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A fully charged capacitor $C$ with initial charge $q_0$ is connected to a coil of selfinductance $L$ at $t=0$. The time at which the energy is stored equally between the electric and the magnetic field is
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The correct answer is:
$\frac{\pi}{4} \sqrt{L C}$
In $L C$-oscillation, maximum stored energy in inductor coil $L$ is $\frac{1}{2} L I_0^2$ and maximum stored energy in capacitor $(C)$ is $\frac{q_0^2}{2 C}$.
Since, energy is equally divided in $L$ and $C$. Hence, total stored energy
$$
\begin{aligned}
& E=\frac{1}{2} \times \text { maximum stored energy in either } L \text { o } \\
& \Rightarrow \frac{1}{2} L I^2=\frac{1}{2} \times \frac{1}{2} L I_0^2 \Rightarrow I^2=\left(\frac{I_0}{\sqrt{2}}\right)^2 \Rightarrow I=\frac{I_0}{\sqrt{2}}
\end{aligned}
$$
We know that, $I=I_0 \sin \omega t$
$$
\begin{aligned}
& \Rightarrow \quad \frac{I_0}{\sqrt{2}}=I_0 \sin \omega t \Rightarrow \sin \frac{\pi}{4}=\sin \frac{2 \pi}{T} \cdot t \\
& \Rightarrow \quad \frac{\pi}{4}=\frac{2 \pi}{T} \cdot t \Rightarrow t=\frac{T}{8}=\frac{2 \pi \sqrt{L C}}{8}=\frac{\pi \sqrt{L C}}{4}
\end{aligned}
$$
Since, energy is equally divided in $L$ and $C$. Hence, total stored energy
$$
\begin{aligned}
& E=\frac{1}{2} \times \text { maximum stored energy in either } L \text { o } \\
& \Rightarrow \frac{1}{2} L I^2=\frac{1}{2} \times \frac{1}{2} L I_0^2 \Rightarrow I^2=\left(\frac{I_0}{\sqrt{2}}\right)^2 \Rightarrow I=\frac{I_0}{\sqrt{2}}
\end{aligned}
$$
We know that, $I=I_0 \sin \omega t$
$$
\begin{aligned}
& \Rightarrow \quad \frac{I_0}{\sqrt{2}}=I_0 \sin \omega t \Rightarrow \sin \frac{\pi}{4}=\sin \frac{2 \pi}{T} \cdot t \\
& \Rightarrow \quad \frac{\pi}{4}=\frac{2 \pi}{T} \cdot t \Rightarrow t=\frac{T}{8}=\frac{2 \pi \sqrt{L C}}{8}=\frac{\pi \sqrt{L C}}{4}
\end{aligned}
$$
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