Question: What is the electric potential energy stored in a capacitor of capacitance 2 µF charged to 12 V?

Options:

1. 0.144 mJ
2. 0.024 mJ
3. 0.288 mJ
4. 0.072 mJ

Correct Answer: 0.144 mJ

Solution:

Electric potential energy (U) is given by U = 1/2 CV^2. Here, U = 1/2 * 2 µF * (12 V)^2 = 0.144 mJ.