Energy stored in a capacitor

1. Overview

A capacitor stores energy by separating opposite charges onto its plates, which establishes an electric field in the dielectric material or space between them. The process of charging a capacitor is not instantaneous; it requires work done by an external source (such as a battery) to push electrons onto a plate that is already becoming negatively charged, and to pull electrons away from a plate that is becoming positively charged. This work done against the electrostatic repulsion of existing charges is stored as electric potential energy within the electric field. When the capacitor is connected to a load, this stored energy is released as the charges flow back together, neutralizing the plates.

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Key Definitions

• Capacitance ($C$): The charge stored on one plate of a capacitor per unit potential difference across the plates. It is defined by the ratio $C = \frac{Q}{V}$ and is measured in farads (F).
• Electric Potential Energy ($W$): The total work done in transferring charge from one plate to the other during the charging process. This energy is stored in the electric field between the plates.
• Potential Difference ($V$): The work done per unit charge in moving a small test charge between two points in an electric field. In the context of a capacitor, it is the voltage across the plates.

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Content

3.1 The Potential–Charge ($V$–$Q$) Graph

To determine the energy stored in a capacitor, we must analyze the relationship between the potential difference ($V$) across the plates and the charge ($Q$) stored on them.

For any capacitor, the capacitance $C$ is a constant determined by its physical geometry. From the definition $C = \frac{Q}{V}$, we can see that $V$ is directly proportional to $Q$: $$V = \frac{1}{C} \times Q$$

This linear relationship means that as more charge is added to the plates, the potential difference increases at a constant rate.

The Area Under the Graph If we plot a graph with Charge ($Q$) on the x-axis and Potential Difference ($V$) on the y-axis, the resulting line is a straight line passing through the origin.

• The gradient of a $V$–$Q$ graph represents $\frac{1}{C}$.
• The area under the line represents the total work done (energy stored).

Consider adding a very small increment of charge $\Delta Q$ to a capacitor that already has a potential difference $V$. The small amount of work done $\Delta W$ is: $$\Delta W = V \Delta Q$$

As the capacitor charges from zero to a total charge $Q$, the total work done $W$ is the sum of all these small increments. Geometrically, this is the area of the triangle formed under the $V$–$Q$ line.

[DIAGRAM DESCRIPTION: A graph showing Potential Difference $V$ on the vertical y-axis and Charge $Q$ on the horizontal x-axis. A straight line starts at $(0,0)$ and rises to a point $(Q, V)$. The area under this line is shaded as a right-angled triangle. The area of this triangle is labeled "Energy Stored ($W$)".]

3.2 Deriving the Energy Formulae

Since the area under the $V$–$Q$ graph is a triangle, we use the formula for the area of a triangle: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ $$\text{Area} = \frac{1}{2} \times Q \times V$$

This gives us our primary equation for the energy stored ($W$): $W = \frac{1}{2}QV$

By using the capacitance definition ($Q = CV$), we can derive two other useful forms of this equation to suit different problem contexts.

1. Energy in terms of Capacitance and Voltage: Substitute $Q = CV$ into the primary equation: $$W = \frac{1}{2}(CV) \times V$$ $W = \frac{1}{2}CV^2$ (This is the most common form used in exams, as $C$ and $V$ are usually the known variables.)

2. Energy in terms of Charge and Capacitance: Substitute $V = \frac{Q}{C}$ into the primary equation: $$W = \frac{1}{2}Q \times \left(\frac{Q}{C}\right)$$ $W = \frac{Q^2}{2C}$ (This form is particularly useful when capacitors are in series, as they store the same charge $Q$.)

3.3 Energy Dissipation during Charging

A critical concept in the 9702 syllabus is the "Energy Paradox" of capacitor charging. When a battery of constant e.m.f. $V$ charges a capacitor to that same potential difference $V$, the battery must move a total charge $Q$.

1. Work done by the battery: The battery provides a constant potential difference. The work done by the battery is $W_{battery} = QV$.
2. Energy stored in the capacitor: As derived above, the energy stored is $W_{capacitor} = \frac{1}{2}QV$.
3. The Energy Gap: There is a discrepancy of $\frac{1}{2}QV$.

Where does the missing energy go? Exactly 50% of the energy supplied by the battery is dissipated as thermal energy (heat) in the connecting wires and the internal resistance of the battery. This energy loss occurs regardless of the resistance of the circuit; a higher resistance simply means the charging process takes longer, but the total energy dissipated remains $\frac{1}{2}QV$.

3.4 Worked Examples

Worked Example 1 — Basic Energy Calculation

A camera flash unit uses a $120\text{ }\mu\text{F}$ capacitor charged to a potential difference of $250\text{ V}$. Calculate the energy stored in the capacitor.

• Step 1: Identify knowns and convert to SI units.
  • $C = 120\text{ }\mu\text{F} = 120 \times 10^{-6}\text{ F}$
  • $V = 250\text{ V}$
• Step 2: Select the appropriate equation.
  • $W = \frac{1}{2}CV^2$
• Step 3: Substitute and calculate.
  • $W = 0.5 \times (120 \times 10^{-6}) \times (250)^2$
  • $W = 0.5 \times 120 \times 10^{-6} \times 62500$
  • $W = 3.75\text{ J}$
• Answer: $W = 3.8\text{ J}$ (to 2 s.f.)

Worked Example 2 — Energy Change from Voltage Increase

A capacitor of capacitance $4.7\text{ nF}$ is initially charged to $10\text{ V}$. The potential difference is then increased to $30\text{ V}$. Calculate the additional work done to increase the charge.

• Step 1: Identify knowns.
  • $C = 4.7 \times 10^{-9}\text{ F}$
  • $V_1 = 10\text{ V}$
  • $V_2 = 30\text{ V}$
• Step 2: Set up the energy change equation.
  • $\Delta W = W_{final} - W_{initial}$
  • $\Delta W = \frac{1}{2}CV_2^2 - \frac{1}{2}CV_1^2 = \frac{1}{2}C(V_2^2 - V_1^2)$
• Step 3: Substitute and calculate.
  • $\Delta W = 0.5 \times (4.7 \times 10^{-9}) \times (30^2 - 10^2)$
  • $\Delta W = 0.5 \times 4.7 \times 10^{-9} \times (900 - 100)$
  • $\Delta W = 0.5 \times 4.7 \times 10^{-9} \times 800$
  • $\Delta W = 1.88 \times 10^{-6}\text{ J}$
• Answer: $\Delta W = 1.9 \times 10^{-6}\text{ J}$ (or $1.9\text{ }\mu\text{J}$)

Worked Example 3 — Finding Potential Difference from Energy

A $2200\text{ }\mu\text{F}$ capacitor stores $0.50\text{ J}$ of energy. Determine the potential difference across its plates.

• Step 1: Identify knowns.
  • $C = 2200 \times 10^{-6}\text{ F}$
  • $W = 0.50\text{ J}$
• Step 2: Rearrange the energy equation for $V$.
  • $W = \frac{1}{2}CV^2 \rightarrow V^2 = \frac{2W}{C} \rightarrow V = \sqrt{\frac{2W}{C}}$
• Step 3: Substitute and calculate.
  • $V = \sqrt{\frac{2 \times 0.50}{2200 \times 10^{-6}}}$
  • $V = \sqrt{\frac{1}{0.0022}}$
  • $V = \sqrt{454.54...}$
  • $V = 21.32\text{ V}$
• Answer: $V = 21\text{ V}$ (to 2 s.f.)

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Key Equations

Equation	Description	Data Sheet?
$W = \frac{1}{2}QV$	Energy stored using Charge and P.D.	Yes
$W = \frac{1}{2}CV^2$	Energy stored using Capacitance and P.D.	Yes
$W = \frac{Q^2}{2C}$	Energy stored using Charge and Capacitance	No (Derive)
$W = QV$	Total work done by a battery/power supply	No (Recall)
Area under $V$–$Q$ graph	Represents the Electric Potential Energy	No (Recall)

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Common Mistakes to Avoid

• ❌ Wrong: Using $W = QV$ for the energy stored in a capacitor.
  • ✓ Right: $W = QV$ is the energy supplied by the battery. The capacitor only stores half of this ($W = \frac{1}{2}QV$).
• ❌ Wrong: Forgetting to square the potential difference in $W = \frac{1}{2}CV^2$.
  • ✓ Right: This is a very common calculation error. Always write the formula clearly before substituting.
• ❌ Wrong: Calculating energy change as $\Delta W = \frac{1}{2}C(V_2 - V_1)^2$.
  • ✓ Right: You must square the voltages before subtracting: $\Delta W = \frac{1}{2}C(V_2^2 - V_1^2)$.
• ❌ Wrong: Misinterpreting the gradient of a $Q$–$V$ graph.
  • ✓ Right: On a $Q$ (y-axis) vs $V$ (x-axis) graph, the gradient is $C$. On a $V$ (y-axis) vs $Q$ (x-axis) graph, the gradient is $1/C$. However, the area represents energy in both cases.
• ❌ Wrong: Ignoring unit prefixes like $\text{p}$ (pico) or $\text{n}$ (nano).
  • ✓ Right: Always convert to standard SI units immediately: $1\text{ pF} = 10^{-12}\text{ F}$, $1\text{ nF} = 10^{-9}\text{ F}$, $1\text{ }\mu\text{F} = 10^{-6}\text{ F}$.

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Exam Tips

1. The "Why" of the 1/2: If asked why the energy stored is $\frac{1}{2}QV$ and not $QV$, explain that the potential difference across the capacitor increases linearly from $0$ to $V$ as it charges. Therefore, the average potential difference during the process is $\frac{1}{2}V$. Work done = Average P.D. $\times$ Total Charge = $\frac{1}{2}VQ$.
2. Proportionality Reasoning: Remember that $W \propto V^2$. If an exam question states the voltage is doubled, the energy stored increases by a factor of four ($2^2$). If the voltage is tripled, the energy increases by a factor of nine ($3^2$).
3. Graph Questions: If you are given a graph of $V$ against $Q$, and it is a curve rather than a straight line (which can happen with non-ohmic components, though rare for standard capacitors), you must count squares to find the area, as the $\frac{1}{2}QV$ triangle formula only applies to linear relationships.
4. Efficiency: If a question asks for the efficiency of the charging process, the theoretical maximum is $50\%$.
5. Significant Figures: Cambridge 9702 is strict on significant figures. Look at the data provided in the question. If the voltage is "$12\text{ V}$" (2 s.f.) and capacitance is "$470\text{ }\mu\text{F}$" (2 s.f.), your final answer must be to 2 s.f.
6. Discharging: When a capacitor discharges through a resistor, all the stored energy ($\frac{1}{2}CV^2$) is eventually converted into thermal energy in that resistor. This is a common way to link Topic 19 (Capacitance) with Topic 14 (Temperature/Heat).