NECO 2002 Physics · Question 27

The energy stored in a capacitor of capacitance 5 μF is 40J. What is the voltage applied across its terminals?

Options

A.4V
B.8V
C.16V
D.200V
E.4000V

Answer: E. 4000V

Explanation written with AI and checked against the answer key

Analyze the question We're given a capacitor with C = 5 μF = 5 × 10⁻⁶ F and stored energy W = 40 J. We need the voltage V across it. Concept The energy stored in a capacitor is: W = ½CV² so V = √(2W / C) Solution V² = 2W / C = (2 × 40) / (5 × 10⁻⁶) V² = 80 / (5 × 10⁻⁶) = 16 × 10⁶ = 1.6 × 10⁷ V = √(1.6 × 10⁷) ≈ 4000 V Check: 4000² = 16 × 10⁶, which matches exactly, so V = 4000 V. Answer The voltage is 4000 V, which is option E. Tip Always convert μF to F (× 10⁻⁶) before substituting. Forgetting this is the most common mistake and leads to wrong options like 4 V. Also remember the three equivalent energy formulas: W = ½CV² = ½QV = Q²/2C. Pick the one that matches the values you're given.

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