(a) Total energy is initially in the form of electric field within the plates of charged capacitor.
$$
U_E=\frac{Q^2}{2 C}=\frac{\left(10 \times 10^{-3}\right)^2}{2 \times 50 \times 10^{-6}}=1 \mathrm{~J}
$$
If we neglect the losses due to resistance of connecting wires, the total energy remain consumed during $L C$ oscillations.
(b) Natural frequency of the circuit
$f=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \times \sqrt{20 \times 10^{-3} \times 50 \times 10^{-6}}} \mathrm{~Hz}=\frac{500}{\pi}$
(c) Instantaneous electrical energy
$$
U_E=\frac{q_0^2 \cos ^2 \omega t}{2 C}
$$
(i) At $\omega t=0, \pi, 2 \pi, 3 \pi \ldots$ the energy is completely electrical.
$$
t=\frac{n \pi}{2 \pi f}=\frac{n}{2 f}=\frac{n \pi}{1000} \sec n=0,1,2,3,4 \ldots
$$
or $t=0, T / 2, T, 3 T / 2 \ldots$
(ii) Instantaneous magnetic energy
$$
U_B=\frac{1}{2} L q_0^2 \omega^2 \sin ^2 \omega t
$$
or $U_B=\frac{q_0^2}{2 C} \sin ^2 \omega t$
so at $\omega \mathrm{t}=\pi / 2,3 \pi / 2,5 \pi / 2 \ldots$
The energy is completely magnetic
$$
t=\frac{(2 n+1) \pi}{2(2 \pi f)}=\frac{(2 n+1)}{4 f}=\frac{(2 n+1) \pi}{2000} \mathrm{sec}
$$
where $n=0,1,2,3,4 \ldots \quad$ or $t=T / 4,3 T / 4,5 T / 4 \ldots$
(d) timings for energy shared equally between inductor and capacitor.
$$
\begin{aligned}
&U_B=U_E \\
&\frac{q_0^2}{2 C} \sin ^2 \omega t=\frac{q_0^2}{2 C} \cos ^2 \omega t \\
&\tan ^2 \omega t=1 \text { or } \tan \omega t=\tan \pi / 4 \\
&t=\frac{\pi}{4 \omega}, \frac{3 \pi}{4 \omega}, \frac{5 \pi}{4 \omega} \ldots \text { or } t=\frac{T}{8}, \frac{3 T}{8}, \frac{5 T}{8} \ldots
\end{aligned}
$$
(e) When a resistor in inserted in the circuit, eventually all the energy will be lost as heat across resistence. The oscillation will be damped.