Energy

What you'll learn

How a capacitor stores energy as it is charged.
Why the area under a p.d.–charge graph gives the energy stored.
How to use $\frac{1}{2}QV$ , $\frac{1}{2}\frac{Q^2}{C}$ and $\frac{1}{2}V^2 C$ .
Why capacitors are useful when energy needs to be released quickly.

Starting point: charge, p.d. and capacitance

A capacitor is a component that stores charge. In its simplest form, it has two conducting plates separated by an insulator. When connected to a supply, one plate becomes positively charged and the other becomes negatively charged.

The charge stored on one plate is written as $Q$ and is measured in coulombs, C.

Definition

Potential difference

Potential difference, $V$ , is the energy transferred per unit charge. One volt means one joule per coulomb: $C−11\ \text{V} = 1\ \text{J C}^{-1}$ .

If a charge $Q$ moves through a constant potential difference $V$ , the energy transferred is:

W = Q V

where $W$ is the work done or energy transferred, measured in joules, J.

Definition

Capacitance

Capacitance, $C$ , is the charge stored per unit potential difference across the capacitor:

\frac{Q}{V}

It is measured in farads, F.

For a capacitor with constant capacitance, rearranging $\frac{Q}{V}$ gives:

\frac{Q}{C}

So as more charge is stored, the p.d. across the capacitor increases.

Common Mistake

The two meanings of C

The symbol $C$ can mean capacitance, but C is also the unit symbol for coulomb. Look at the context: $220\ \mu\text{F}$ is capacitance, while $220\ \mu\text{C}$ is charge.

Why charging a capacitor takes increasing work

At the start, an uncharged capacitor has zero p.d. across it. It is relatively easy to move the first small amount of charge onto the plates.

As charge builds up, the p.d. across the capacitor increases. Later charge has to be pushed on against a larger p.d., so more work is needed per coulomb.

This is the key reason the stored energy is not simply $Q V$ .

Key Idea

The p.d. is not constant while charging

For a capacitor charging from zero to final p.d. $V$ , the p.d. rises from zero to $V$ . The average p.d. during the charging process is therefore $V2\frac{V}{2}$ .

The p.d.–charge graph

A p.d.–charge graph plots the potential difference $V$ across the capacitor against the charge $Q$ stored on it.

For a capacitor with constant capacitance:

\frac{Q}{C}

This is the equation of a straight line through the origin. The gradient is:

gradient=VQ=1C\text{gradient} = \frac{V}{Q} = \frac{1}{C}

So a larger capacitance gives a smaller gradient: the capacitor can store more charge for the same p.d.

The energy stored is the area under the p.d.–charge graph.

p.d.–charge graph for a capacitor showing the triangular area under the graph as stored energy

Because the graph is a triangle:

\frac{1}{2}QV

where $Q$ is the final charge and $V$ is the final p.d.

Example

Finding energy from a p.d.–charge graph

A capacitor is charged until the charge stored is $4.0 \times 10^{-3}\ \text{C}$ and the p.d. across it is 12 V. Find the energy stored and the capacitance.

Use the triangular area under the p.d.–charge graph for the energy stored:
$\frac{1}{2}QV$
Substitute the final charge and final p.d.:
$\frac{1}{2}(4.0 \times 10^{-3}\ \text{C})(12\ \text{V})$
Calculate the energy:
$\times 10^{-2}\ \text{J}$
Use $\frac{Q}{V}$ to find the capacitance:
$\frac{4.0 \times 10^{-3}\ \text{C}}{12\ \text{V}} = 3.3 \times 10^{-4}\ \text{F}$

The energy equations for a capacitor

The main capacitor energy equation is:

\frac{1}{2}QV

You can combine this with $\frac{Q}{V}$ to get two other useful forms.

Since $Q = C V$ :

\frac{1}{2}V^2 C

Since $\frac{Q}{C}$ :

\frac{1}{2}\frac{Q^2}{C}

So the three OCR equations are:

\frac{1}{2}QV \qquad W = \frac{1}{2}\frac{Q^2}{C} \qquad W = \frac{1}{2}V^2 C

Use the version that matches the quantities you are given.

Tip

Choosing the right equation

Use $\frac{1}{2}QV$ if you know charge and p.d.; use $\frac{1}{2}V^2 C$ if you know capacitance and p.d.; use $\frac{1}{2}\frac{Q^2}{C}$ if you know charge and capacitance.

Example

Calculating stored energy from capacitance and voltage

A 2200 µF capacitor is charged to 12.0 V. Calculate the energy stored.

Convert the capacitance into farads:
$F2200\ \mu\text{F} = 2200 \times 10^{-6}\ \text{F} = 2.20 \times 10^{-3}\ \text{F}$
Choose the equation using capacitance and p.d.:
$\frac{1}{2}V^2 C$
Substitute the values, keeping units in the calculation:
$\frac{1}{2}(12.0\ \text{V})^2(2.20 \times 10^{-3}\ \text{F})$
Calculate and quote sensibly:
$0.158\ \text{J} \approx 0.16\ \text{J}$

How energy depends on voltage and capacitance

From:

\frac{1}{2}V^2 C

the energy stored is proportional to capacitance and proportional to the square of the p.d.

So:

doubling the capacitance doubles the energy, if $V$ is unchanged;
doubling the p.d. makes the energy four times larger, if $C$ is unchanged;
tripling the p.d. makes the energy nine times larger, if $C$ is unchanged.

Common Mistake

Forgetting the square on voltage

The energy stored in a capacitor does not just double when the p.d. doubles. Because $\propto V^2$ , doubling $V$ gives four times the energy.

Example

Comparing energy at two voltages

A capacitor stores 0.50 J when charged to 5.0 V. Find the energy stored when the same capacitor is charged to 10.0 V.

Identify what stays constant: the same capacitor is used, so $C$ is unchanged.
Use the proportionality from $\frac{1}{2}V^2 C$ :
$\propto V^2$
Compare the voltages:
$V=2\frac{10.0\ \text{V}}{5.0\ \text{V}} = 2$
Square the voltage ratio to find the energy ratio:
$2^2 = 4$
Multiply the original energy by 4:
$4(0.50\ \text{J}) = 2.0\ \text{J}$

Why the factor of one-half appears

A common question is: if energy transferred by charge through p.d. is $W = Q V$ , why is capacitor energy $\frac{1}{2}QV$ ?

The reason is that the p.d. across the capacitor is not equal to the final value for the whole charging process. It rises from zero to $V$ .

For a constant-capacitance capacitor, the average p.d. during charging is:

0+V2=V2\frac{0 + V}{2} = \frac{V}{2}

So the energy stored is:

\left(\frac{V}{2}\right) = \frac{1}{2}QV

Key Idea

Area gives the work done

On a p.d.–charge graph, area represents energy because a small amount of work is given by p.d. multiplied by a small amount of charge. Adding all those small contributions gives the total work done charging the capacitor.

Uses of capacitors as energy stores

Capacitors are especially useful when energy must be stored briefly and then released quickly.

Unlike a battery, a capacitor does not store energy mainly through chemical changes. It stores energy in the electric field between its plates. This means it can often discharge very rapidly, producing a large current for a short time.

Common uses include:

Camera flashes — energy is stored slowly, then released quickly through a flash lamp.
Defibrillators — a capacitor stores energy and then releases a controlled pulse through the patient.
Power supply smoothing — capacitors store energy when the supply voltage is high and release it when the voltage falls.
Backup power for memory circuits — a capacitor can keep a small circuit powered for a short time if the main supply is interrupted.

Example

Estimating energy for a camera flash

A camera flash uses a 470 µF capacitor charged to 300 V. Estimate the energy stored before the flash fires.

Convert the capacitance into farads:
$F470\ \mu\text{F} = 470 \times 10^{-6}\ \text{F} = 4.70 \times 10^{-4}\ \text{F}$
Use the equation involving capacitance and voltage:
$\frac{1}{2}V^2 C$
Substitute the values:
$\frac{1}{2}(300\ \text{V})^2(4.70 \times 10^{-4}\ \text{F})$
Calculate the energy:
$21.15\ \text{J} \approx 21\ \text{J}$

Common Mistake

High p.d. can be dangerous

Even a small capacitor can store a significant amount of energy if charged to a high p.d., because the energy depends on $V^2$ . Large charged capacitors should be treated as potentially dangerous.

Exam technique

In the exam

For a p.d.–charge graph, use the area under the graph for energy stored; for a straight-line capacitor graph, this is a triangle.
Always convert prefixes before calculating: µF means $F10^{-6}\ \text{F}$ , mC means $C10^{-3}\ \text{C}$ .
Check whether voltage has been squared. If the question compares the same capacitor at different p.d.s, use $\propto V^2$ .

Self review

Check yourself

Why is the energy stored by a capacitor $12QV\frac{1}{2}QV$ rather than $Q V$ ?
What does the gradient of a p.d.–charge graph represent?
If the p.d. across a capacitor is doubled, what happens to the stored energy?

What you'll learn

How a capacitor stores energy as it is charged.
Why the area under a p.d.–charge graph gives the energy stored.
How to use $\frac{1}{2}QV$ , $\frac{1}{2}\frac{Q^2}{C}$ and $\frac{1}{2}V^2 C$ .
Why capacitors are useful when energy needs to be released quickly.

Starting point: charge, p.d. and capacitance

The charge stored on one plate is written as $Q$ and is measured in coulombs, C.

Definition

Potential difference

Potential difference, $V$ , is the energy transferred per unit charge. One volt means one joule per coulomb: $C−11\ \text{V} = 1\ \text{J C}^{-1}$ .

If a charge $Q$ moves through a constant potential difference $V$ , the energy transferred is:

W = Q V

where $W$ is the work done or energy transferred, measured in joules, J.

Definition

Capacitance

Capacitance, $C$ , is the charge stored per unit potential difference across the capacitor:

\frac{Q}{V}

It is measured in farads, F.

For a capacitor with constant capacitance, rearranging $\frac{Q}{V}$ gives:

\frac{Q}{C}

So as more charge is stored, the p.d. across the capacitor increases.

Common Mistake

The two meanings of C

The symbol $C$ can mean capacitance, but C is also the unit symbol for coulomb. Look at the context: $220\ \mu\text{F}$ is capacitance, while $220\ \mu\text{C}$ is charge.

Why charging a capacitor takes increasing work

At the start, an uncharged capacitor has zero p.d. across it. It is relatively easy to move the first small amount of charge onto the plates.

As charge builds up, the p.d. across the capacitor increases. Later charge has to be pushed on against a larger p.d., so more work is needed per coulomb.

This is the key reason the stored energy is not simply $Q V$ .

Key Idea

The p.d. is not constant while charging

For a capacitor charging from zero to final p.d. $V$ , the p.d. rises from zero to $V$ . The average p.d. during the charging process is therefore $V2\frac{V}{2}$ .

The p.d.–charge graph

A p.d.–charge graph plots the potential difference $V$ across the capacitor against the charge $Q$ stored on it.

For a capacitor with constant capacitance:

\frac{Q}{C}

This is the equation of a straight line through the origin. The gradient is:

gradient=VQ=1C\text{gradient} = \frac{V}{Q} = \frac{1}{C}

So a larger capacitance gives a smaller gradient: the capacitor can store more charge for the same p.d.

The energy stored is the area under the p.d.–charge graph.

p.d.–charge graph for a capacitor showing the triangular area under the graph as stored energy

Because the graph is a triangle:

\frac{1}{2}QV

where $Q$ is the final charge and $V$ is the final p.d.

Example

Finding energy from a p.d.–charge graph

A capacitor is charged until the charge stored is $4.0 \times 10^{-3}\ \text{C}$ and the p.d. across it is 12 V. Find the energy stored and the capacitance.

Use the triangular area under the p.d.–charge graph for the energy stored:
$\frac{1}{2}QV$
Substitute the final charge and final p.d.:
$\frac{1}{2}(4.0 \times 10^{-3}\ \text{C})(12\ \text{V})$
Calculate the energy:
$\times 10^{-2}\ \text{J}$
Use $\frac{Q}{V}$ to find the capacitance:
$\frac{4.0 \times 10^{-3}\ \text{C}}{12\ \text{V}} = 3.3 \times 10^{-4}\ \text{F}$

The energy equations for a capacitor

The main capacitor energy equation is:

\frac{1}{2}QV

You can combine this with $\frac{Q}{V}$ to get two other useful forms.

Since $Q = C V$ :

\frac{1}{2}V^2 C

Since $\frac{Q}{C}$ :

\frac{1}{2}\frac{Q^2}{C}

So the three OCR equations are:

\frac{1}{2}QV \qquad W = \frac{1}{2}\frac{Q^2}{C} \qquad W = \frac{1}{2}V^2 C

Use the version that matches the quantities you are given.

Tip

Choosing the right equation

Use $\frac{1}{2}QV$ if you know charge and p.d.; use $\frac{1}{2}V^2 C$ if you know capacitance and p.d.; use $\frac{1}{2}\frac{Q^2}{C}$ if you know charge and capacitance.

Example

Calculating stored energy from capacitance and voltage

A 2200 µF capacitor is charged to 12.0 V. Calculate the energy stored.

Convert the capacitance into farads:
$F2200\ \mu\text{F} = 2200 \times 10^{-6}\ \text{F} = 2.20 \times 10^{-3}\ \text{F}$
Choose the equation using capacitance and p.d.:
$\frac{1}{2}V^2 C$
Substitute the values, keeping units in the calculation:
$\frac{1}{2}(12.0\ \text{V})^2(2.20 \times 10^{-3}\ \text{F})$
Calculate and quote sensibly:
$0.158\ \text{J} \approx 0.16\ \text{J}$

How energy depends on voltage and capacitance

From:

\frac{1}{2}V^2 C

the energy stored is proportional to capacitance and proportional to the square of the p.d.

So:

doubling the capacitance doubles the energy, if $V$ is unchanged;
doubling the p.d. makes the energy four times larger, if $C$ is unchanged;
tripling the p.d. makes the energy nine times larger, if $C$ is unchanged.

Common Mistake

Forgetting the square on voltage

The energy stored in a capacitor does not just double when the p.d. doubles. Because $\propto V^2$ , doubling $V$ gives four times the energy.

Example

Comparing energy at two voltages

A capacitor stores 0.50 J when charged to 5.0 V. Find the energy stored when the same capacitor is charged to 10.0 V.

Identify what stays constant: the same capacitor is used, so $C$ is unchanged.
Use the proportionality from $\frac{1}{2}V^2 C$ :
$\propto V^2$
Compare the voltages:
$V=2\frac{10.0\ \text{V}}{5.0\ \text{V}} = 2$
Square the voltage ratio to find the energy ratio:
$2^2 = 4$
Multiply the original energy by 4:
$4(0.50\ \text{J}) = 2.0\ \text{J}$

Why the factor of one-half appears

A common question is: if energy transferred by charge through p.d. is $W = Q V$ , why is capacitor energy $\frac{1}{2}QV$ ?

The reason is that the p.d. across the capacitor is not equal to the final value for the whole charging process. It rises from zero to $V$ .

For a constant-capacitance capacitor, the average p.d. during charging is:

0+V2=V2\frac{0 + V}{2} = \frac{V}{2}

So the energy stored is:

\left(\frac{V}{2}\right) = \frac{1}{2}QV

Key Idea

Area gives the work done

Uses of capacitors as energy stores

Capacitors are especially useful when energy must be stored briefly and then released quickly.

Common uses include:

Camera flashes — energy is stored slowly, then released quickly through a flash lamp.
Defibrillators — a capacitor stores energy and then releases a controlled pulse through the patient.
Power supply smoothing — capacitors store energy when the supply voltage is high and release it when the voltage falls.
Backup power for memory circuits — a capacitor can keep a small circuit powered for a short time if the main supply is interrupted.

Example

Estimating energy for a camera flash

A camera flash uses a 470 µF capacitor charged to 300 V. Estimate the energy stored before the flash fires.

Convert the capacitance into farads:
$F470\ \mu\text{F} = 470 \times 10^{-6}\ \text{F} = 4.70 \times 10^{-4}\ \text{F}$
Use the equation involving capacitance and voltage:
$\frac{1}{2}V^2 C$
Substitute the values:
$\frac{1}{2}(300\ \text{V})^2(4.70 \times 10^{-4}\ \text{F})$
Calculate the energy:
$21.15\ \text{J} \approx 21\ \text{J}$

Common Mistake

High p.d. can be dangerous

Even a small capacitor can store a significant amount of energy if charged to a high p.d., because the energy depends on $V^2$ . Large charged capacitors should be treated as potentially dangerous.

Exam technique

In the exam

For a p.d.–charge graph, use the area under the graph for energy stored; for a straight-line capacitor graph, this is a triangle.
Always convert prefixes before calculating: µF means $F10^{-6}\ \text{F}$ , mC means $C10^{-3}\ \text{C}$ .
Check whether voltage has been squared. If the question compares the same capacitor at different p.d.s, use $\propto V^2$ .

Self review

Check yourself

Why is the energy stored by a capacitor $12QV\frac{1}{2}QV$ rather than $Q V$ ?
What does the gradient of a p.d.–charge graph represent?
If the p.d. across a capacitor is doubled, what happens to the stored energy?

What you'll learn

Starting point: charge, p.d. and capacitance

Why charging a capacitor takes increasing work

The p.d.–charge graph

The energy equations for a capacitor

How energy depends on voltage and capacitance

Why the factor of one-half appears

Uses of capacitors as energy stores

Review Energy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Energy

What you'll learn

Starting point: charge, p.d. and capacitance

Why charging a capacitor takes increasing work

The p.d.–charge graph

The energy equations for a capacitor

How energy depends on voltage and capacitance

Why the factor of one-half appears

Uses of capacitors as energy stores

Review Energy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Energy

What you'll learn

Starting point: charge, p.d. and capacitance

Why charging a capacitor takes increasing work

The p.d.–charge graph

The energy equations for a capacitor

How energy depends on voltage and capacitance

Why the factor of one-half appears

Uses of capacitors as energy stores

Review Energy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Energy Revision Guide

Energy

What you'll learn

Starting point: charge, p.d. and capacitance

Why charging a capacitor takes increasing work

The p.d.–charge graph

The energy equations for a capacitor

How energy depends on voltage and capacitance

Why the factor of one-half appears

Uses of capacitors as energy stores

Review Energy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Energy Revision Guide