A capacity of capacity `C_(1)` is charged up to `V` volt and then connected to an uncharged capacitor of capacity `C_(2)`. Then final potential difference across each will be

Options

A. `(C_(2))/(C_(1)+C_(2))V_(0)`
B. `(C_(1))/(C_(1)+C_(2))V_(0)`
C. `(C_(1)+C_(2))/(C_(2))`
D. `(C_(1)+C_(2))/(C_(1))V_(0)`

Answer

To solve the problem step by step, we will analyze the situation where a charged capacitor \( C_1 \) is connected to an uncharged capacitor \( C_2 \).

### Step-by-Step Solution:

1. **Initial Charge on Capacitor \( C_1 \)**:
- The capacitor \( C_1 \) is charged to a potential \( V \).
- The charge \( Q \) on capacitor \( C_1 \) can be calculated using the formula:
\[ Q = C_1 \cdot V \]

2. **Connecting Capacitors**:
- When capacitor \( C_1 \) is connected to the uncharged capacitor \( C_2 \), the total charge \( Q \) remains constant, but it will redistribute between the two capacitors.

3. **Final Voltage Across Both Capacitors**:
- Let the final potential difference across both capacitors be \( V_f \). Since they are in parallel, the voltage across both capacitors will be the same.
- The total charge after connection is still \( Q \), which is now distributed across both capacitors. The relationship can be expressed as:
\[ Q = C_1 \cdot V_f + C_2 \cdot V_f \]
- This can be rearranged to:
\[ Q = V_f (C_1 + C_2) \]

4. **Substituting for Charge \( Q \)**:
- Substitute \( Q = C_1 \cdot V \) into the equation:
\[ C_1 \cdot V = V_f (C_1 + C_2) \]

5. **Solving for Final Voltage \( V_f \)**:
- Rearranging the equation to solve for \( V_f \):
\[ V_f = \frac{C_1 \cdot V}{C_1 + C_2} \]

6. **Conclusion**:
- The final potential difference across each capacitor after they are connected is:
\[ V_f = \frac{C_1 \cdot V}{C_1 + C_2} \]

### Final Answer:
The final potential difference across each capacitor will be:
\[ V_f = \frac{C_1 \cdot V}{C_1 + C_2} \]