A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is connected to another battery and is charged to potential difference 2V. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. what is the final energy of the configuration?

3CV² / 2

Net charge between capacitors, Q' = Q₁- Q₂ = C₂V₂ - C₁V₁ = (2C)(2V) - CV = 3CV

in parallel, C' = C₁ + C₂ = 3C

so $ U = \frac{Q'2}{2C'} = \frac{9C^2V^2}{6C}= \frac{3}{2} CV^2$