When a capacitor’s capacitance changes, check whether its battery is still connected. An ideal connected battery fixes the potential difference. If the capacitor is disconnected and isolated, its charge stays fixed. These two cases can give opposite changes in stored energy.
The 2027 H2 Physics 9478 syllabus includes using expressions for energy stored by a capacitor. This example doubles the capacitance under two different conditions.
- Decide whether charge or potential difference stays fixed.
- Choose the energy equation containing that quantity.
- Calculate the initial and final stored energies.
- State the direction of change for each condition.
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Start with the same charged capacitor
Take a 4.0 μF capacitor charged to 6.0 V. Its charge magnitude is Q = CV = 24 μC. Its stored energy is ½CV² = 72 μJ. Imagine doubling the capacitance to 8.0 μF by changing the capacitor’s arrangement. Consider two separate versions of this change.
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Worked example | The battery remains connected
The ideal battery holds the voltage at 6.0 V. The final charge is 8.0 μF × 6.0 V = 48 μC. Using U = ½CV², the stored energy becomes ½ × 8.0 × 6.0² = 144 μJ. Doubling capacitance doubles stored energy here because the voltage stays fixed.
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Work the same example with an isolated capacitor
Disconnect the battery before changing the capacitance. With no conducting path for charge to leave, the charge remains 24 μC. The final voltage is Q/C = 24 ÷ 8.0 = 3.0 V. Using U = ½QV gives 36 μJ. Equivalently, U = Q²/(2C) shows directly why doubling capacitance halves stored energy at fixed charge.
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Frequently asked questions
Can I use ½CV² in the isolated case?
Yes. Use the new voltage of 3.0 V. Substituting 6.0 V would assume that the disconnected capacitor is still held at the battery voltage.
Where can the extra charge come from in the connected case?
Charge can move through the circuit linking the capacitor to the battery. This allows the capacitor to carry more charge at the same voltage.