Solution:

When the distance between the plates of a charged capacitor is decreased after disconnecting the battery, the following happens:

1. Charge remains constant: Since the battery is disconnected, the charge
   $Q$ 

   on the plates does not change.

2. Capacitance increases: The capacitance of a parallel plate capacitor is given by: 

   $$C = \frac{\varepsilon_0 A}{d}$$where

   $d$is the distance between the plates. As

   $d$decreases,

   $C$increases.

3. Potential difference decreases: The voltage
   $V$ 

   across the capacitor is related by: 

   $$V = \frac{Q}{C}$$Since

   $Q$is constant and

   $C$increases,

   $V$decreases.

4. Potential energy decreases: The energy stored in a capacitor is: 

   $$U = \frac{1}{2} \frac{Q^2}{C}$$As

   $C$increases,

   $U$decreases because

   $Q^2$is constant.

Thus, the potential energy decreases when the distance between the plates is reduced.