A parallel plate capacitor is charged and then isolated. On increasing the plate separation

Options

A. `{:(" Charge Potential Difference Capacitance"),("remains constant remains constant decreases"):}`
B. `{:(" Charge Potential Difference Capacitance"),("remains constant increases decreases"):}`
C. `{:(" Charge Potential Difference Capacitance"),("remains constant decreases increases"):}`
D. `{:(" Charge Potential Difference Capacitance"),("increases decreases increases"):}`

Answer

To solve the problem regarding the behavior of a parallel plate capacitor when the plate separation is increased after it has been charged and isolated, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- A parallel plate capacitor is charged and then isolated. This means that the charge (Q) on the capacitor plates remains constant after isolation.

2. **Capacitance Formula**:
- 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 separation between the plates.

3. **Effect of Increasing Plate Separation**:
- When the plate separation \(d\) is increased, the capacitance \(C\) decreases because capacitance is inversely proportional to the distance between the plates.

4. **Charge Remains Constant**:
- Since the capacitor is isolated, the charge \(Q\) on the plates does not change. Therefore, we have:
\[ Q = C \cdot V \]
where \(V\) is the potential difference across the capacitor.

5. **Relating Charge, Capacitance, and Voltage**:
- Rearranging the equation \(Q = C \cdot V\), we can express the potential difference as:
\[ V = \frac{Q}{C} \]
- Since \(Q\) remains constant and \(C\) decreases (due to increased plate separation), it follows that \(V\) must increase.

6. **Conclusion**:
- Therefore, when the plate separation is increased:
- The charge \(Q\) remains constant.
- The capacitance \(C\) decreases.
- The potential difference \(V\) increases.

### Final Answer:
- Charge remains constant, capacitance decreases, and potential difference increases.