determine the electric potential energy stored in a capacitor from the area under the potential–charge graph

Capacitors & Capacitance – A‑Level Physics 9702

What you’ll learn today

  • How to read a potential–charge (V–Q) graph for a capacitor.

  • Why the area under the V–Q curve equals the electric potential energy stored.

  • How to calculate that area for both linear and non‑linear graphs.

  • Exam‑style questions and tips for tackling them.

1️⃣ Capacitor Basics

A capacitor stores electric charge on two conductive plates separated by an insulator (dielectric). The key relationship is

\$ Q = C V \$

where \$Q\$ is charge (C), \$V\$ is voltage (V), and \$C\$ is capacitance (F).

Think of a capacitor as a water tank that can hold a certain amount of water (charge) for a given pressure (voltage). The tank’s size is the capacitance.

2️⃣ Potential–Charge Graph

On a V–Q graph, the x‑axis is charge \$Q\$, the y‑axis is voltage \$V\$.

For an ideal capacitor, the graph is a straight line through the origin because \$V = Q/C\$.

📐 Area under the curve represents the work done (energy) to charge the capacitor. Mathematically:

\$ U = \int_0^{Q} V \, dQ \$

This integral is the same as the area under the V–Q plot.

3️⃣ Calculating Energy – Linear Case

For a straight‑line V–Q graph, the area is a right triangle:

  • Base = final charge \$Q\$

  • Height = final voltage \$V\$

So,

\$ U = \frac{1}{2} Q V \$

Using the capacitor equation \$V = Q/C\$, we get the familiar result:

\$ U = \frac{1}{2} C V^2 = \frac{Q^2}{2C} \$

4️⃣ Calculating Energy – Non‑Linear Case

If the V–Q relationship is not linear (e.g., a dielectric that changes with voltage), you must integrate the actual curve:

\$ U = \int0^{Q{\text{max}}} V(Q) \, dQ \$

Example: Suppose \$V = k Q^2\$ (a quadratic relationship). Then

\$ U = \int0^{Q{\text{max}}} k Q^2 \, dQ = k \frac{Q_{\text{max}}^3}{3} \$

Always check the graph to identify the correct functional form before integrating.

5️⃣ Example Problem

  1. A capacitor has a capacitance of \$10\,\mu\text{F}\$. It is charged to \$5\,\text{V}\$. Find the stored energy.

  2. Sketch the V–Q graph and verify that the area equals your calculated energy.

Solution:

Using \$U = \frac{1}{2} C V^2\$:

\$ U = \frac{1}{2} (10 \times 10^{-6}\,\text{F}) (5\,\text{V})^2 = 0.125\,\text{J} \$

The graph is a straight line from (0,0) to (50 µC, 5 V). The area of the triangle is

\$ \frac{1}{2} \times 50\,\mu\text{C} \times 5\,\text{V} = 0.125\,\text{J} \$

??

The two methods agree.

6️⃣ Exam Tips Box

⚡️ Remember:

  • Energy = area under the V–Q curve.

  • For linear graphs, use the triangle formula.

  • Always check units – J = C·V.

  • When a graph is non‑linear, set up the integral before calculating.

  • Show all steps in your answer; partial credit is awarded for correct reasoning.

7️⃣ Quick Reference Table

CaseEnergy FormulaGraph Shape
Linear (ideal capacitor)\$U = \frac{1}{2} C V^2 = \frac{Q^2}{2C}\$Straight line through origin
Quadratic (e.g., \$V = kQ^2\$)\$U = \frac{k Q_{\text{max}}^3}{3}\$Parabolic curve
General case\$U = \int0^{Q{\text{max}}} V(Q) \, dQ\$Depends on material properties

8️⃣ Summary

The electric potential energy stored in a capacitor is directly linked to the area under its V–Q graph.

• For linear graphs, use the triangle area formula.

• For non‑linear graphs, set up and evaluate the integral.

• Always double‑check units and show your work – examiners love clear reasoning.

Keep practicing with different graph shapes, and you’ll master this topic in no time! 🚀