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. increases
B. decrease
C. does not charge
D. becomes zero

Answer

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-by-Step Solution:

1. **Understanding the Capacitor**:
A parallel plate capacitor consists of two plates separated by a distance \( d \), with one plate positively charged and the other negatively charged. The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{A \epsilon_0}{d} \]
where \( A \) is the area of the plates, \( \epsilon_0 \) is the permittivity of free space, and \( d \) is the distance between the plates.

**Hint**: Remember that capacitance depends on the area of the plates and the distance between them.

2. **Initial Conditions**:
The capacitor is charged to a potential difference \( V \) volts. After charging, the battery is disconnected, meaning the charge \( Q \) on the capacitor remains constant.

3. **Relationship Between Charge, Capacitance, and Voltage**:
The relationship between charge, capacitance, and voltage is given by:
\[ Q = C \cdot V \]
Rearranging this gives:
\[ V = \frac{Q}{C} \]

**Hint**: Keep in mind that when the battery is disconnected, the charge \( Q \) remains constant.

4. **Effect of Increasing Distance**:
When the distance \( d \) between the plates is increased, the capacitance \( C \) decreases because capacitance is inversely proportional to the distance \( d \):
\[ C \propto \frac{1}{d} \]
Therefore, as \( d \) increases, \( C \) decreases.

**Hint**: Think about how changing the distance affects the capacitance.

5. **Impact on Voltage**:
Since \( Q \) is constant and \( C \) is decreasing, we can analyze the effect on \( V \):
- If \( C \) decreases, then according to the equation \( V = \frac{Q}{C} \), the voltage \( V \) must increase.

**Hint**: Consider the implications of a constant charge with a changing capacitance.

6. **Conclusion**:
As a result of increasing the distance between the plates of the capacitor, the potential difference \( V \) between the plates increases.

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