Two identical capacitors, have the same capacitance C. One of them is charged to potential V₁ and the other to V₂. The negative ends are also connected, the decrease in energy of the combined system is :

$\frac{1}{4} C\left(V_1-V_2\right)^2$

ΔU = decrease in potential energy

$=U_i-U_f$

$=\frac{1}{2} C\left(V_1^2+V_2^2\right) -\frac{1}{2}(2 C)\left(\frac{V_1+V_2}{2}\right)^2$

$=\frac{1}{4} C\left(V_1-V_2\right)^2$