A parallel plate capacitor is charged by a battery after charging the capacitor, battery is disconnected and if a dielectric plate is inserted between the place of plates. Then which one of the following statements is not correct :

Options

A. increase in the stored energy
B. decrease in the potential difference
C. decrease in the electric field
D. increase in the capacitance

Answer

To solve the problem, we need to analyze the behavior of a parallel plate capacitor when a dielectric is inserted after it has been charged and disconnected from the battery. We will go through the steps systematically.

### Step-by-Step Solution:

1. **Understanding the Initial Setup:**
- A parallel plate capacitor is charged by a battery. Let the initial capacitance be \( C \), the voltage across the capacitor be \( V \), and the charge stored in the capacitor be \( Q \).
- The relationship is given by:
\[ Q = C \cdot V \]

2. **Disconnecting the Battery:**
- After charging, the battery is disconnected. The charge \( Q \) remains constant because there is no external circuit to allow charge to flow.

3. **Inserting the Dielectric:**
- A dielectric material with dielectric constant \( K \) is inserted between the plates of the capacitor. The new capacitance \( C' \) becomes:
\[ C' = K \cdot C \]
- Since \( K > 1 \), the capacitance increases.

4. **Charge Conservation:**
- The charge \( Q \) remains constant after the battery is disconnected. Therefore, we can express the relationship with the new capacitance and new voltage \( V' \):
\[ Q = C' \cdot V' \]
- Since \( C' > C \), to keep \( Q \) constant, \( V' \) must decrease:
\[ V' = \frac{Q}{C'} \]

5. **Effect on Electric Field:**
- The electric field \( E \) between the plates is related to the voltage and the distance \( d \) between the plates:
\[ E = \frac{V}{d} \]
- Since the voltage decreases when the dielectric is inserted, the electric field \( E' \) also decreases:
\[ E' = \frac{V'}{d} \]

6. **Stored Energy Calculation:**
- The energy stored in the capacitor is given by:
\[ U = \frac{Q^2}{2C} \]
- After inserting the dielectric, the new stored energy \( U' \) is:
\[ U' = \frac{Q^2}{2C'} \]
- Since \( C' > C \), it follows that \( U' < U \). Therefore, the stored energy decreases.

7. **Conclusion:**
- The statements we need to evaluate are:
- The capacitance increases.
- The potential difference decreases.
- The electric field decreases.
- The stored energy increases.
- The statement that is **not correct** is that the stored energy increases, as we have shown that it actually decreases.

### Final Answer:
The statement that is not correct is: **The stored energy increases.**