There are two indetical capacitor , the first one is uncharged and filled with dielectric of constant K while the other one is charged to potential v having air between its plates, if two capacitors are joined end to end, the common potential will be

Options

A. `(V)/(K - 1)`
B. `(KV)/(K + 1)`
C. `(KV)/(K - 1)`
D. `(V)/(K + 1)`

Answer

To solve the problem, we need to find the common potential when two identical capacitors are connected end to end, one of which is uncharged and filled with a dielectric, while the other is charged.

### Step-by-Step Solution:

1. **Identify the Capacitors**:
- Let \( C_1 \) be the capacitor filled with a dielectric of constant \( K \) (initially uncharged).
- Let \( C_2 \) be the capacitor filled with air (charged to potential \( V \)).

2. **Calculate the Capacitance**:
- The capacitance of the first capacitor \( C_1 \) (with dielectric) is given by:
\[ C_1 = \frac{K \cdot A \cdot \epsilon_0}{D} \]
- The capacitance of the second capacitor \( C_2 \) (with air) is given by:
\[ C_2 = \frac{A \cdot \epsilon_0}{D} \]

3. **Identify Initial Conditions**:
- The potential across the first capacitor \( V_1 = 0 \) (since it is uncharged).
- The potential across the second capacitor \( V_2 = V \) (since it is charged).

4. **Use the Formula for Common Potential**:
- The formula for the common potential \( V_C \) when two capacitors are connected in series is:
\[ V_C = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \]

5. **Substitute Values**:
- Substituting \( V_1 \) and \( V_2 \):
\[ V_C = \frac{C_1 \cdot 0 + C_2 \cdot V}{C_1 + C_2} \]
- This simplifies to:
\[ V_C = \frac{C_2 \cdot V}{C_1 + C_2} \]

6. **Substitute Capacitance Values**:
- Substitute \( C_1 \) and \( C_2 \):
\[ V_C = \frac{\left(\frac{A \cdot \epsilon_0}{D}\right) V}{\left(\frac{K \cdot A \cdot \epsilon_0}{D} + \frac{A \cdot \epsilon_0}{D}\right)} \]

7. **Simplify the Expression**:
- Factor out \( \frac{A \cdot \epsilon_0}{D} \):
\[ V_C = \frac{\left(\frac{A \cdot \epsilon_0}{D}\right) V}{\left(\frac{A \cdot \epsilon_0}{D}\right) (K + 1)} \]
- The \( \frac{A \cdot \epsilon_0}{D} \) cancels out:
\[ V_C = \frac{V}{K + 1} \]

8. **Final Result**:
- Therefore, the common potential \( V_C \) when the two capacitors are connected end to end is:
\[ V_C = \frac{V}{K + 1} \]