Answer
(a) $C = 7.1\times 10^{-6}~F$
(b) $\Delta V = 1.1\times 10^4~V$
Work Step by Step
(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:
$E = \frac{Q~\Delta V}{2}$
$\Delta V = \frac{2~E}{Q}$
$\Delta V = \frac{(2)(450~J)}{8.0\times 10^{-2}~C}$
$\Delta V = 1.1\times 10^4~V$
