A parallel plate air capacitor is charged to a potential difference of `V` volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an isulating handle. As a result the potential difference between the plates

Options

A. decreases
B. does not change
C. becomes zero
D. increases

Explanation

To solve the problem, we need to analyze the behavior of a parallel plate capacitor when the distance between the plates is increased after it has been charged and disconnected from the battery.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- Initially, the capacitor is charged to a potential difference \( V \) volts.
- The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\varepsilon_0 A}{d} \]
where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the distance between the plates.

2. **Disconnecting the Battery**:
- Once the capacitor is charged and the battery is disconnected, the charge \( Q \) on the capacitor plates remains constant.
- The relationship between charge, capacitance, and potential difference is given by:
\[ Q = C \cdot V \]
- Since \( Q \) is constant after disconnecting the battery, we can express this as:
\[ V = \frac{Q}{C} \]

3. **Increasing the Distance Between Plates**:
- When the distance \( d \) between the plates is increased, the capacitance \( C \) decreases because capacitance is inversely proportional to the distance between the plates.
- Therefore, as \( d \) increases, \( C \) decreases.

4. **Effect on Potential Difference**:
- Since \( Q \) is constant and \( C \) decreases, the potential difference \( V \) must increase. This is because:
\[ V = \frac{Q}{C} \]
- If \( C \) decreases, then \( V \) must increase to keep the equation balanced since \( Q \) remains constant.

5. **Conclusion**:
- As a result of increasing the distance between the plates after disconnecting the battery, the potential difference between the plates increases.

### Final Answer:
The potential difference between the plates of the capacitor increases.