Answer
The charge on the capacitor increases.
Work Step by Step
We can write the general equation for the capacitance of a parallel-plate capacitor:
$C = \frac{\epsilon_0~A}{d}$
Note that $d$ is the plate separation.
If $d$ decreases, then the capacitance increases.
Since the capacitor is connected to a battery, the potential difference across the capacitor remains the same.
In general: $~~q = CV$
Since $C$ increases while $V$ remains the same, the charge $q$ on the capacitor increases.
