Answer
(a) The potential difference between the plates remains at 12 V.
(b) The electric field increases.
(c) The magnitude of the charge on the plates increases.
Work Step by Step
(a) Since the battery remains connected, the potential difference between the plates will be the same as the battery, which is 12 V.
(b) $E = \frac{\Delta V}{d}$
Since the potential difference remains the same while the distance decreases, the electric field $E$ increases.
(c) $Q = C~\Delta V$
$Q = (\frac{A~\epsilon_0}{d})~\Delta V$
Since the distance $d$ decreases while the other values remain the same, the magnitude of the charge on the plates must increase.
