Answer
(a) $Q = 0.135~C$
(b) The average power is $3.0\times 10^5~W$
Work Step by Step
(a) We can find the charge:
$Q = C~\Delta V$
$Q = (15\times 10^{-6}~F)(9.0\times 10^3~V)$
$Q = 0.135~C$
(b) We can find the energy:
$U = \frac{1}{2}C~(\Delta V)^2$
$U = \frac{1}{2}(15\times 10^{-6}~F)~(9.0\times 10^3~V)^2$
$U = 607.5~J$
We can find the average power:
$P = \frac{E}{t}$
$P = \frac{607.5~J}{2.0\times 10^{-3}~s}$
$P = 3.0\times 10^5~W$
The average power is $3.0\times 10^5~W$
