## University Physics with Modern Physics (14th Edition)

The capacitance is halved. For a parallel-plate capacitor, $C=\frac{\epsilon_oA}{d}$. The electric field stays the same, because for a parallel-plate capacitor, $E=\frac{Q}{\epsilon_oA }$. The potential difference is doubled. The electric field stays the same, and the potential difference is proportional to the electric field strength (same) and the distance between the parallel plates (doubled). The stored energy is doubled, because $U=\frac{1}{2}CV^2$ and the capacitance is halved while the potential difference increased by a factor of 2.