College Physics (4th Edition)

(a) The potential difference between the plates is $667~V$ (b) The potential difference doubles to $1330~V$.
(a) We can find the potential difference: $\Delta V = \frac{Q}{C}$ $\Delta V = \frac{0.800\times 10^{-6}~C}{1.20\times 10^{-9}~F}$ $\Delta V = 667~V$ The potential difference between the plates is $667~V$ (b) The capacitance is $C = \frac{\epsilon_0~A}{d}$ If the plate separation $d$ is doubled, then the new capacitance is $\frac{C}{2}$ Then the new potential difference is $\Delta V = \frac{Q}{C/2} = \frac{2~Q}{C}$ The new potential difference is $\Delta V = (2)(667~V) = 1330~V$ The potential difference doubles to $1330~V$.