Question

when a 12 V battery is connected to a 6 uF capacitor, how much energy is stored

Answer 1

The energy stored in a capacitor is

E = (1/2) · (capacitance) · (voltage)²

E = (1/2) · (6 x 10⁻⁶ F) · (12 V)²

E = (3 x 10⁻⁶ F) · (144 V²)

E = 4.32 x 10⁻⁴ Joule

(That's 0.000432 of a Joule)