A capacitor having a capacitance of 2.0 F is charged to a potential difference of 12 V . Subsequently, it is connected to an uncharged capacitor of capacitance 4.0 F as shown in the figure. Determine the charge on each of the two capacitors after they are connected.

Step-by-step solution

The initial charge on the first capacitor is: Q₁ = C₁ V₁ = 2.0 12 = 24 C The second capacitor is initially uncharged, so Q₂ = 0 . When the capacitors are connected, they are in parallel. The common potential V across the combination is: V = Total charge Total capacitance = Q₁ + Q₂ C₁ + C₂ V = 24 + 0 2.0 + 4.0 = 24 6.0 = 4 V The final charge on the 2.0 F capacitor is: Q₁' = C₁ V = 2.0 4 = 8 C The final charge on the 4.0 F capacitor is: Q₂' = C₂ V = 4.0 4 = 16 C