As the distance between the plates of a parallel plate capacitor decreased

Options

A. If both assertion and reason are ture and reason is the correct explanation of assertion.
B. If both assertin and reason are ture but reason is not the correct explanation of assertion .
C. If assertion is true but reason is false.
D. If both assertion and reason are false.

Explanation

To solve the question regarding the effect of decreasing the distance between the plates of a parallel plate capacitor, we can analyze it step by step.

### Step-by-Step Solution:

1. **Understanding Capacitance Formula**:
The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\varepsilon_0 A}{D} \]
where:
- \( C \) is the capacitance,
- \( \varepsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates, and
- \( D \) is the distance between the plates.

2. **Effect of Decreasing Distance**:
From the formula, we can see that capacitance \( C \) is inversely proportional to the distance \( D \). Therefore, if the distance \( D \) decreases, the capacitance \( C \) increases. This can be expressed mathematically as:
\[ \text{If } D \text{ decreases, then } C \text{ increases.} \]

3. **Charge on the Plates**:
If the charge \( Q \) on the plates remains constant, we can analyze the electric field \( E \) between the plates. The electric field \( E \) is given by:
\[ E = \frac{Q}{\varepsilon_0 A} \]
and it can also be expressed in terms of potential difference \( V \) and distance \( D \):
\[ E = \frac{V}{D} \]

4. **Constant Potential Difference**:
If we assume that the potential difference \( V \) remains constant while the distance \( D \) decreases, we can derive the following:
- As \( D \) decreases, \( E = \frac{V}{D} \) implies that the electric field \( E \) must increase.

5. **Breakdown of the Capacitor**:
As the electric field \( E \) increases, there is a limit to how much the electric field can increase before the dielectric material (or air) breaks down. Thus, if the electric field exceeds a certain threshold, the capacitor may experience dielectric breakdown, leading to failure.

6. **Conclusion**:
In summary, as the distance between the plates of a parallel plate capacitor decreases:
- The capacitance \( C \) increases.
- The electric field \( E \) increases.
- There is an increased risk of dielectric breakdown if the electric field becomes too high.