Answer
a.) $t = 2\pi\cdot10^{-7}$ seconds
b.) $I(2\pi\cdot10^{-7}) = 0.5$ mA
Work Step by Step
a.)
Chapter 30.9 gives the equations for both the current of the circuit and charge of the capacitor as $I(t) = I_{max}\sin(\omega t) = \omega Q_0 \sin(\omega t) = \displaystyle \frac{Q_0}{\sqrt{LC}}\sin(\frac{t}{\sqrt{LC}})\qquad\qquad$ and $Q(t) = Q_0\cos(\omega t) = Q_0\cos(\displaystyle \frac{t}{\sqrt{LC}})$, respectively.
The capacitor is fully discharged when the charge on the capacitor plates is zero. Thus,
$Q(t) = 0 = Q_0\cos(\displaystyle \frac{t}{\sqrt{LC}}) \qquad\rightarrow\qquad \cos(\frac{t}{\sqrt{LC}}) = 0$
$\displaystyle \cos^{-1}(0) = \frac{t}{\sqrt{LC}}$
$\displaystyle \frac{\pi}{2} = \frac{t}{\sqrt{LC}}$
$t = \displaystyle \frac{\sqrt{LC}}{2}\cdot \pi = \frac{\sqrt{(20\cdot10^{-3})(8.0\cdot10^{-12})}}{2} \cdot \pi= \frac{4\cdot10^{-7}}{2}\cdot \pi$
$t = 2\pi\cdot10^{-7} \approx 6.28\cdot10^{-7}$ s
b.) From above, we know $I(t) = \displaystyle \frac{Q_0}{\sqrt{LC}}\sin(\frac{t}{\sqrt{LC}})$
The initial charge on the capacitor is $Q_0 = C\Delta V = (8.0\cdot10^{-12})(25) = 2\cdot10^{-10}$ C
$I(2\pi\cdot10^{-7}) = \displaystyle \frac{2\cdot10^{-10}}{\sqrt{(8.0\cdot10^{-12})(20\cdot10^{-3})}}\sin(\frac{2\pi\cdot10^{-7}}{\sqrt{(8.0\cdot10^{-12})(20\cdot10^{-3})}})$
$I(2\pi\cdot10^{-7}) = \displaystyle \frac{1}{2000} = 5\cdot10^{-4}$ A = $0.5$ mA
