Answer
(a) $I_0 = 1.88~mA$
(b) $I_0 = 1.88~A$
Work Step by Step
We can write an expression for the peak current:
$I_0 = \omega ~C~\epsilon_0 = 2\pi~f ~C~\epsilon_0$
(a)We can find the peak current:
$I_0 = 2\pi~f ~C~\epsilon_0$
$I_0 = (2\pi)~(100~Hz) (0.30\times 10^{-6}~F)~(10.0~V)$
$I_0 = 1.88\times 10^{-3}~A$
$I_0 = 1.88~mA$
(b) We can find the peak current:
$I_0 = 2\pi~f ~C~\epsilon_0$
$I_0 = (2\pi)~(100\times 10^3~Hz) (0.30\times 10^{-6}~F)~(10.0~V)$
$I_0 = 1.88~A$