Answer
(a) $\Delta V = 630~V$
(b) $Q = 0.063~C$
Work Step by Step
(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:
$Q = C~\Delta V$
$Q = (100.0\times 10^{-6}~F)(630~V)$
$Q = 0.063~C$
