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.