A parallel plate capacitor is charged by a battery after charging the capacitor, battery is disconnected and decrease the distance between the plates then which following statement is correct ?

Options

A. electric field is not constant
B. potential difference is increased
C. decrease the capacitance
D. decrease the stored energy

Answer

To solve the problem regarding the behavior of a parallel plate capacitor when the distance between its plates is decreased after disconnection from a battery, we can follow these steps:

### Step 1: Understand the Initial Conditions
When the capacitor is connected to a battery, it charges up to a certain voltage \( V \). The charge \( Q \) on the capacitor can be expressed as:
\[ Q = C \cdot V \]
where \( C \) is the capacitance of the capacitor.

### Step 2: Determine the Capacitance
The capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{A \epsilon_0}{D} \]
where:
- \( A \) = area of the plates,
- \( \epsilon_0 \) = permittivity of free space,
- \( D \) = distance between the plates.

### Step 3: Analyze the Situation After Disconnecting the Battery
After disconnecting the battery, the charge \( Q \) on the capacitor remains constant because there is no path for charge to flow away from the plates.

### Step 4: Electric Field Calculation
The electric field \( E \) between the plates is given by:
\[ E = \frac{Q}{A \epsilon_0} \]
Since \( Q \) is constant and \( A \) and \( \epsilon_0 \) do not change, the electric field \( E \) remains constant even if \( D \) changes.

### Step 5: Voltage Across the Capacitor
The voltage \( V \) across the capacitor is related to the electric field and the distance between the plates:
\[ V = E \cdot D \]
Since \( E \) is constant and \( D \) is decreasing, the voltage \( V \) will increase as \( D \) decreases.

### Step 6: Capacitance Change
The capacitance \( C \) will change when \( D \) changes:
\[ C = \frac{A \epsilon_0}{D} \]
As \( D \) decreases, \( C \) increases.

### Step 7: Energy Stored in the Capacitor
The energy \( U \) stored in the capacitor is given by:
\[ U = \frac{Q^2}{2C} \]
Since \( Q \) is constant and \( C \) is increasing (because \( D \) is decreasing), the stored energy \( U \) will decrease.

### Conclusion
Based on the analysis, the correct statement is that the stored energy decreases when the distance between the plates is decreased after disconnecting the battery.