## College Physics (4th Edition)

(a) The electric field between the plates can be expressed as $E = \frac{Q/A}{\epsilon_0}$ The area does not change and obviously the constant $\epsilon_0$ does not change. Since the capacitor is disconnected from the battery, the charge $Q$ on the plates can not change. Therefore, the electric field does not change. (b) $V = E~d$ Since the distance $d$ increases while the electric field does not change, the potential difference increases.