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

### Step 1: Understand the Initial Conditions
- A parallel plate capacitor is charged to a potential difference \( V \) volts.
- The charge \( Q \) on the capacitor can be expressed as:
\[ Q = C \cdot V \]
where \( C \) is the capacitance of the capacitor.

### Step 2: Disconnect the Battery
- After charging, the battery is disconnected. This means that the charge \( Q \) on the capacitor remains constant because there is no external circuit to allow charge to flow in or out.

### Step 3: Increase the Distance Between the Plates
- The distance \( D \) between the plates is increased using an insulating handle. This action affects the capacitance \( C \) of the capacitor.
- The formula for the capacitance of a parallel plate capacitor is:
\[ C = \frac{A \epsilon_0}{D} \]
where \( A \) is the area of the plates and \( \epsilon_0 \) is the permittivity of free space.

### Step 4: Analyze the Effect of Increasing Distance
- When the distance \( D \) is increased, the capacitance \( C \) decreases because capacitance is inversely proportional to the distance between the plates.
- Since the charge \( Q \) remains constant (as the battery is disconnected), we can use the relationship:
\[ Q = C \cdot V \]
to analyze how the potential difference \( V \) changes.

### Step 5: Relate Charge, Capacitance, and Voltage
- Since \( Q \) is constant and \( C \) decreases due to the increase in \( D \), we can deduce that:
\[ Q = C \cdot V \implies V = \frac{Q}{C} \]
- If \( C \) decreases, then \( V \) must increase to keep the product \( C \cdot V \) constant (since \( Q \) is constant).

### Conclusion
- Therefore, as the distance between the plates of the capacitor is increased, the potential difference \( V \) between the plates **increases**.

### Final Answer
The potential difference between the plates **increases**.

---