Parallel plate capacitor (A-level only)

What you'll learn:

How the physical dimensions of a capacitor (area and separation) determine its capacitance.
What relative permittivity (the dielectric constant) means.
The microscopic "dielectric action" of polar molecules in an electric field.
How to apply the equation $\frac{A\varepsilon_0\varepsilon_r}{d}$ in exam questions.

The geometry of a capacitor

Most of the capacitors you have looked at so far have just been a symbol in a circuit diagram. Now, we are going to look at how they are actually built.

A simple capacitor consists of two parallel, conductive metal plates separated by an insulating gap. This gap can be a complete vacuum, filled with air, or filled with a solid insulating material (which we call a dielectric).

Diagram of a parallel plate capacitor

The capacitance $C$ (the amount of charge it can store per volt) depends on three physical factors:

The area of the plates ( $A$ ): A larger overlapping area means more physical space to store electrons. Capacitance is directly proportional to area.
The separation of the plates ( $d$ ): If the plates are closer together, the positive and negative charges on opposite plates pull on each other more strongly, making it easier to pack more charge onto the plates. Capacitance is inversely proportional to separation distance.
The material between the plates: Different insulators affect the electric field between the plates in different ways.

If the gap between the plates is a completely empty vacuum, the capacitance is given by:

\frac{A \varepsilon_0}{d}

Here, $ε0\varepsilon_0$ is the permittivity of free space (a fundamental constant provided on your data sheet, $m−1\approx 8.85 \times 10^{-12}\text{ F m}^{-1}$ ).

Key Idea

Air behaves like a vacuum

For A-level Physics, the permittivity of air is so incredibly close to the permittivity of a vacuum that you can treat an air-filled capacitor exactly as if it were a vacuum-filled one.

Introducing the dielectric

To make a capacitor physically smaller while keeping its capacitance high, manufacturers insert a solid insulating material between the plates. This material is called a dielectric. Examples include waxed paper, ceramics, or plastics.

Every material has a property called its relative permittivity (sometimes called the dielectric constant).

Definition

Relative permittivity ( $ε_{r}$ )

The ratio of the charge stored by a capacitor with the dielectric in place to the charge stored when the capacitor has a vacuum between its plates (for the same potential difference).

Because it is a ratio, $εr\varepsilon_r$ has no units.

When a dielectric is inserted, the formula for the parallel plate capacitor becomes:

\frac{A \varepsilon_0 \varepsilon_r}{d}

If the material is a vacuum (or air), $εr=1\varepsilon_r = 1$ , and the equation neatly simplifies back to the vacuum version. For any other insulator, $εr\varepsilon_r$ is greater than 1, meaning the dielectric increases the capacitance.

Common Mistake

Mixing up permittivity constants

Do not confuse the absolute permittivity of free space ( $ε0\varepsilon_0$ , which has units of $m−1\text{F m}^{-1}$ ) with relative permittivity ( $εr\varepsilon_r$ , which is just a dimensionless ratio). The actual permittivity of the material is the product of the two: $ε=ε0εr\varepsilon = \varepsilon_0 \varepsilon_r$ .

How to use the equation

Let's look at a typical calculation you might face in an exam.

Example

Worked Example: Calculating plate area

A parallel plate capacitor is constructed using a sheet of paper of thickness $mm0.05\text{ mm}$ as a dielectric. The relative permittivity of the paper is $3.5$ . Calculate the plate area required to produce a capacitance of $nF4.7\text{ nF}$ . (Permittivity of free space $m−1\varepsilon_0 = 8.85 \times 10^{-12}\text{ F m}^{-1}$ )

Extract the given values and convert them to standard SI units:

$\times 10^{-9}\text{ F}$
$\times 10^{-3}\text{ m}$
$εr=3.5\varepsilon_r = 3.5$
$m−1\varepsilon_0 = 8.85 \times 10^{-12}\text{ F m}^{-1}$

State the relevant formula:

\frac{A \varepsilon_0 \varepsilon_r}{d}

Rearrange the formula to make Area ( $A$ ) the subject: Multiply both sides by $d$ and divide by $ε0εr\varepsilon_0 \varepsilon_r$ :

\frac{C \cdot d}{\varepsilon_0 \varepsilon_r}

Substitute the values:

\begin{aligned} A &= \frac{(4.7 \times 10^{-9}) \cdot (0.05 \times 10^{-3})}{(8.85 \times 10^{-12}) \cdot 3.5} \\ A &= \frac{2.35 \times 10^{-13}}{3.0975 \times 10^{-11}} \end{aligned}

Calculate the final answer with units:

0.00759\text{ m}^2

Dielectric action: The microscopic view

You need to be able to explain why inserting a dielectric increases the capacitance. This is a very common written question in AQA exams.

Dielectric materials are made of molecules. Some of these are polar molecules. This means that although the molecule is electrically neutral overall, one end of the molecule is slightly positive and the other end is slightly negative. It has a permanent dipole.

When there is no electric field, the thermal movement of these molecules means they point in random directions. Their electric fields cancel each other out.

Diagram showing polar molecules aligning in an electric field between two capacitor plates

When a potential difference is applied across the capacitor plates, an electric field is created between them. Here is the step-by-step logic of what happens next (memorise this sequence!):

The positive end of each polar molecule is repelled by the positive plate and attracted to the negative plate (and vice versa).
This creates a turning force (torque), causing the polar molecules to rotate and align themselves anti-parallel to the applied electric field.
Because the molecules are now aligned, they produce their own, weaker electric field that opposes the applied electric field.
This opposing field reduces the overall net electric field between the plates.
Since the potential difference is directly proportional to the electric field ( $V = E d$ ), reducing the net electric field reduces the potential difference required to store the same amount of charge.
Finally, because $\frac{Q}{V}$ , a smaller $V$ for a given $Q$ results in a larger capacitance.

Tip

Step-by-step logic

Whenever AQA asks you to "explain the action of a dielectric", walk exactly through that 6-step logical chain: Alignment $→\to$ Opposing field $→\to$ Net field drops $→\to$ $V$ drops $→\to$ $C$ increases.

Example

Worked Example: Explaining dielectric action

A capacitor with a vacuum between its plates is fully charged and then disconnected from the power supply. A slab of dielectric material is then inserted between the plates. Explain what happens to the potential difference across the plates. (4 marks)

State the microscopic effect on the molecules: The polar molecules in the dielectric align themselves with their positive ends pointing towards the negative plate.
Describe the effect on the electric field: The aligned molecules produce their own electric field which opposes the applied electric field from the plates, reducing the net electric field.
Link the electric field to the potential difference: The potential difference decreases, because potential difference is proportional to electric field strength ( $V = E d$ ).
Confirm the conservation of charge: Because the capacitor was disconnected, the charge $Q$ on the plates must remain constant, so the potential difference strictly drops as $C$ increases ( $V = Q / C$ ).

Investigating the parallel plate capacitor

You may be asked how to determine the relative permittivity of a material experimentally.

To do this, you would use a parallel plate capacitor where the distance $d$ can be varied, and a capacitance meter to measure $C$ .

Measure the diameter of the circular plates to calculate the overlapping area $A$ .
Measure the separation $d$ using vernier callipers.
Record the capacitance $C$ for various separations $d$ .

Because $(A\varepsilon_0\varepsilon_r) \cdot \frac{1}{d}$ , if you plot a graph of Capacitance ( $C$ on the y-axis) against $1d\frac{1}{d}$ (on the x-axis), you will get a straight line through the origin. The gradient of this line will be equal to $Aε0εrA\varepsilon_0\varepsilon_r$ . By dividing your gradient by $A$ and $ε0\varepsilon_0$ , you can find the relative permittivity $εr\varepsilon_r$ .

Exam technique

In the exam

Check your prefixes: Plate areas are often given in $cm2\text{cm}^2$ or $mm2\text{mm}^2$ . Remember that $m21\text{ cm}^2 = 10^{-4}\text{ m}^2$ and $m21\text{ mm}^2 = 10^{-6}\text{ m}^2$ . Separations are usually in $mm\text{mm}$ , so don't forget the $×10−3\times 10^{-3}$ multiplier.
Identify isolated vs connected capacitors: If a capacitor is connected to a battery while a dielectric is inserted, the potential difference $V$ is fixed, so the charge $Q$ must increase. If the capacitor is charged and disconnected before the dielectric is inserted, the charge $Q$ is fixed, so the potential difference $V$ must decrease.
Keywords for dielectric explanations: Always use the words "polar molecules", "rotate/align", and "opposing electric field". These are heavily rewarded on AQA mark schemes.

Self review

Check yourself

What happens to the capacitance if the distance between the plates is halved?
Does relative permittivity have units?
In terms of electric fields, why does a dielectric decrease the potential difference required to hold a specific amount of charge?

Parallel plate capacitor (A-level only)

What you'll learn:

How the physical dimensions of a capacitor (area and separation) determine its capacitance.
What relative permittivity (the dielectric constant) means.
The microscopic "dielectric action" of polar molecules in an electric field.
How to apply the equation $\frac{A\varepsilon_0\varepsilon_r}{d}$ in exam questions.

The geometry of a capacitor

Most of the capacitors you have looked at so far have just been a symbol in a circuit diagram. Now, we are going to look at how they are actually built.

Diagram of a parallel plate capacitor

The capacitance $C$ (the amount of charge it can store per volt) depends on three physical factors:

The area of the plates ( $A$ ): A larger overlapping area means more physical space to store electrons. Capacitance is directly proportional to area.
The separation of the plates ( $d$ ): If the plates are closer together, the positive and negative charges on opposite plates pull on each other more strongly, making it easier to pack more charge onto the plates. Capacitance is inversely proportional to separation distance.
The material between the plates: Different insulators affect the electric field between the plates in different ways.

If the gap between the plates is a completely empty vacuum, the capacitance is given by:

\frac{A \varepsilon_0}{d}

Here, $ε0\varepsilon_0$ is the permittivity of free space (a fundamental constant provided on your data sheet, $m−1\approx 8.85 \times 10^{-12}\text{ F m}^{-1}$ ).

Key Idea

Air behaves like a vacuum

For A-level Physics, the permittivity of air is so incredibly close to the permittivity of a vacuum that you can treat an air-filled capacitor exactly as if it were a vacuum-filled one.

Introducing the dielectric

Every material has a property called its relative permittivity (sometimes called the dielectric constant).

Definition

Relative permittivity ( $ε_{r}$ )

The ratio of the charge stored by a capacitor with the dielectric in place to the charge stored when the capacitor has a vacuum between its plates (for the same potential difference).

Because it is a ratio, $εr\varepsilon_r$ has no units.

When a dielectric is inserted, the formula for the parallel plate capacitor becomes:

\frac{A \varepsilon_0 \varepsilon_r}{d}

Common Mistake

Mixing up permittivity constants

How to use the equation

Let's look at a typical calculation you might face in an exam.

Example

Worked Example: Calculating plate area

Extract the given values and convert them to standard SI units:

$\times 10^{-9}\text{ F}$
$\times 10^{-3}\text{ m}$
$εr=3.5\varepsilon_r = 3.5$
$m−1\varepsilon_0 = 8.85 \times 10^{-12}\text{ F m}^{-1}$

State the relevant formula:

\frac{A \varepsilon_0 \varepsilon_r}{d}

Rearrange the formula to make Area ( $A$ ) the subject: Multiply both sides by $d$ and divide by $ε0εr\varepsilon_0 \varepsilon_r$ :

\frac{C \cdot d}{\varepsilon_0 \varepsilon_r}

Substitute the values:

\begin{aligned} A &= \frac{(4.7 \times 10^{-9}) \cdot (0.05 \times 10^{-3})}{(8.85 \times 10^{-12}) \cdot 3.5} \\ A &= \frac{2.35 \times 10^{-13}}{3.0975 \times 10^{-11}} \end{aligned}

Calculate the final answer with units:

0.00759\text{ m}^2

Dielectric action: The microscopic view

You need to be able to explain why inserting a dielectric increases the capacitance. This is a very common written question in AQA exams.

When there is no electric field, the thermal movement of these molecules means they point in random directions. Their electric fields cancel each other out.

Diagram showing polar molecules aligning in an electric field between two capacitor plates

When a potential difference is applied across the capacitor plates, an electric field is created between them. Here is the step-by-step logic of what happens next (memorise this sequence!):

The positive end of each polar molecule is repelled by the positive plate and attracted to the negative plate (and vice versa).
This creates a turning force (torque), causing the polar molecules to rotate and align themselves anti-parallel to the applied electric field.
Because the molecules are now aligned, they produce their own, weaker electric field that opposes the applied electric field.
This opposing field reduces the overall net electric field between the plates.
Since the potential difference is directly proportional to the electric field ( $V = E d$ ), reducing the net electric field reduces the potential difference required to store the same amount of charge.
Finally, because $\frac{Q}{V}$ , a smaller $V$ for a given $Q$ results in a larger capacitance.

Tip

Step-by-step logic

Example

Worked Example: Explaining dielectric action

State the microscopic effect on the molecules: The polar molecules in the dielectric align themselves with their positive ends pointing towards the negative plate.
Describe the effect on the electric field: The aligned molecules produce their own electric field which opposes the applied electric field from the plates, reducing the net electric field.
Link the electric field to the potential difference: The potential difference decreases, because potential difference is proportional to electric field strength ( $V = E d$ ).
Confirm the conservation of charge: Because the capacitor was disconnected, the charge $Q$ on the plates must remain constant, so the potential difference strictly drops as $C$ increases ( $V = Q / C$ ).

Investigating the parallel plate capacitor

You may be asked how to determine the relative permittivity of a material experimentally.

To do this, you would use a parallel plate capacitor where the distance $d$ can be varied, and a capacitance meter to measure $C$ .

Measure the diameter of the circular plates to calculate the overlapping area $A$ .
Measure the separation $d$ using vernier callipers.
Record the capacitance $C$ for various separations $d$ .

Exam technique

In the exam

Check your prefixes: Plate areas are often given in $cm2\text{cm}^2$ or $mm2\text{mm}^2$ . Remember that $m21\text{ cm}^2 = 10^{-4}\text{ m}^2$ and $m21\text{ mm}^2 = 10^{-6}\text{ m}^2$ . Separations are usually in $mm\text{mm}$ , so don't forget the $×10−3\times 10^{-3}$ multiplier.
Identify isolated vs connected capacitors: If a capacitor is connected to a battery while a dielectric is inserted, the potential difference $V$ is fixed, so the charge $Q$ must increase. If the capacitor is charged and disconnected before the dielectric is inserted, the charge $Q$ is fixed, so the potential difference $V$ must decrease.
Keywords for dielectric explanations: Always use the words "polar molecules", "rotate/align", and "opposing electric field". These are heavily rewarded on AQA mark schemes.

Self review

Check yourself

What happens to the capacitance if the distance between the plates is halved?
Does relative permittivity have units?
In terms of electric fields, why does a dielectric decrease the potential difference required to hold a specific amount of charge?

Parallel plate capacitor (A-level only)

The geometry of a capacitor

Introducing the dielectric

How to use the equation

Dielectric action: The microscopic view

Investigating the parallel plate capacitor

Parallel plate capacitor (A-level only) Revision Guide

Parallel plate capacitor (A-level only)

The geometry of a capacitor

Introducing the dielectric

How to use the equation

Dielectric action: The microscopic view

Investigating the parallel plate capacitor

Parallel plate capacitor (A-level only) Revision Guide