The capacities of two capacitors are C₁ and C2 and their respective potentials are V₁ and V2. If they are connected with a wire, then the loss of energy is given by:

The correct answer is:
$\frac{C_1{C}_2{(V_1-V_2)}^2}{2(C_1+C_2)}$

Step-by-Step Derivation and Logical Reasoning:

1. Initial State:
Before the two capacitors are connected, they are charged independently to potentials $V_1$ and $V_2$ respectively. The electrostatic potential energy stored in a capacitor of capacity C charged to potential V is given by $U=\frac{1}{2}C{V}^2$.
Therefore, the initial total energy ($U_i$) stored in the system is the sum of the energies stored in individual capacitors:
$U_i=\frac{1}{2}{C}_1{V}_1^2+\frac{1}{2}{C}_2{V}_2^2$

2. Common Potential after Connection:
When the two capacitors are connected in parallel with a wire, charge flows from the capacitor at higher potential to the one at lower potential until they both reach a common potential, V. By the law of conservation of charge, the total charge remains constant.
Initial total charge is:
$Q=q_1+q_2=C_1{V}_1+C_2{V}_2$
The equivalent capacitance of the parallel combination is $C_{\text{eq}}=C_1+C_2$.
Thus, the common potential V is:
$V=\frac{Q}{C_{\text{eq}}}=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$

3. Final State:
The total final potential energy ($U_f$) of the system at the common potential V is:
$U_f=\frac{1}{2}(C_1+C_2)V^2$
Substituting the expression for the common potential V:
$U_f=\frac{1}{2}(C_1+C_2){\left[\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}\right]}^2=\frac{{(C_1{V}_1+C_2{V}_2)}^2}{2(C_1+C_2)}$

4. Calculating the Loss of Energy ($\Delta U$):
The loss of energy during charge redistribution (dissipated primarily as heat in the connecting wire and electromagnetic radiation) is given by:
$\Delta U=U_i-U_f$
Substituting $U_i$ and $U_f$:
$\Delta U=[\frac{1}{2}{C}_1{V}_1^2+\frac{1}{2}{C}_2{V}_2^2]-\frac{{(C_1{V}_1+C_2{V}_2)}^2}{2(C_1+C_2)}$
Taking $2(C_1+C_2)$ as the common denominator:
$\Delta U=\frac{(C_1{V}_1^2+C_2{V}_2^2)(C_1+C_2)-{(C_1{V}_1+C_2{V}_2)}^2}{2(C_1+C_2)}$
Expanding the numerator terms:
$\text{Numerator}=(C_1^2{V}_1^2+C_1{C}_2{V}_1^2+C_1{C}_2{V}_2^2+C_2^2{V}_2^2)-(C_1^2{V}_1^2+C_2^2{V}_2^2+2C_1{C}_2{V}_1{V}_2)$
Simplifying the numerator by canceling out the common terms:
$\text{Numerator}=C_1{C}_2{V}_1^2+C_1{C}_2{V}_2^2-2C_1{C}_2{V}_1{V}_2$
Factoring out $C_1{C}_2$:
$\text{Numerator}=C_1{C}_2(V_1^2+V_2^2-2V_1{V}_2)=C_1{C}_2{(V_1-V_2)}^2$
Substituting this back into the expression for $\Delta U$ yields the final relation:
$\Delta U=\frac{C_1{C}_2{(V_1-V_2)}^2}{2(C_1+C_2)}$