Answer
(a) $C = 9.55\times 10^{-11}~F$
(b) $I_0 = 660~\mu A$
Work Step by Step
(a) We can find the capacitance:
$I_0 = \omega ~V_C ~C$
$C = \frac{I_0}{\omega ~V_C}$
$C = \frac{I_0}{2\pi~f~ V_C}$
$C = \frac{330\times 10^{-6}~A}{(2\pi)~(250\times 10^3~Hz) (2.2~V)}$
$C = 9.55\times 10^{-11}~F$
(b) We can find the peak current:
$I_0 = \omega ~V_C ~C$
$I_0 = 2\pi ~f ~V_C ~C$
$I_0 = (2\pi) (500\times 10^3~Hz) (2.2~V) (9.55\times 10^{-11}~F)$
$I_0 = 6.60\times 10^{-4}~A$
$I_0 = 660~\mu A$
