A capacitor of capacitance 5.00 is charged to 24.0 V , while a...

Initial charge on the first capacitor, $Q_1 = C_1 V_1 = 5.00 \times 24.0 = 120 \mu\text{C}$

Initial charge on the second capacitor, $Q_2 = C_2 V_2 = 6.0 \times 12.0 = 72 \mu\text{C}$

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 $Q = Q_1 - Q_2 = 120 - 72 = 48 \mu\text{C}$

The equivalent capacitance of the parallel combination is $C = C_1 + C_2 = 5.00 + 6.0 = 11.0 \mu\text{F}$

The common potential across the capacitors is $V = \dfrac{Q}{C} = \dfrac{48}{11.0} \text{ V}$

New charge on the first capacitor, $Q_1' = C_1 V = 5.00 \times \dfrac{48}{11.0} = \dfrac{240}{11} \approx 21.8 \mu\text{C}$

New charge on the second capacitor, $Q_2' = C_2 V = 6.0 \times \dfrac{48}{11.0} = \dfrac{288}{11} \approx 26.2 \mu\text{C}$