Calculate the work done in charging a capacitor of capacitance 20µF through a potential difference of 1000V.

Analyze the question: We need the work done in charging a capacitor, which equals the energy stored in it.

Concept: The energy stored in a capacitor is

W = ½CV²

(equivalently ½QV or Q²/2C).

Solution:

• C = 20 µF = 20 × 10⁻⁶ F = 2.0 × 10⁻⁵ F
• V = 1000 V = 10³ V

W = ½ × (2.0 × 10⁻⁵) × (10³)² W = ½ × 2.0 × 10⁻⁵ × 10⁶ W = ½ × 20 W = 10 J

Answer: 10.00 J, which is option E.

Common mistakes:

• Forgetting the ½ gives 20 J, which is not an option.
• Forgetting to convert µF to F (1 µF = 10⁻⁶ F) leads to wildly wrong values.
• Using W = QV instead of ½QV gives double the correct answer. Charge flows in at a gradually increasing voltage, so the average voltage is V/2.

Exam tip: Whenever you see "work done in charging" or "energy stored" for a capacitor, go straight to ½CV². Convert units first. Here, 10⁻⁵ × 10⁶ = 10, which makes the arithmetic quick.