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. The final energy of the configuration is?

Let's denote the two capacitors as C₁ and C₂. We have C₁ = C and C₂ = 2C. Initially, they are charged to V₁ = V and V₂ = 2V respectively.

The initial charges on the capacitors are:
Q₁ = C₁V₁ = CV
Q₂ = C₂V₂ = 2C(2V) = 4CV

When the capacitors are connected in parallel with opposite polarities, the equivalent capacitance is still C₁ + C₂ = 3C. However, the total charge is Q = Q₂ - Q₁ = 4CV - CV = 3CV.

The final potential difference across the parallel combination is:
V_f = Q / (C₁ + C₂) = (3CV) / (3C) = V

The final energy stored in the parallel combination is:
E_f = 1/2 (C₁ + C₂) V_f² = 1/2 (3C) V² = (3/2)CV² = 3CV²/2

Therefore, the final energy of the configuration is 3CV²/2.