A capacitor of capacitance 2.0 F is charged to a potential difference of 12 V . It is then connected to an uncharged capacitor with a capacitance of 4.0 F . Calculate the electrostatic energy stored in each of the two capacitors after the connection is established.

Step-by-step solution

Initial charge on the 2.0 F capacitor is: Q = C₁ V₁ = 2.0 F 12 V = 24 C When it is connected to the uncharged 4.0 F capacitor, the total charge is conserved and they reach a common potential V . V = Total Charge Total Capacitance = Q C₁ + C₂ V = 24 C 2.0 F + 4.0 F = 24 6.0 V = 4 V The electrostatic energy stored in the 2.0 F capacitor is: U₁ = 1 2 C₁ V^2 = 1 2 2.0 F (4 V )^2 = 16 J The electrostatic energy stored in the 4.0 F capacitor is: U₂ = 1 2 C₂ V^2 = 1 2 4.0 F (4 V )^2 = 32 J