What is the energy stored in a 10μ F capacitor if the potential difference between the plates is 40 V?

Analyze the question: You're given a capacitance (C = 10 μF) and a potential difference (V = 40 V), and you're asked for the energy stored in the capacitor.

Concept: The energy stored in a charged capacitor is

E = ½CV²

(Equivalent forms: E = ½QV = Q²/2C.)

Solution:

• C = 10 μF = 10 × 10⁻⁶ F = 1.0 × 10⁻⁵ F
• V = 40 V

E = ½ × (1.0 × 10⁻⁵) × (40)²

E = ½ × (1.0 × 10⁻⁵) × 1600

E = 0.5 × 1.6 × 10⁻²

E = 8.0 × 10⁻³ J

Tip: The most common mistakes are forgetting to square the voltage (which gives 2.0 × 10⁻⁴ J, option A) and forgetting the ½ (which gives 1.6 × 10⁻² J, option E). Always convert μF to F first (1 μF = 10⁻⁶ F), then apply ½CV² carefully.