A capacitor of capacitance 5.00 F is charged to 24.0 V , while a second capacitor of capacitance 6.0 F is charged to 12.0 V . Determine the energy stored in each capacitor.

Step-by-step solution

The energy stored in a capacitor is given by the formula U = 1 2 CV^2 . For the first capacitor: U₁ = 1 2 C₁ V₁^2 U₁ = 1 2 5.00 10⁻⁶ (24.0)^2 U₁ = 1440 10⁻⁶ J = 1.44 mJ For the second capacitor: U₂ = 1 2 C₂ V₂^2 U₂ = 1 2 6.0 10⁻⁶ (12.0)^2 U₂ = 432 10⁻⁶ J = 0.432 mJ