A parallel plate capacitor is charged by connecting it to a battery. Which of the following will remain constant if the distance between the plates of the capacitor is increased in this situation?

Options

A. Energy stored
B. Electric field
C. Potential difference
D. Capacitance

Explanation

To solve the problem, we need to analyze the behavior of a parallel plate capacitor when it is connected to a battery and the distance between the plates is increased.

### Step-by-Step Solution:

1. **Understanding the Capacitor Setup**:
A parallel plate capacitor consists of two conductive plates separated by a distance (D). When connected to a battery, the capacitor gets charged, and a potential difference (V) is established across the plates.

**Hint**: Remember that a battery maintains a constant potential difference across the capacitor.

2. **Effect of Increasing Distance (D)**:
When the distance between the plates (D) is increased while the capacitor is connected to the battery, the capacitance (C) of the capacitor is affected. The formula for capacitance of a parallel plate capacitor is given by:
\[ C = \frac{K \epsilon_0 A}{D} \]
where \(K\) is the dielectric constant, \(\epsilon_0\) is the permittivity of free space, and \(A\) is the area of the plates. As \(D\) increases, \(C\) decreases.

**Hint**: Consider how the formula for capacitance changes with distance.

3. **Potential Difference (V)**:
Since the capacitor is connected to a battery, the potential difference (V) across the plates remains constant, regardless of the change in distance.

**Hint**: Think about the role of the battery in maintaining voltage.

4. **Energy Stored in the Capacitor**:
The energy (U) stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
Since \(C\) decreases as \(D\) increases and \(V\) remains constant, the energy stored in the capacitor will also decrease.

**Hint**: Analyze how energy changes with capacitance and voltage.

5. **Electric Field (E)**:
The electric field (E) between the plates of a capacitor is given by:
\[ E = \frac{V}{D} \]
As \(D\) increases, and since \(V\) is constant, the electric field \(E\) will decrease.

**Hint**: Relate electric field to voltage and distance.

6. **Conclusion**:
From the analysis, we find that:
- The potential difference (V) remains constant.
- The capacitance (C) decreases.
- The energy stored (U) decreases.
- The electric field (E) decreases.

Therefore, the only quantity that remains constant when the distance between the plates is increased while connected to a battery is the **potential difference (V)**.

### Final Answer:
The potential difference (V) remains constant.