Solution:

Let's break down the situation step by step:

Given:

• A capacitor is charged using a battery and then disconnected (so no current can flow after disconnection).
• A dielectric slab is inserted between the plates of the capacitor after disconnecting the battery.

Key concepts:

• Capacitance with Dielectric: When a dielectric slab is inserted, the capacitance of the capacitor increases. The new capacitance $C'$ is related to the original capacitance $C$ by the dielectric constant $K$:

  $$C' = K \cdot C$$where $K$ is the dielectric constant of the material.

• Charge on the Plates: Since the capacitor is disconnected from the battery, no additional charge can flow onto the plates. Thus, the charge $Q$ remains the same, given by:

  $$Q = C \cdot V$$where $V$ is the potential difference across the plates. Since the charge remains constant, the equation becomes:

  $$Q = C' \cdot V'$$where $V'$ is the new potential difference across the plates.

• Potential Difference: Since the capacitance increases and the charge stays constant, the potential difference $V'$ must decrease (because $Q = C' \cdot V'$ and $C' > C$).
• Stored Energy: The energy stored in a capacitor is given by:

  $$U = \frac{Q^2}{2C}$$Since the capacitance increases and the charge is constant, the stored energy $U$ decreases, as it is inversely proportional to the capacitance.

Conclusion:

• Decrease in potential difference: The potential difference across the plates decreases because the capacitance increases while the charge remains constant.
• Reduction in stored energy: The energy stored in the capacitor decreases because the capacitance increases.
• No change in charge: The charge on the plates remains the same since the capacitor is disconnected from the battery.

Thus, the correct answer is: "Decrease in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates."