Final answer:
The decrease in energy when two capacitors with capacitance C charged to voltages V₁ and V₂ are connected together is (1/4)C(V₁ - V₂)².
Step-by-step explanation:
When two identical capacitors with capacitance C are charged to different voltages, V₁ and V₂, and are then connected with like terminals together, the total energy before the connection is the sum of the energies stored in each capacitor individually.
The initial energy (U₁) in the first capacitor is given by, U₁ = (1/2)C(V₁)². Similarly, the energy (U₂) stored in the second capacitor is U₂ = (1/2)C(V₂)². Therefore, the total initial energy Uₙ₀ = U₁ + U₂ = (1/2)C(V₁)² + (1/2)C(V₂)².
After connecting the capacitors together, they will share the same charge and the voltage across both capacitors will be the average of V₁ and V₂. The final energy (Uₙₒ) stored in the system is Uₙₒ = (1/2)C((V₁ + V₂)/2)².
The decrease in energy (ΔU) of the system is ΔU = Uₙ₀ - Uₙₒ = (1/2)C(V₁)² + (1/2)C(V₂)² - (1/2)C((V₁ + V₂)/2)² = (1/4)C(V₁ - V₂)², which corresponds to option A.