A dielectric slab is placed between the plates of a parallel plate capacitor. Its capacitance

Options

A. becomes zero
B. remains the same
C. decreases
D. increases

Answer

To solve the problem regarding the capacitance of a parallel plate capacitor with a dielectric slab inserted, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Basic Formula for Capacitance**:
The capacitance \( C \) of a parallel plate capacitor without any dielectric is given by the formula:
\[ C = \frac{\epsilon_0 A}{d} \]
where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( A \) is the area of one of the plates,
- \( d \) is the separation between the plates.

2. **Introduce the Dielectric Constant**:
When a dielectric material is introduced between the plates, the capacitance increases. The new capacitance \( C' \) can be expressed as:
\[ C' = K \cdot C \]
where \( K \) is the dielectric constant of the material.

3. **Substitute the Original Capacitance**:
Substitute the expression for \( C \) into the equation for \( C' \):
\[ C' = K \cdot \frac{\epsilon_0 A}{d} \]

4. **Final Expression for Capacitance with Dielectric**:
Thus, the capacitance of the capacitor with the dielectric slab becomes:
\[ C' = \frac{K \epsilon_0 A}{d} \]

5. **Conclusion**:
The presence of the dielectric slab increases the capacitance by a factor of \( K \). Therefore, the capacitance of the parallel plate capacitor with the dielectric slab is greater than the capacitance without it.