Two identical capacitors, have the same capacitance C. One of them is charged to potential `V_1` and the other `V_2`. The negative ends of the capacitors are connected together. When the poistive ends are also connected, the decrease in energy of the combined system is

Options

A. `1/4C(V_(1)^(2)-V_(2)^(2))`
B. `1/4C(V_(1)^(2)+V_(2)^(2))`
C. `1/4C(V_(1)-V_(2))^(2)`
D. `1/4C(V_(1)+V_(2))^(2)`

Explanation

To solve the problem of finding the decrease in energy when two identical capacitors are connected, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Energy (U1)**:
The initial energy stored in the two capacitors can be calculated using the formula for the energy stored in a capacitor:
\[ U_1 = \frac{1}{2} C V_1^2 + \frac{1}{2} C V_2^2 \]
This can be simplified to:
\[ U_1 = \frac{C}{2} (V_1^2 + V_2^2) \]

2. **Determine Final Voltage (V)**:
When the positive ends of the capacitors are connected, the total charge is conserved. The common potential \( V \) across both capacitors can be calculated as:
\[ V = \frac{C V_1 + C V_2}{C + C} = \frac{V_1 + V_2}{2} \]

3. **Calculate Final Energy (U2)**:
After connecting the capacitors in parallel, the final energy stored in the system can be calculated as:
\[ U_2 = \frac{1}{2} (C + C) \left(\frac{V_1 + V_2}{2}\right)^2 \]
Simplifying this gives:
\[ U_2 = C \left(\frac{V_1 + V_2}{2}\right)^2 = \frac{C}{4} (V_1 + V_2)^2 \]

4. **Calculate the Decrease in Energy (ΔU)**:
The decrease in energy, denoted as \( \Delta U \), is given by the difference between the initial and final energies:
\[ \Delta U = U_1 - U_2 \]
Substituting the expressions for \( U_1 \) and \( U_2 \):
\[ \Delta U = \frac{C}{2} (V_1^2 + V_2^2) - \frac{C}{4} (V_1 + V_2)^2 \]

5. **Simplify ΔU**:
To simplify \( \Delta U \):
\[ \Delta U = \frac{C}{2} (V_1^2 + V_2^2) - \frac{C}{4} (V_1^2 + 2V_1V_2 + V_2^2) \]
This can be rewritten as:
\[ \Delta U = \frac{C}{4} (2V_1^2 + 2V_2^2 - V_1^2 - 2V_1V_2 - V_2^2) \]
Simplifying further gives:
\[ \Delta U = \frac{C}{4} (V_1^2 - 2V_1V_2 + V_2^2) = \frac{C}{4} (V_1 - V_2)^2 \]

6. **Final Result**:
Thus, the decrease in energy of the combined system is:
\[ \Delta U = \frac{C}{4} (V_1 - V_2)^2 \]