A capacitor or capacitance `C_(1)` is charge to a potential V and then connected in parallel to an uncharged capacitor of capacitance `C_(2)`. The fianl potential difference across each capacitor will be

Options

A. `(C_(1)V)/(C_(1)+C_(2))`
B. `(C_(2)V)/(C_(1)+C_(2))`
C. `1+(C_(2))/(C_(1))`
D. `1-(C_(2))/(C_(1))`

Explanation

To find the final potential difference across each capacitor when a charged capacitor \( C_1 \) is connected in parallel to an uncharged capacitor \( C_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Capacitor \( C_1 \) is charged to a potential \( V \).
- The charge on capacitor \( C_1 \) can be calculated using the formula:
\[ Q_1 = C_1 \cdot V \]
- Capacitor \( C_2 \) is initially uncharged, so:
\[ Q_2 = 0 \]

2. **Connect the Capacitors in Parallel**:
- When \( C_1 \) is connected in parallel to \( C_2 \), the charge will redistribute between the two capacitors until they reach a common potential \( V'' \).

3. **Apply Charge Conservation**:
- The total charge before connecting the capacitors must equal the total charge after they are connected. Therefore:
\[ Q_{\text{total}} = Q_1 + Q_2 = C_1 \cdot V + 0 = C_1 \cdot V \]

4. **Express the Final Common Potential**:
- After connecting the capacitors, the total charge will be shared between \( C_1 \) and \( C_2 \) at the common potential \( V'' \):
\[ Q_{\text{total}} = (C_1 + C_2) \cdot V'' \]

5. **Set Up the Equation**:
- By equating the total charge before and after the connection, we have:
\[ C_1 \cdot V = (C_1 + C_2) \cdot V'' \]

6. **Solve for the Common Potential \( V'' \)**:
- Rearranging the equation gives:
\[ V'' = \frac{C_1 \cdot V}{C_1 + C_2} \]

### Final Answer:
The final potential difference across each capacitor after they are connected in parallel will be:
\[ V'' = \frac{C_1 \cdot V}{C_1 + C_2} \]