Answer
See explanation.
Work Step by Step
The capacitance is halved. For a parallel-plate capacitor, $C=\frac{\epsilon_oA}{d}$.
The electric field is halved, because for a parallel-plate capacitor, $E=\frac{V}{d}$.
The charge Q is halved. $Q=CV$ and the capacitance is halved while the potential difference stays the same.
The stored energy is halved, because $U=\frac{1}{2}CV^2$ and the capacitance is halved while the potential difference stays the same.
