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."