A parallel plate condenser with oil between the plates (dielectric constant of oil K = 2 ) has a capacitance C. if the oil is removed , then capacitance of the capacitor becomes

Options

A. `(C )/(sqrt(2))`
B. 2C
C. `sqrt(2)`C
D. `(C )/(2)`

Answer

To solve the problem, we need to understand how the capacitance of a parallel plate capacitor changes when a dielectric material is removed.

### Step-by-Step Solution:

1. **Understand the Capacitance with Dielectric:**
The formula for the capacitance \( C \) of a parallel plate capacitor with a dielectric material is given by:
\[ C = K \cdot \frac{\varepsilon_0 A}{d} \]
where:
- \( K \) is the dielectric constant,
- \( \varepsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates,
- \( d \) is the separation between the plates.

2. **Substituting the Given Values:**
In this case, the dielectric constant \( K \) of the oil is given as 2. Therefore, the capacitance with oil is:
\[ C = 2 \cdot \frac{\varepsilon_0 A}{d} \]

3. **Capacitance without Dielectric:**
When the oil is removed, the dielectric constant \( K \) becomes 1 (since air or vacuum has a dielectric constant of 1). Thus, the new capacitance \( C' \) is:
\[ C' = 1 \cdot \frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 A}{d} \]

4. **Relating the Two Capacitances:**
Now we can relate \( C' \) to \( C \):
\[ C' = \frac{\varepsilon_0 A}{d} \]
From the earlier equation for \( C \):
\[ C = 2 \cdot \frac{\varepsilon_0 A}{d} \]
Therefore, we can express \( C' \) in terms of \( C \):
\[ C' = \frac{1}{2} C \]

5. **Final Result:**
Thus, when the oil is removed, the capacitance of the capacitor becomes:
\[ C' = \frac{C}{2} \]

### Conclusion:
The capacitance of the capacitor after the oil is removed is \( \frac{C}{2} \).