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, decrease, decrease
B. increase, decreases, decreases
C. constant, decreases, increases
D. constant, increases, decreases.

Explanation

To solve the problem step by step, we will analyze the effects of increasing the plate separation in a parallel plate capacitor that has been charged and isolated.

### Step 1: Understand the relationship between capacitance, charge, and potential
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,
- \( D \) is the separation between the plates.

### Step 2: Analyze the effect of increasing plate separation
Since the capacitor is isolated, the charge \( Q \) on the capacitor remains constant. When the plate separation \( D \) is increased, we can see from the formula that capacitance \( C \) is inversely proportional to \( D \):
\[ C \propto \frac{1}{D} \]
Thus, if \( D \) increases, \( C \) will decrease.

### Step 3: Determine the effect on potential
The relationship between charge, capacitance, and potential \( V \) is given by:
\[ V = \frac{Q}{C} \]
Since the charge \( Q \) is constant and we have established that capacitance \( C \) decreases when the plate separation \( D \) increases, it follows that the potential \( V \) must increase:
\[ \text{If } C \text{ decreases, then } V \text{ increases.} \]

### Step 4: Summarize the effects
Now we can summarize the effects of increasing the plate separation on charge, potential, and capacitance:
- **Charge \( Q \)**: Remains constant (since the capacitor is isolated).
- **Potential \( V \)**: Increases (as capacitance decreases).
- **Capacitance \( C \)**: Decreases (as plate separation increases).

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