## College Physics (4th Edition)

(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.