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

Capacitors and Capacitance – Cambridge IGCSE/A‑Level (9702)

Learning Objective

Use the area under a potential–charge (V‑Q) graph to determine the electric potential energy stored in a capacitor. Relate this method to the syllabus topics of capacitance, dielectrics, series/parallel combinations and energy density.

1. Core Definitions (Topic 19.1 a‑d)

• Capacitance:
  

  Definition in words – the ability of a system to store charge per unit potential difference.
  

  Formula – \(C=\dfrac{Q}{V}\) [farads, F]

• Parallel‑plate capacitor: \(C=\dfrac{\varepsilon{0}A}{d}\) where \(\varepsilon{0}=8.85\times10^{-12}\,\text{F m}^{-1}\).
• Isolated spherical conductor (Topic 19.1 b):
  

  \(C=4\pi\varepsilon_{0}r\).
  

  Example: For a metal sphere of radius \(r=5.0\;\text{cm}\),
  

  \(C=4\pi(8.85\times10^{-12})(0.05)=5.6\times10^{-12}\,\text{F}=5.6\;\text{pF}\).

• Dielectric effect (Topic 19.2 b): Inserting a dielectric of constant \(\kappa\) multiplies the capacitance by \(\kappa\):
  

  \(C{\text{new}}=\kappa C{\text{old}}\).

2. The V‑Q Graph and Stored Energy (Topic 19.1 c)

On a graph of potential \(V\) (vertical) against charge \(Q\) (horizontal) the area under the curve from \(Q=0\) to a final charge \(Q_f\) equals the work done in moving the charge – i.e. the stored electric potential energy \(U\):

\[

U=\int{0}^{Qf}V\,\mathrm{d}Q

\]

2.1 Ideal (linear) capacitor – constant \(C\)

For a constant capacitance, \(V=\dfrac{Q}{C}\); the graph is a straight line through the origin.

\[

U=\int{0}^{Qf}\frac{Q}{C}\,\mathrm{d}Q

=\frac{1}{C}\left[\frac{Q^{2}}{2}\right]{0}^{Qf}

=\frac{Q_f^{2}}{2C}

=\frac{1}{2}CV_f^{2}

=\frac{1}{2}QfVf

\]

Graphically this is the area of a right‑angled triangle (base = \(Qf\), height = \(Vf\)).

2.2 Linear V‑Q with offset – non‑zero intercept

If the line does not pass through the origin, the equation is \(V=aQ+b\) (with \(b\neq0\)). The area consists of a triangle plus a rectangle (a trapezoid):

\[

U=\frac{a}{2}Qf^{2}+bQf

\]

2.3 Non‑linear (voltage‑dependent) capacitance – curved graph

When the capacitance varies with voltage, the V‑Q relationship is a curve \(V=f(Q)\). The stored energy is obtained by direct integration:

\[

U=\int{0}^{Qf}f(Q)\,\mathrm{d}Q

\]

In practice the curve may be given analytically (e.g. \(V=kQ^{2}\) or \(V=5\sqrt{Q}\)) or as tabulated data (use the trapezoidal rule).

3. Quick Reference – Links to the Rest of the Syllabus

4. Unit‑Check Reminder

• Capacitance: \(\mu\text{F}=10^{-6}\,\text{F},\; \text{nF}=10^{-9}\,\text{F},\; \text{pF}=10^{-12}\,\text{F}\).
• Energy: \(1\ \text{J}=1\ \text{V·C}\).
• When using \(\tfrac12 QV\), ensure \(Q\) is in coulombs and \(V\) in volts.

5. Step‑by‑Step Procedure for Energy from a V‑Q Graph

1. Read the final charge or voltage from the graph (\(Qf\) and/or \(Vf\)).
2. Identify the shape of the curve

   • Straight line through the origin → ideal capacitor (triangle).
   • Straight line with intercept → trapezoid (triangle + rectangle).
   • Curved line → write the functional form \(V=f(Q)\) or use numerical integration.

3. Calculate the area

   • Triangle: \(U=\tfrac12 QfVf\).
   • Trapezoid: \(U=\tfrac12 (V1+V2)Q_f\) (average height × base).
   • Curve: evaluate \(\displaystyle U=\int{0}^{Qf}f(Q)\,\mathrm{d}Q\).

4. Convert to joules and, if required, express per unit volume using \(u=\tfrac12\varepsilon E^{2}\).

6. Summary Tables

6.1 Energy Formulas by V‑Q Relationship

6.2 Series & Parallel Combination Rules (Topic 19.1 d)

7. Worked Examples

Example 1 – Ideal Capacitor (Triangle)

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

1. Convert: \(C=5.0\times10^{-6}\;\text{F}\).
2. Charge: \(Qf=CVf=5.0\times10^{-6}\times200=1.0\times10^{-3}\;\text{C}\).
3. Energy (area): \(U=\tfrac12 QfVf=\tfrac12(1.0\times10^{-3})(200)=0.10\;\text{J}\).

Example 2 – Linear V‑Q with Offset (Trapezoid)

The graph is a line from \((0,0)\) to \((2.0\;\text{C},40\;\text{V})\) and then stays horizontal at \(40\;\text{V}\) up to \(Q=3.0\;\text{C}\).

• Triangle (0–2 C): \(U_1=\tfrac12(2)(40)=40\;\text{J}\).
• Rectangle (2–3 C): \(U_2=(1)(40)=40\;\text{J}\).
• Total \(U=U1+U2=80\;\text{J}\).

Example 3 – Voltage‑Dependent Capacitance (Curved Graph)

Capacitor obeys \(V=5\sqrt{Q}\) (V in volts, Q in coulombs). Find \(U\) for \(Q_f=4\;\text{C}\).

\[

U=\int_{0}^{4}5Q^{1/2}\,\mathrm{d}Q

=5\left[\frac{2}{3}Q^{3/2}\right]_{0}^{4}

=\frac{10}{3}\times8=\frac{80}{3}\;\text{J}\approx26.7\;\text{J}

\]

Example 4 – Isolated Spherical Conductor (Topic 19.1 b)

A metal sphere of radius \(r=8.0\;\text{cm}\) is charged to a potential of \(V=500\;\text{V}\). Find the stored energy.

1. Capacitance: \(C=4\pi\varepsilon_{0}r=4\pi(8.85\times10^{-12})(0.08)=8.9\times10^{-12}\;\text{F}\).
2. Charge: \(Q=CV=8.9\times10^{-12}\times500=4.45\times10^{-9}\;\text{C}\).
3. Energy: \(U=\tfrac12 QV=\tfrac12(4.45\times10^{-9})(500)=1.11\times10^{-6}\;\text{J}\).

8. Practice Questions (with marks)

1. (2 marks) A \(10\;\mu\text{F}\) capacitor is charged to \(150\;\text{V}\). Compute the stored energy using the V‑Q area method.
2. (3 marks) Three capacitors are connected in series: \(C1=2\;\mu\text{F}\), \(C2=3\;\mu\text{F}\), \(C_3=6\;\mu\text{F}\). The combination is then charged to \(120\;\text{V}\). Find the total energy stored.
3. (5 marks) A parallel‑plate capacitor has plate area \(A=0.02\;\text{m}^2\) and separation \(d=1.0\;\text{mm}\). A dielectric with \(\kappa=4\) fills the space. Determine:

   • its capacitance,
   • the energy stored when the voltage across it is \(250\;\text{V}\), and
   • the energy density inside the dielectric.

4. (2 marks) The V‑Q graph of a device is given by \(V=3Q^{2}\) (V in volts, Q in coulombs) up to \(Q=0.5\;\text{C}\). Calculate the energy stored.
5. (2 marks) A spherical conductor of radius \(r=5.0\;\text{cm}\) is at a potential of \(400\;\text{V}\). Find the energy stored.

9. Answers – Teacher’s Mark Scheme

1. \(Q=CV=10\times10^{-6}\times150=1.5\times10^{-3}\;\text{C}\);
   

   \(U=\tfrac12QV=0.1125\;\text{J}\).

2. Series equivalent: \(\displaystyle\frac{1}{C{\text{eq}}}= \frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1\;\Rightarrow\;C{\text{eq}}=1\;\mu\text{F}\).
   

   Energy: \(U=\tfrac12C_{\text{eq}}V^{2}= \tfrac12(1\times10^{-6})(120)^{2}=7.2\times10^{-3}\;\text{J}\).

3. • Without dielectric: \(C{0}= \varepsilon{0}A/d=8.85\times10^{-12}\times0.02/1.0\times10^{-3}=1.77\times10^{-10}\;\text{F}\).
   • With dielectric: \(C=\kappa C_{0}=4\times1.77\times10^{-10}=7.08\times10^{-10}\;\text{F}\).
   • Energy: \(U=\tfrac12CV^{2}=0.5\times7.08\times10^{-10}\times(250)^{2}=2.21\times10^{-5}\;\text{J}\).
   • Electric field: \(E=V/d=250/(1.0\times10^{-3})=2.5\times10^{5}\;\text{V m}^{-1}\).
   • Energy density: \(u=\tfrac12\varepsilon E^{2}=0.5\,(4\varepsilon_{0})\,(2.5\times10^{5})^{2}=1.11\times10^{-2}\;\text{J m}^{-3}\).

4. \(U=\int{0}^{0.5}3Q^{2}\,\mathrm{d}Q=3\left[\frac{Q^{3}}{3}\right]{0}^{0.5}=0.125\;\text{J}\).
5. Capacitance: \(C=4\pi\varepsilon_{0}r=4\pi(8.85\times10^{-12})(0.05)=5.6\times10^{-12}\;\text{F}\).
   

   Charge: \(Q=CV=5.6\times10^{-12}\times400=2.24\times10^{-9}\;\text{C}\).
   

   Energy: \(U=\tfrac12QV=\tfrac12(2.24\times10^{-9})(400)=4.48\times10^{-7}\;\text{J}\).

Concept‑Check Box

Q1. In your own words, state what capacitance measures.

Q2. Why does the area under a V‑Q graph represent energy? (Hint: think of work = force × distance, but for charge.)