Energy stored in a capacitor

A flash in a blink

A camera flash is charged slowly, over a few seconds, by a small battery. Then it gives out all its energy in a few thousandths of a second. The lamp is very bright for that instant. The energy was waiting in a capacitor.

How much energy can a capacitor hold? We find it by asking how much work it takes to charge it.

The work of charging

Suppose the capacitor already holds a charge $q$. Its voltage is $v = q/C$. To move one more small packet of charge $dq$ from one plate to the other, we must work against that voltage. The work is $dW = v\,dq$.

At the start, $v$ is zero and the packets are easy to move. Near the end, $v$ is nearly $V$ and each packet costs more. So the work for each packet grows as the capacitor fills.

On a graph of $v$ against $q$, the work for one packet is a thin strip of height $v$ and width $dq$. All the strips together fill the triangle under the line. Its area is half of base times height:

Now use $Q = CV$ to write the energy in three ways:

Which one should you use? Use the one that holds the two quantities you know. If you know $C$ and $V$, use $\tfrac{1}{2}CV^2$. If you know $Q$ and $C$, use $\dfrac{Q^2}{2C}$.

Why a half and not $QV$? The first charge is moved when the voltage is almost zero, and the last when it is $V$. The average voltage is only $\tfrac{1}{2}V$.

A worked example. A flash capacitor of $20\ \mu\text{F}$ is charged to $300\ \text{V}$.

1. The charge. $Q = CV = 20\ \mu\text{F} \times 300\ \text{V} = 6000\ \mu\text{C} = 6.0\ \text{mC}$.
2. The energy. $U = \tfrac{1}{2}CV^2 = \tfrac{1}{2} \times 20 \times 10^{-6}\ \text{F} \times (300\ \text{V})^2 = 0.90\ \text{J}$.
3. The flash. If all of it leaves in $2.0\ \text{ms}$, the average power is $\dfrac{0.90\ \text{J}}{2.0 \times 10^{-3}\ \text{s}} = 450\ \text{W}$. A small battery could never give that much power directly.

A second example. A $4.0\ \mu\text{F}$ capacitor holds $80\ \mu\text{C}$. What is the energy? We know $Q$ and $C$, so

Check. The voltage is $\dfrac{80\ \mu\text{C}}{4.0\ \mu\text{F}} = 20\ \text{V}$, and $\tfrac{1}{2} \times 4.0 \times 10^{-6}\ \text{F} \times (20\ \text{V})^2 = 0.80\ \text{mJ}$.

The battery does twice the work

A battery of voltage $V$ moves a charge $Q$ round the circuit. Every unit of charge gains $V$ of energy from it, so the battery does work $QV$. That is twice the energy $\tfrac{1}{2}QV$ that the capacitor keeps.

Where does the other half go? It turns into heat in the wires and the resistor. This is true for any resistance, large or small. The half that is lost does not depend on it. The simulation shows the two curves.

The red curve is always twice the blue one. Now set the voltage to $100\ \text{V}$, and then to $200\ \text{V}$. By what factor does the stored energy change? Why?

Energy in series and in parallel

Two capacitors of $2.0\ \mu\text{F}$ and $3.0\ \mu\text{F}$ are joined to a $10\ \text{V}$ battery. Let us see where the energy sits.

In parallel. Both have the full $10\ \text{V}$, so $U = \tfrac{1}{2}CV^2$ is proportional to $C$.

• $U_1 = \tfrac{1}{2} \times 2.0\ \mu\text{F} \times (10\ \text{V})^2 = 100\ \mu\text{J}$.
• $U_2 = \tfrac{1}{2} \times 3.0\ \mu\text{F} \times (10\ \text{V})^2 = 150\ \mu\text{J}$.
• The total is $250\ \mu\text{J}$. The bigger capacitor stores more.

In series. Now the charge is the same. The equivalent is $1.2\ \mu\text{F}$ and $Q = 1.2\ \mu\text{F} \times 10\ \text{V} = 12\ \mu\text{C}$. So $U = \dfrac{Q^2}{2C}$ is proportional to $\dfrac{1}{C}$.

• $U_1 = \dfrac{(12\ \mu\text{C})^2}{2 \times 2.0\ \mu\text{F}} = 36\ \mu\text{J}$.
• $U_2 = \dfrac{(12\ \mu\text{C})^2}{2 \times 3.0\ \mu\text{F}} = 24\ \mu\text{J}$.
• The total is $60\ \mu\text{J}$, which equals $\tfrac{1}{2} \times 1.2\ \mu\text{F} \times (10\ \text{V})^2$. The smaller capacitor stores more.

The rule: in parallel the energy is shared in the ratio of the capacitances. In series it is shared in the inverse ratio. The same battery stores $250\ \mu\text{J}$ in parallel and only $60\ \mu\text{J}$ in series.

Where is the energy kept?

Take a parallel-plate capacitor with plates of area $A$ and gap $d$. Its capacitance is $C = \dfrac{\varepsilon_0A}{d}$, and the field between the plates is $E = V/d$. Put these in $U = \tfrac{1}{2}CV^2$:

The product $Ad$ is the volume between the plates. So the energy for each unit of volume is

This is the energy density of an electric field. It was found for a capacitor, but it holds for any electric field in empty space. The energy is stored in the field itself, in the space between the plates.

A worked example. The field in a region of air is $2.0 \times 10^5\ \text{V/m}$. Then

The pull between the plates

The plates carry opposite charges, so they attract each other. We can find the force from the energy. Take an isolated capacitor with charge $Q$, and pull the plates a little further apart, by $dx$. Since $C = \dfrac{\varepsilon_0A}{x}$, the energy is $U = \dfrac{Q^2x}{2\varepsilon_0A}$. It grows by $dU = \dfrac{Q^2}{2\varepsilon_0A}\,dx$. This is the work done by the force that holds the plates, $F\,dx$. So

where $E = \dfrac{Q}{\varepsilon_0A}$ is the field between the plates. Why a half? The field $E$ between the plates comes from both plates together. A plate does not pull on itself. It feels only the field of the other plate, which is $\tfrac{1}{2}E$.

A worked example. The plates of an isolated capacitor have an area of $0.020\ \text{m}^2$ and a charge of $0.40\ \mu\text{C}$.

1. The field. $E = \dfrac{Q}{\varepsilon_0A} = \dfrac{0.40 \times 10^{-6}\ \text{C}}{8.85 \times 10^{-12}\ \text{C}^2/(\text{N}\,\text{m}^2) \times 0.020\ \text{m}^2} = 2.26 \times 10^6\ \text{V/m}$.
2. The force. $F = \tfrac{1}{2}QE = \tfrac{1}{2} \times 0.40 \times 10^{-6}\ \text{C} \times 2.26 \times 10^6\ \text{V/m} = 0.45\ \text{N}$.

Practise with new numbers

Try it in Python

A computer can add up the work of charging, packet by packet. The program does it with a few steps, and then with many.

Common mistakes

• Writing $U = QV$. The energy stored is $\tfrac{1}{2}QV$. The work done by a battery is $QV$.
• Forgetting to square the voltage. In $\tfrac{1}{2}CV^2$ the voltage is squared. Doubling $V$ makes the energy four times larger.
• Mixing units. With $C$ in $\mu\text{F}$ and $V$ in $\text{V}$, the result of $\tfrac{1}{2}CV^2$ is in $\mu\text{J}$. Convert before you compare with joules.
• Sharing the energy equally. In parallel the energy goes in proportion to $C$. In series it goes in proportion to $\dfrac{1}{C}$.
• Taking the force as $QE$. The force on one plate is $\tfrac{1}{2}QE$, because a plate does not feel its own field.

Remember these

• Energy stored: $U = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \dfrac{Q^2}{2C}$.
• Battery work: $QV$, which is twice $U$. The other half is lost as heat.
• Two capacitors on one battery: in parallel $U \propto C$, and in series $U \propto \dfrac{1}{C}$.
• Energy density: $u = \tfrac{1}{2}\varepsilon_0E^2$.
• Force between the plates: $F = \dfrac{Q^2}{2\varepsilon_0A} = \tfrac{1}{2}QE$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A $40\ \mu\text{F}$ capacitor is charged to $25\ \text{V}$. How much energy does it store?

2. The voltage across a capacitor is doubled. By what factor does the stored energy change?

3. A capacitor holds $60\ \mu\text{C}$ and stores $90\ \mu\text{J}$. What is its capacitance?

4. A battery charges a capacitor through a resistor. The battery does $0.50\ \text{J}$ of work. How much energy is stored, and where does the rest go?

5. Two capacitors, $C$ and $3C$, are joined in series across a battery. Which one stores more energy?