Calculate energy stored in a capacitor
Find capacitor energy from the area under a potential–charge graph and use the equivalent charge–voltage formulas.
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 19. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use charge, potential difference, current, resistance, graph area, energy and SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the shape of the area under a linear V–Q charging graph from the origin?
What you will learn
- Determine stored energy from the area under a potential–charge graph
- Use W = ½QV
- Use W = ½CV²
Energy is the area under the charging graph
For a capacitor whose Q–V relationship is linear from the origin, the stored electric potential energy is the area under the potential–charge graph. The graph is a triangle, so W = ½QV at the final charge and potential difference.
Use Q in coulombs and V in volts to obtain joules because C V = J. On a graph with V vertical and Q horizontal, read the final coordinates and calculate the triangular area.
Use capacitance to express the same energy
Since Q = CV, substituting into W = ½QV gives W = ½CV². Both formulas describe the same stored energy. Choose the form that matches the quantities supplied, and convert microfarads before using SI units.
The straight-line graph means potential difference rises in proportion to charge. The energy is not the rectangular area QV, because the potential difference increases from zero as the capacitor charges.
Pause and explain
A 4.0 μF capacitor is charged to 10 V. What is its stored energy?
Worked example
A 20 μF capacitor is charged to 12 V. Calculate its stored energy.
Show the worked solution
- Use W = ½CV².
- Convert 20 μF to 20 × 10⁻⁶ F.
- Substitute the voltage squared and state joules.
Answer W = ½ × 20 × 10⁻⁶ × 12² = 1.44 × 10⁻³ J.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A capacitor stores Q = 60 μC at V = 12 V. Calculate its energy from the graph area formula.
Give me a hint
Use the triangular area W = ½QV.
Compare my reasoning
- W = ½ × 60 × 10⁻⁶ × 12.
- W = 360 × 10⁻⁶ J.
- Write the result in joules.
W = 3.6 × 10⁻⁴ J.
Look for these in your work
- Triangle area used
- Charge converted
- Energy unit stated
Calculate the energy stored by C = 50 μF at V = 8.0 V.
Give me a hint
Use W = ½CV² and convert C to farads.
Compare my reasoning
- C = 50 × 10⁻⁶ F.
- W = ½ × 50 × 10⁻⁶ × 8.0².
- Evaluate in joules.
W = 1.6 × 10⁻³ J.
Look for these in your work
- Capacitance converted
- Voltage squared
- Energy calculated
A student calculates capacitor energy as QV from the final graph coordinates. Explain and correct the error.
Give me a hint
The potential difference rises from zero during charging, so use the area under the graph.
Compare my reasoning
- The Q–V relationship is a straight line from the origin.
- The area under the V–Q graph is triangular, not rectangular.
- Use W = ½QV; the student's result is twice the stored energy.
The correct energy is ½QV because the area is a triangle; QV is twice too large.
Look for these in your work
- Graph shape used
- Correct area identified
- Error factor stated
Common mistakes
- Using rectangular area QV instead of triangular area ½QV
- Forgetting to square V in ½CV²
- Using microfarads as farads
- Reporting energy in coulombs or farads instead of joules
What can you explain now?
What is the shape of the area under a linear V–Q charging graph from the origin?
Compare with the explanation
A triangle
Potential difference rises linearly from zero to its final value.
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Make it stick
Why does the area under the charging graph give ½QV rather than QV?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.