Answer
The maximum rate at which the current changes is $~~1020~A/s$
Work Step by Step
We can write an expression for the current at time $t$:
$i(t) = -\omega Q~sin~(\omega t+\phi)$
We can take the derivative to find $\frac{di}{dt}$:
$\frac{di}{dt} = -\omega^2 Q ~cos~(\omega t+\phi)$
In part (a), we found that $~~\omega = 1.364\times 10^5~rad/s~~$ and $~~Q = 55.0~nC$
We can find the maximum rate at which the current changes:
$\omega^2~Q = (1.364\times 10^5~rad/s)^2(55.0\times 10^{-9}~C) = 1020~A/s$
The maximum rate at which the current changes is $~~1020~A/s$
