## College Physics (4th Edition)

Since the battery is still connected, the potential difference $V$ does not change when the distance between the plates increases. The electric field $E$ is equal to $\frac{V}{d}$. Since $d$ increases while $V$ remains constant, the electric field decreases. The magnitude of the charge on each plate is $Q = CV = \frac{\epsilon_0~A~V}{d}$ Since $d$ increases while the other values remain constant, the charge on the plates decreases. The correct answer is: (d) Both the electric field and the charge on the plates decrease.