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

Options

A. `sqrt(2)C`
B. `2c`
C. `(C)/(sqrt(2))`
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 formula for capacitance with a dielectric**:
The capacitance \( C \) of a parallel plate capacitor with a dielectric material is given by the formula:
\[ C = k \cdot \frac{\epsilon_0 \cdot A}{d} \]
where:
- \( k \) is the dielectric constant,
- \( \epsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates,
- \( d \) is the separation between the plates.

2. **Identify the given values**:
We are given that the capacitance with oil (dielectric constant \( k = 2 \)) is \( C \). Therefore, we can express this as:
\[ C = 2 \cdot \frac{\epsilon_0 \cdot A}{d} \]

3. **Calculate the capacitance without the dielectric**:
When the oil is removed, the dielectric constant \( k \) becomes 1 (since air or vacuum has a dielectric constant of 1). The new capacitance \( C' \) can be expressed as:
\[ C' = 1 \cdot \frac{\epsilon_0 \cdot A}{d} = \frac{\epsilon_0 \cdot A}{d} \]

4. **Relate the new capacitance to the original capacitance**:
From the first equation, we know:
\[ C = 2 \cdot \frac{\epsilon_0 \cdot A}{d} \]
Thus, we can express \( \frac{\epsilon_0 \cdot A}{d} \) in terms of \( C \):
\[ \frac{\epsilon_0 \cdot A}{d} = \frac{C}{2} \]

5. **Substituting back to find the new capacitance**:
Now substituting this back into the equation for \( C' \):
\[ C' = \frac{\epsilon_0 \cdot A}{d} = \frac{C}{2} \]

6. **Conclusion**:
Therefore, the capacitance of the capacitor after the oil is removed becomes:
\[ C' = \frac{C}{2} \]

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

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