## College Physics (4th Edition)

(a) $C = 7.1\times 10^{-6}~F$ (b) $\Delta V = 1.1\times 10^4~V$
(a) We can find the capacitance: $E = \frac{Q^2}{2C}$ $C = \frac{Q^2}{2E}$ $C = \frac{(8.0\times 10^{-2}~C)^2}{(2)(450~J)}$ $C = 7.1\times 10^{-6}~F$ (b) We can find the potential difference: $E = \frac{Q~\Delta V}{2}$ $\Delta V = \frac{2~E}{Q}$ $\Delta V = \frac{(2)(450~J)}{8.0\times 10^{-2}~C}$ $\Delta V = 1.1\times 10^4~V$