Answer
(a) decrease
(b) decrease
(c) decrease
(d) decrease
Work Step by Step
(a) We know that $E=\frac{-\Delta V}{\Delta S}$. This equation shows that with an increase in the separation of the plates, the electric field decreases.
(b) As we can see in part (a), the electric field decreases and we know that the charge on the plates will also decrease, since the appearance of the electric field is due to the charge.
(c) $C=\frac{\epsilon_{\circ}A}{d}$. This equation shows that the capacitance will decrease if we increase the separation d.
(d) We know that $U\propto Q^{2}$, so as the charges on the plates are decreasing, the stored energy in the capacitor will also decrease.
