## College Physics (4th Edition)

(a) $Q = 1.44\times 10^{-8}~C$ (b) The magnitude of the charge on the plates is halved. The electric field is halved.
(a) We can find the magnitude of the charge on each plate: $Q = C~\Delta V$ $Q = (1.20\times 10^{-9}~F)(12~V)$ $Q = 1.44\times 10^{-8}~C$ (b) Since the battery remains connected, the potential difference between the plates will be the same as the battery, which is 12 V. $Q = C~\Delta V$ $Q = (\frac{A~\epsilon_0}{d})~\Delta V$ Since the distance $d$ doubles while the other values remain the same, the magnitude of the charge on the plates is halved. $E = \frac{\Delta V}{d}$ Since the potential difference remains the same while the distance doubles, the electric field $E$ is halved.