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 . The positive plate of the first capacitor is subsequently connected to the negative plate of the second, and vice versa. Determine the new charges on the capacitors.

Step-by-step solution

Initial charge on the first capacitor, Q₁ = C₁ V₁ = 5.00 24.0 = 120 C Initial charge on the second capacitor, Q₂ = C₂ V₂ = 6.0 12.0 = 72 C Since the positive plate of the first capacitor is connected to the negative plate of the second, the net charge on the combined plates is Q = Q₁ - Q₂ = 120 - 72 = 48 C The equivalent capacitance of the parallel combination is C = C₁ + C₂ = 5.00 + 6.0 = 11.0 F The common potential across the capacitors is V = Q C = 48 11.0 V New charge on the first capacitor, Q₁' = C₁ V = 5.00 4