A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential and capacitance respectively are

Options

A. constant, decreases, increases
B. constant, decreases, decreases
C. constant, increases, decreases
D. increases, decreases, decreases

Answer

To analyze the effect of increasing the plate separation on charge, potential, and capacitance of a parallel plate capacitor that has been charged and then isolated, we can follow these steps:

### Step 1: Understanding the Capacitor
A parallel plate capacitor consists of two conductive plates separated by a distance \(D\). When charged, one plate accumulates positive charge \(+Q\) and the other accumulates negative charge \(-Q\).

### Step 2: Capacitance Formula
The capacitance \(C\) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\varepsilon_0 A}{D} \]
where:
- \(C\) is the capacitance,
- \(\varepsilon_0\) is the permittivity of free space,
- \(A\) is the area of the plates,
- \(D\) is the separation between the plates.

### Step 3: Isolated Condition
Once the capacitor is charged and isolated, it means that there is no external circuit connected to it. Therefore, the charge \(Q\) on the capacitor remains constant.

### Step 4: Effect of Increasing Plate Separation
Now, if we increase the plate separation \(D\):
- From the capacitance formula, if \(D\) increases, \(C\) decreases because capacitance is inversely proportional to the distance \(D\).

### Step 5: Charge on the Capacitor
Since the capacitor is isolated, the charge \(Q\) remains constant regardless of the change in plate separation.

### Step 6: Potential Difference
The potential difference \(V\) across the capacitor is given by:
\[ V = \frac{Q}{C} \]
Since \(Q\) is constant and \(C\) is decreasing (as established in Step 4), the potential \(V\) must increase. This is because if the denominator of a fraction decreases while the numerator remains constant, the overall value of the fraction increases.

### Summary of Effects
1. **Charge \(Q\)**: Constant (remains the same).
2. **Potential \(V\)**: Increases (because capacitance decreases).
3. **Capacitance \(C\)**: Decreases (due to increased plate separation).

### Final Answer
- Charge: Constant
- Potential: Increases
- Capacitance: Decreases