The capacities of two capacitors are $C_1$ and $C_2$ and their respective potentials are $V_1$ and $V_2$. If they are connected with a wire, then the loss of energy is given by:

$\frac{C_1 C_2\left(V_1+V_2\right)}{2\left(C_1+C_2\right)}$ $\frac{C_1 C_2\left(V_1-V_2\right)}{2\left(C_1+C_2\right)}$ $\frac{C_1 C_2\left(V_1-V_2\right)^2}{2\left(C_1+C_2\right)}$ $\frac{\left(C_1+C_2\right)\left(V_1-V_2\right)}{C_1 C_2}$

$\frac{C_1 C_2\left(V_1-V_2\right)^2}{2\left(C_1+C_2\right)}$

The correct answer is Option (3) → $\frac{C_1 C_2\left(V_1-V_2\right)^2}{2\left(C_1+C_2\right)}$