## College Physics (4th Edition)

(a) $\Delta V = 630~V$ (b) $Q = 0.063~C$
(a) We can find the energy supplied: $E = P~t$ $E = (10.0\times 10^3~W)(2.0\times 10^{-3}~s)$ $E = 20~J$ We can find the initial potential difference: $U = \frac{1}{2}C~(\Delta V)^2$ $(\Delta V)^2 = \frac{2~U}{C}$ $\Delta V = \sqrt{\frac{2~U}{C}}$ $\Delta V = \sqrt{\frac{(2)(20~J)}{100.0\times 10^{-6}~F}}$ $\Delta V = 630~V$ (b) We can find the initial charge: $Q = C~\Delta V$ $Q = (100.0\times 10^{-6}~F)(630~V)$ $Q = 0.063~C$