Energy: (c) uses of capacitors as storage of energy.

The idea

A capacitor stores energy in the electric field between its plates, so it can act as a temporary energy store that is charged slowly and discharged quickly. The energy stored is W = ½QV, and substituting Q = CV gives W = ½CV² or W = ½Q²/C. For example, a 470 µF capacitor charged to 12 V stores ½ × 470 × 10⁻⁶ F × (12 V)² = 0.034 J. Uses exploit this: camera flashes charge a capacitor then release energy rapidly through a tube; defibrillators deliver a controlled pulse; smoothing circuits in power supplies hold charge between peaks; and backup or memory-hold circuits keep data alive briefly when power is interrupted.

What earns the marks

• States that a capacitor stores energy in the electric field between its plates.
• Recalls and applies W = ½QV, W = ½CV² or W = ½Q²/C correctly.
• Converts capacitance to farads and squares the voltage when using W = ½CV².
• Links a named use, such as a camera flash or defibrillator, to rapid discharge of stored energy.
• Recognises that a capacitor is a temporary store, not a source of continuous power.

Common mistakes and how to put them right

• Using W = CV² and omitting the factor of ½.

  Put it right: The correct expression is W = ½CV².

• Treating capacitance in microfarads as farads without multiplying by 10⁻⁶, which makes the energy a million times too large.

• Assuming a capacitor can supply steady power indefinitely.

  Put it right: It discharges as charge is removed, so its voltage falls.

Exam tips

• Write the energy equation before substituting numbers so the factor of ½ is not lost.
• Convert µF to F and check the power of ten in the answer.
• When a question names a device, state the energy transfer and why rapid discharge matters.