Question

Find the maximum value of $I$.

Answer 1

The is a problem of $L-C$ oscillations.
Charge stored in the capacitor oscillates simple harmonicaly as
$Q=Q_0\sin (\omega t\pm \varphi )$
Here, $Q_0=$ maximum value of $Q=200\mu C$
$=2\times {10}^{-4}C$
$\omega =\frac{1}{\sqrt{LC}}$
$\frac{1}{\sqrt{(2\times {10}^{-3})(5.0\times {10}^{-6})}}={10}^4{s}^{-1}$
Let at $t=0,Q=Q_0$
then $Q\left(t\right)=Q_0\cos \omega t.......\left(i\right)$
$I\left(t\right)=\frac{dQ}{dt}=-Q_0\omega \sin \omega t.....\left(ii\right)$
and $\frac{dI\left(t\right)}{dt}=-Q_0{\omega }^2\cos \left(\omega t\right).......\left(iii\right)$
$I\left(t\right)=-Q_0\omega \sin \omega t$
$\therefore$ Maximum value of $I$ is $Q_0\omega$
$I_{max}=Q_0\omega$
$=(2.0\times {10}^{-4})\left({10}^4\right)$
$I_{max}=2.0A$