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

Capacitors and Capacitance

Learning Objective

Determine the electric potential energy stored in a capacitor by evaluating the area under the potential–charge (V‑Q) graph.

Key Concepts

• Capacitance \$C\$ is the ratio of charge \$Q\$ on a conductor to the potential difference \$V\$ across it: \$C = \dfrac{Q}{V}\$.
• For a linear (ideal) capacitor \$C\$ is constant, giving a straight‑line V‑Q relationship passing through the origin.
• The electric potential energy \$U\$ stored in a capacitor is the work required to move charge from one plate to the other.

V‑Q Graph and Energy

On a \$V\$ (vertical) versus \$Q\$ (horizontal) graph, the area under the curve from \$Q=0\$ to \$Q=Q_f\$ represents the work done, i.e. the stored energy \$U\$:

\$U = \int{0}^{Qf} V \, dQ\$

For an ideal capacitor where \$V = \dfrac{Q}{C}\$, the integral becomes:

\$\$U = \int{0}^{Qf} \frac{Q}{C}\, dQ = \frac{1}{C}\left[\frac{Q^{2}}{2}\right]{0}^{Qf}

= \frac{Qf^{2}}{2C} = \frac{1}{2} C Vf^{2} = \frac{1}{2} Qf Vf\$\$

Step‑by‑Step Procedure

1. Identify the final charge \$Qf\$ (or final voltage \$Vf\$) on the capacitor.
2. Confirm whether the capacitor behaves linearly (constant \$C\$). If not, the V‑Q curve will be non‑linear and the integral must be evaluated accordingly.
3. Calculate the area under the V‑Q curve:

   • For a straight line: use the area of a triangle, \$U = \tfrac{1}{2} Qf Vf\$.
   • For a curved line: perform the integral \$U = \int V\,dQ\$ analytically or numerically.

4. Express the result in joules (J). Remember \$1\ \text{J} = 1\ \text{V·C}\$.

Example Calculation (Ideal Capacitor)

Given: \$C = 5.0\ \mu\text{F}\$, final voltage \$V_f = 200\ \text{V}\$.

First find the final charge:

\$Qf = C Vf = 5.0 \times 10^{-6}\,\text{F} \times 200\,\text{V} = 1.0 \times 10^{-3}\,\text{C}\$

Energy stored (area of triangle):

\$U = \frac{1}{2} Qf Vf = \frac{1}{2} (1.0 \times 10^{-3}\,\text{C})(200\,\text{V}) = 0.10\ \text{J}\$

Non‑Linear Example (Variable Capacitance)

If \$C\$ varies with voltage such that \$V = k Q^{2}\$ (where \$k\$ is a constant), then:

\$\$U = \int{0}^{Qf} k Q^{2}\, dQ = k \left[\frac{Q^{3}}{3}\right]{0}^{Qf}

= \frac{k Q_f^{3}}{3}\$\$

In this case the V‑Q graph is a curve, and the area must be evaluated using the appropriate integral.

Common Pitfalls

• Confusing the area under a \$Q\$‑V graph (which gives energy) with the slope of the graph (which gives capacitance).
• Using \$U = \tfrac{1}{2} C V^{2}\$ without checking that \$C\$ is constant over the charge range.
• Neglecting units: ensure \$\mu\text{F}\$ is converted to farads before calculations.

Summary Table

Practice Questions

1. A \$10\ \mu\text{F}\$ capacitor is charged to \$150\ \text{V}\$. Calculate the stored energy using the area under the V‑Q graph.
2. The V‑Q graph of a device is a straight line from \$(0,0)\$ to \$(2.0\ \text{C}, 40\ \text{V})\$ and then continues horizontally at \$40\ \text{V}\$ up to \$Q = 3.0\ \text{C}\$. Find the total energy stored when \$Q = 3.0\ \text{C}\$.
3. For a capacitor whose voltage varies with charge as \$V = 5 Q^{0.5}\$ (with \$V\$ in volts and \$Q\$ in coulombs), determine the energy stored when \$Q = 4\ \text{C}\$.

Answers (for teacher reference)

• 1. \$Q = CV = 10\times10^{-6}\times150 = 1.5\times10^{-3}\ \text{C}\$; \$U = \tfrac{1}{2}QV = 0.1125\ \text{J}\$.
• 2. Energy = area of triangle (0–2 C) + area of rectangle (2–3 C). Triangle: \$\tfrac12(2)(40)=40\ \text{J}\$. Rectangle: \$(1)(40)=40\ \text{J}\$. Total \$U = 80\ \text{J}\$.
• 3. \$U = \int{0}^{4}5 Q^{0.5}\,dQ = 5\left[\frac{2}{3}Q^{3/2}\right]{0}^{4}= \frac{10}{3}(8)=\frac{80}{3}\ \text{J}\approx26.7\ \text{J}\$.