The energy added to an already charged capacitor is the area under the voltage–charge graph between the starting and final charges. For a fixed capacitance, that section is a trapezium. If the charge rises from 8.0 μC to 20 μC as the voltage rises from 2.0 V to 5.0 V, the capacitor stores 42 μJ more energy.
The capacitor already has stored energy at 8.0 μC. If you use an energy formula, calculate the energy at both states and subtract the starting value from the final value.
- Place charge on the horizontal axis and potential difference on the vertical axis.
- Use the average of the endpoint voltages only when the graph is straight over the interval.
- Check the result by subtracting the two values of ½QV.
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Identify the area that represents the change
For a fixed capacitance, V = Q/C gives a straight line through the origin on a graph of V against Q. The area from zero charge to a given charge Q represents the stored energy at that charge.
If the initial charge is already positive, mark both charge values on the axis. The area from the origin to the initial charge represents energy stored before the change. The area between the two charges represents the increase.
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Worked example | Charge a 4.0 μF capacitor from 8.0 μC to 20 μC
The starting voltage is 8.0 ÷ 4.0 = 2.0 V and the final voltage is 20 ÷ 4.0 = 5.0 V. The charge interval is 12 μC. Multiply it by the mean voltage across this straight section, (2.0 + 5.0) ÷ 2 = 3.5 V.
The energy increase is 12 μC × 3.5 V = 42 μJ. A volt times a coulomb gives a joule, so volts times microcoulombs gives microjoules.
Check using the two triangle areas. The starting energy is ½ × 8.0 × 2.0 = 8.0 μJ. The final energy is ½ × 20 × 5.0 = 50 μJ. The difference is 42 μJ.
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Apply the same reasoning to partial discharge
Suppose the same capacitor then falls from 20 μC to 12 μC. The final voltage is 12 ÷ 4.0 = 3.0 V. The stored energy falls from 50 μJ to ½ × 12 × 3.0 = 18 μJ.
The change in stored energy is −32 μJ. The magnitude of the energy lost by the capacitor is 32 μJ. Read the question carefully to decide whether it asks for a signed change or the amount released.
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Frequently asked questions
Can I use ½ × change in Q × change in V?
That finds only the triangular part of the trapezium. In the charging example, it omits the rectangle with height 2.0 V and width 12 μC, which contributes another 24 μJ.
Why does the battery’s energy transfer sometimes differ from the increase in stored energy?
The graph calculation is for the capacitor. Other circuit elements, such as a resistor, can dissipate energy during charging.
Is the trapezium method exact for a curved graph?
For a curved trace, the simple trapezium using its two endpoints generally gives an estimate. Use the actual area or the method specified in the question.
Contents · Official syllabus and next practice
Official syllabus and next practice
Read the official SEAB 2027 syllabus
Compare this with capacitor energy with the battery connected or disconnected.