What is the energy stored in that 20 μF capacitor at the initial state with a charge of 30 μC and voltage 1.5 V?

• A. 2.25×10^-6 J
• B. 2.25 J
• C. 2.25×10^-5 J
• D. 2.25×10^-4 J

Answer: (C)

Energy stored in a capacitor comes from how much charge and how strong the voltage is across it. It follows U = (1/2) C V^2. Here C = 20 μF = 20 × 10^-6 F and V = 1.5 V. So U = 0.5 × (20 × 10^-6) × (1.5)^2 = 0.5 × 20 × 10^-6 × 2.25 = 10 × 10^-6 × 2.25 = 2.25 × 10^-5 J.

You can check with U = Q^2 / (2C) using Q = 30 μC = 30 × 10^-6 C: Q^2 = 9 × 10^-10 C^2, 2C = 4 × 10^-5 F, so U = (9 × 10^-10) / (4 × 10^-5) = 2.25 × 10^-5 J. The energy stored is 2.25 × 10^-5 joules.