Assertion : A parallel plate capacitor is first charged and then distance between the plates is increased. In this process, electric field between the plates remains the same, while potential difference gets increased.
Reason: `E=q/(Aepsilon_0)` and `V=qd/(Aepsilon_0)` Since `q`, remain same, `E` will remain same while V will increase.

Options

A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C. If Assertion is true, but the Reason is false.
D. If Assertion is false but the Reason is true.

Explanation

To solve the question, we need to analyze the assertion and the reason provided regarding the behavior of a parallel plate capacitor when the distance between its plates is increased after it has been charged.

### Step-by-Step Solution:

1. **Understanding the Assertion**:
The assertion states that when a parallel plate capacitor is charged and then the distance between the plates is increased, the electric field (E) remains the same while the potential difference (V) increases.

2. **Electric Field in a Parallel Plate Capacitor**:
The electric field (E) between the plates of a parallel plate capacitor is given by the formula:
\[ E = \frac{Q}{A \epsilon_0} \]
where:
- \(Q\) is the charge on the plates,
- \(A\) is the area of the plates,
- \(\epsilon_0\) is the permittivity of free space.

3. **Potential Difference in a Parallel Plate Capacitor**:
The potential difference (V) between the plates is given by:
\[ V = E \cdot d \]
where \(d\) is the distance between the plates.

4. **Analyzing the Situation**:
- Initially, the capacitor is charged, meaning \(Q\) is constant.
- When the distance \(d\) is increased, the area \(A\) and charge \(Q\) remain unchanged.
- Since \(E\) depends only on \(Q\) and \(A\), it remains constant as \(d\) increases.

5. **Effect on Potential Difference**:
- As the distance \(d\) increases, the potential difference \(V\) increases because:
\[ V = E \cdot d \quad \text{(with E constant)} \]
- Since \(d\) is increasing and \(E\) is constant, \(V\) must increase.

6. **Conclusion**:
- The assertion is true: the electric field remains the same while the potential difference increases.
- The reason provided is also true and correctly explains the assertion.

### Final Answer:
Both the assertion and the reason are true, and the reason is the correct explanation for the assertion.