D

Step 1: Understanding the system
Let's denote the two identical capacitors as C1 and C2. They have capacitance C. Assume C1 is charged to a voltage V1 and C2 to a voltage V2.
Step 2: Analyze initial charges and energies
The initial charge on C1 is given by:
$Q_1 = C \cdot V_1$
The initial charge on C2 is:
$Q_2 = C \cdot V_2$
When they are connected in parallel, the net charge (Q_net) becomes:
$Q_{net} = Q_1 + Q_2 = C \cdot V_1 + C \cdot V_2 = C(V_1 + V_2)$
Step 3: Energy consideration
The individual energies stored in the capacitors initially are:
Energy in C1: $E_1 = \frac{1}{2} C V_1^2$
Energy in C2: $E_2 = \frac{1}{2} C V_2^2$
The total initial energy is:
$E_{total, initial} = E_1 + E_2 = \frac{1}{2} C V_1^2 + \frac{1}{2} C V_2^2$
If the capacitors are then connected in parallel, the net potential difference across them (V_net) can be found using the conservation of charge. Since charges redistribute, the equivalent voltage across both capacitors becomes:
$V_{net} = \frac{Q_{net}}{2C} = \frac{C(V_1 + V_2)}{2C} = \frac{V_1 + V_2}{2}$
With C being equal for both capacitors, the combined energy will be:
$E_{total, final} = \frac{1}{2} \cdot (2C) \cdot (V_{net})^2 = C \cdot \frac{(V_1 + V_2)^2}{4}$
Through calculations, it can be demonstrated that the energies will actually decrease due to charge redistribution.
Step 4: Conclusion on the options
Option D states that the net potential difference across them equals the sum of the individual initial potential differences ($V_{net} = V_1 + V_2$), which is incorrect. The net potential will be the average ($V_{net} = \frac{V_1 + V_2}{2}$). Therefore, option D is the incorrect statement. Thus, the answer is:
Therefore, D.