Answer
The magnitude of the charge is cut in half.
Work Step by Step
The charge on a capacitor is equal to $$Q=CV$$ Since the capacitance equals $C=\frac{\epsilon_o A}{d}$, the charge becomes $$Q=\frac{V\epsilon_o A}{d}$$ Therefore, as distance between plates doubles, the new charge $Q'$ equals $$Q'=\frac{V\epsilon_o A}{2d}=\frac{1}{2}(\frac{V\epsilon_o A}{d})=\frac{1}{2}Q$$ Therefore, the magnitude of the charge is cut in half.
