pF} = 128 \text{ pF}$$
3. If connected to a battery, $V$ is constant so charge $Q = CV$ increases by a factor of 3.2. If isolated, $Q$ is constant so $V$ drops by a factor of 3.2.

> **tip**
>
> Always confirm first: if the capacitor stays connected to the battery, V is constant. If disconnected (isolated), Q is constant.

## Changes to Capacitance and Related Quantities

Common exam questions ask how capacitance, charge, potential difference and energy stored change when one parameter (plate separation, area, dielectric) is changed. The key is identifying which quantity is held constant.

- Connected to battery: $V$ is constant, $C$ changes with parameters, $Q = CV$ changes proportionally to $C$
- Isolated capacitor: $Q$ is constant, $C$ changes with parameters, $V = Q/C$ changes inversely to $C$

**Worked example:** An isolated air-filled parallel plate capacitor has initial energy $E$. Plate separation is doubled. What is the new energy stored?

1. Isolated capacitor so $Q$ is constant. New capacitance after doubling separation:
2. $$C_\text{new} = \frac{\varepsilon_0 A}{2d} = \frac{C_\text{original}}{2}$$
3. Energy stored for constant Q is $E = \frac{Q^2}{2C}$:
4. $$E_\text{new} = \frac{Q^2}{2 (C/2)} = 2 \times \frac{Q^2}{2C} = 2E$$
5. Energy doubles: work is done pulling the oppositely charged plates apart against the attractive electrostatic force.

## Common pitfalls

- **Wrong:** Forgetting to convert area from cm² to m², using 1 cm² = 0.01 m² instead of 0.0001 m²
  - Why it fails: This leads to a 100× overestimation of capacitance, costing all marks for the calculation
  - Correct: Convert all lengths to meters first, then calculate area from the converted lengths
- **Wrong:** Assuming V is constant for an isolated (disconnected) capacitor
  - Why it fails: An isolated capacitor has no path for charge to flow, so Q is constant, not V
  - Correct: Always start by checking if the capacitor is connected to a battery (V constant) or disconnected (Q constant)
- **Wrong:** Omitting $\varepsilon_r$ when calculating capacitance for a dielectric-filled capacitor
  - Why it fails: Dielectrics increase capacitance, so omitting $\varepsilon_r$ gives a value that is too low by a factor of $\varepsilon_r$
  - Correct: Always multiply by $\varepsilon_r$ if the gap between plates is filled with a dielectric material
- **Wrong:** Confusing diameter and radius for circular plates when calculating area
  - Why it fails: Using diameter instead of radius leads to a 4× error in area and capacitance
  - Correct: Check if the question gives diameter or radius, divide diameter by 2 to get radius before calculating area

## Cheatsheet

| Quantity | Connected to Battery (V constant) | Isolated (Q constant) |
| --- | --- | --- |
| Capacitance | $C = \frac{\varepsilon_0 \varepsilon_r A}{d}$ | $C = \frac{\varepsilon_0 \varepsilon_r A}{d}$ |
| Potential Difference | $V = \text{constant}$ | $V = \frac{Q}{C} \propto 1/C$ |
| Charge | $Q = CV \propto C$ | $Q = \text{constant}$ |
| Energy Stored | $E = \frac{1}{2}CV^2 \propto C$ | $E = \frac{Q^2}{2C} \propto 1/C$ |
| Dielectric-filled C | $C = \varepsilon_r C_0$ | $C = \varepsilon_r C_0$ |

## What's next

Parallel plate capacitors are the foundation for understanding all capacitor concepts in CIE A-level Physics. Mastery of this topic is required for solving problems involving capacitor combinations, energy storage, and charging/discharging in RC circuits, all of which appear regularly in both multiple choice and structured questions. The relationship between permittivity, electric field and capacitance also connects to earlier topics on electrostatics, helping you build a coherent understanding of charge and electric fields. The following topics build directly on the concepts covered here, and you should move on to them next to complete your study of the capacitance unit.

- [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/)
- [Capacitor charging and discharging](https://www.owlsprep.com/study/cie-9702-u22-capacitor-charging-and-discharging/)

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