Energy in a Capacitor Revision Notes | A Level AQA Physics

1. Why Capacitors Store Energy — and Why It Is Not QV

Picture a capacitor being charged from a battery. The first tiny bit of charge flows onto the plates almost effortlessly — there is no opposing electric field yet. But as charge builds up, the field between the plates grows stronger, and each subsequent charge element has to be pushed against an increasing repulsive force. The battery has to do more and more work as charging progresses.

This is the key physical insight: the pd across the capacitor is not constant during charging — it rises from zero to its final value $V$. Because the pd changes, you cannot calculate the total work done by multiplying the final charge $Q$ by the final pd $V$. That would overestimate the energy stored. The correct answer is exactly half of that product.

Understanding why the factor of $\frac{1}{2}$ appears — and being able to derive it from a graph — is precisely what AQA examiners test. The graphical argument is the most elegant route, so that is where we start.

2. The Graphical Derivation — Area Under a Q-V Graph

Plot charge $Q$ on the vertical axis and potential difference $V$ on the horizontal axis for a capacitor. Because $Q = CV$ and $C$ is constant for a given capacitor, this graph is a straight line through the origin with gradient $C$.

The area under this straight line — between $V = 0$ and the final pd $V$ — is a triangle. Its base is $V$ and its height is the final charge $Q$. The area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$, which gives $\frac{1}{2}QV$. This area represents the total work done in charging the capacitor, and therefore equals the energy stored.

This graphical argument is the AQA-required derivation. Each thin vertical strip of the graph represents the small amount of work $\delta W = V \, \delta Q$ done in pushing an infinitesimal charge $\delta Q$ onto the plates against the current pd $V$. Summing all those strips gives the total area — the triangle — confirming $E = \frac{1}{2}QV$.

3. The Three Energy Equations and When to Use Each

Starting from $E = \frac{1}{2}QV$ and using the definition $C = Q/V$ (so $Q = CV$ and $V = Q/C$), you can write the energy in three equivalent forms. Each form is most convenient depending on which two quantities you are given in a problem.

Notice the important asymmetry in $E = \frac{1}{2}CV^2$: energy is proportional to $V^2$. Doubling the pd quadruples the energy stored. This has real consequences — a capacitor charged to 10 V stores four times the energy of one charged to 5 V, even with the same capacitance.

By contrast, $E = \frac{Q^2}{2C}$ shows that for a fixed charge, increasing capacitance reduces the energy stored. This might seem counterintuitive, but a larger capacitor holds the same charge at a lower pd, so less work was done pushing that charge onto the plates.

4. Deriving the Equivalent Forms — Step by Step

AQA mark schemes sometimes award a method mark for showing the substitution that converts one form into another. The derivation is short, but you need to be fluent with it under exam conditions.

Worked example: Deriving $E = \frac{1}{2}CV^2$ and $E = \frac{Q^2}{2C}$ from $E = \frac{1}{2}QV$

Starting from $E = \frac{1}{2}QV$, derive the two alternative expressions for energy stored in a capacitor.

1. $C = \frac{Q}{V}$, so $Q = CV$ and $V = \frac{Q}{C}$ — These two rearrangements are the substitutions needed. Write them out explicitly — this earns the method mark.
2. $E = \frac{1}{2}QV = \frac{1}{2}(CV)V = \frac{1}{2}CV^2$ — Replace $Q$ with $CV$. The result shows energy depends on the square of pd.
3. $E = \frac{1}{2}QV = \frac{1}{2}Q \cdot \frac{Q}{C} = \frac{1}{2} \cdot \frac{Q^2}{C} = \frac{Q^2}{2C}$ — Replace $V$ with $Q/C$. The result shows energy depends on the square of charge.
4. $[C][V^2] = \text{F} \cdot \text{V}^2 = \frac{\text{C}}{\text{V}} \cdot \text{V}^2 = \text{C} \cdot \text{V} = \text{J}$ — Since $1\,\text{F} = 1\,\text{C}\,\text{V}^{-1}$, the unit check confirms the result is in joules. Always verify units in a derivation.

Answer: $E = \frac{1}{2}CV^2$ (substitute $Q = CV$) and $E = \frac{Q^2}{2C}$ (substitute $V = Q/C$), both derived from $E = \frac{1}{2}QV$.

5. Worked Calculations — Applying the Equations

Exam questions on this topic almost always give you two of the three quantities ($Q$, $V$, $C$) and ask you to find the energy stored. The skill is selecting the right form of the equation so you do not need an extra step. Here are two representative examples.

Worked example: Calculate energy stored given capacitance and pd

A capacitor of capacitance $470\,\mu\text{F}$ is charged to a potential difference of $12\,\text{V}$. Calculate the energy stored. [2 marks]

1. $C = 470\,\mu\text{F} = 470 \times 10^{-6}\,\text{F} = 4.70 \times 10^{-4}\,\text{F}$ — Always convert to SI base units (farads) before substituting. Leaving the value in microfarads is a common source of error.
2. $E = \frac{1}{2}CV^2$ — We have $C$ and $V$, so this form avoids needing to calculate $Q$ first. [M1]
3. $E = \frac{1}{2} \times 4.70 \times 10^{-4} \times 12^2$ — Write out the substitution explicitly — this is where the method mark is awarded.
4. $12^2 = 12 \times 12 = 144$ — Expand the square explicitly. Do not skip this step — writing $12^2 = 144$ in one line is fine, but verify it: $12 \times 12 = 144$. Confirmed.
5. $E = \frac{1}{2} \times 4.70 \times 10^{-4} \times 144 = \frac{1}{2} \times 6.768 \times 10^{-2} = 3.384 \times 10^{-2}\,\text{J}$ — $4.70 \times 10^{-4} \times 144 = 6.768 \times 10^{-2}$; halving gives $3.384 \times 10^{-2}\,\text{J}$. [A1]
6. $E = 3.38 \times 10^{-2}\,\text{J}$ (to 3 s.f.) or equivalently $33.8\,\text{mJ}$ — Round to 3 significant figures to match the data given. Include units — omitting them loses the accuracy mark.

Answer: $E = 3.38 \times 10^{-2}\,\text{J}$ (33.8 mJ)

Worked example: Calculate energy stored given charge and capacitance

A capacitor stores a charge of $6.0 \times 10^{-4}\,\text{C}$ and has a capacitance of $200\,\mu\text{F}$. Calculate the energy stored. [2 marks]

1. $C = 200\,\mu\text{F} = 200 \times 10^{-6}\,\text{F} = 2.00 \times 10^{-4}\,\text{F}$ — Convert first to avoid unit errors in the final answer.
2. $E = \frac{Q^2}{2C}$ — We have $Q$ and $C$ but not $V$, so this form is most direct. [M1]
3. $Q^2 = (6.0 \times 10^{-4})^2 = 6.0 \times 10^{-4} \times 6.0 \times 10^{-4} = 36 \times 10^{-8} = 3.6 \times 10^{-7}\,\text{C}^2$ — $6.0 \times 6.0 = 36$ and $10^{-4} \times 10^{-4} = 10^{-8}$, giving $3.6 \times 10^{-7}$. Verified by expanding the multiplication.
4. $E = \frac{3.6 \times 10^{-7}}{2 \times 2.00 \times 10^{-4}} = \frac{3.6 \times 10^{-7}}{4.00 \times 10^{-4}}$ — The denominator is $2C = 2 \times 2.00 \times 10^{-4} = 4.00 \times 10^{-4}$.
5. $E = \frac{3.6}{4.00} \times 10^{-7-(-4)} = 0.90 \times 10^{-3} = 9.0 \times 10^{-4}\,\text{J}$ — $3.6 \div 4.00 = 0.90$ and $10^{-7} \div 10^{-4} = 10^{-3}$. So $E = 9.0 \times 10^{-4}\,\text{J}$. [A1]

Answer: $E = 9.0 \times 10^{-4}\,\text{J}$ (0.90 mJ)

6. Interpreting the Q-V Graph in Exam Questions

AQA questions sometimes present a Q-V graph and ask you to find the energy stored by calculating the area under the line. The approach is identical to finding the area of a triangle, but you must read the coordinates carefully from the axes.

If the graph shows a straight line from the origin to the point $(V_f, Q_f)$, the energy stored is the area of the triangle: $E = \frac{1}{2} \times V_f \times Q_f$. This is exactly $\frac{1}{2}QV$ — the graphical and algebraic approaches agree perfectly, which is a useful self-check.

A subtler question might give you a graph where the capacitor is charged from $V_1$ to $V_2$, and ask for the energy stored during that interval. In that case, the relevant area is a trapezium, not a triangle. You would calculate the area of the trapezium (or subtract the smaller triangle from the larger one) to find the energy transferred during that charging interval.

7. Exam Technique and Common Mistakes

The energy equations are straightforward to apply, but there are several reliable ways to lose marks. The following guidance is drawn directly from AQA mark scheme patterns and examiner reports.

Taken together, the three energy equations — $E = \frac{1}{2}QV$, $E = \frac{1}{2}CV^2$, and $E = \frac{Q^2}{2C}$ — are all equivalent expressions for the same physical quantity, each most useful depending on which pair of variables you are given. The graphical derivation from the area under the Q-V graph underpins all three and is the AQA-required route to understanding why the $\frac{1}{2}$ factor is always present.