A parallel plate capacitor is charged from a cell and then isolated from it. The separation between the plates is now increased

Options

A. The force of attraction between the plates will decrease.
B. The field in the region between the plates will not change.
C. The energy stored in the capacitors will increase.
D. The potential difference between the plates will decrease.

Explanation

To solve the problem regarding the parallel plate capacitor, let's break it down step by step.

### Step 1: Understand the Initial Conditions
A parallel plate capacitor is charged by a cell with an EMF \( V \). After charging, the capacitor holds a charge \( Q \) given by the equation:
\[ Q = C \cdot V \]
where \( C \) is the capacitance of the capacitor.

### Step 2: Remove the Cell
Once the capacitor is fully charged, the cell is removed. The charge \( Q \) on the capacitor remains constant because the capacitor is isolated.

### Step 3: Initial Capacitance
The capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\varepsilon_0 A}{D} \]
where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( D \) is the separation between the plates.

### Step 4: Increase the Plate Separation
Now, the separation between the plates is increased to \( D' \). The new capacitance \( C' \) can be expressed as:
\[ C' = \frac{\varepsilon_0 A}{D'} \]
Since \( D' > D \), it follows that \( C' < C \).

### Step 5: Analyze the Electric Field
The electric field \( E \) between the plates of the capacitor is given by:
\[ E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A \varepsilon_0} \]
Since \( Q \) and \( A \) remain constant, the electric field \( E \) does not change when the plate separation is increased. Therefore, the electric field remains constant.

### Step 6: Analyze the Force of Attraction
The force \( F \) of attraction between the plates can be expressed as:
\[ F = \frac{Q^2}{2A \varepsilon_0} \]
Since \( Q \) and \( A \) are constant, the force of attraction does not depend on the separation distance \( D \). Thus, the force remains unchanged.

### Step 7: Energy Stored in the Capacitor
The energy \( U \) stored in the capacitor is given by:
\[ U = \frac{1}{2} C V^2 = \frac{Q^2}{2C} \]
Since \( C' < C \) and \( Q \) is constant, the energy stored in the capacitor will increase as the capacitance decreases.

### Step 8: Potential Difference Between the Plates
The potential difference \( V' \) across the plates is given by:
\[ V' = \frac{Q}{C'} \]
Since \( C' < C \) and \( Q \) is constant, the potential difference \( V' \) will increase.

### Conclusion
1. The electric field between the plates remains constant.
2. The force of attraction between the plates does not change.
3. The energy stored in the capacitor increases.
4. The potential difference between the plates increases.

### Summary of Answers
- The electric field does not change.
- The force of attraction remains unchanged.
- The energy stored in the capacitor increases.
- The potential difference between the plates increases.