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A fully charged capacitor $C$ with initial charge $q_0$ is connected to a coil of self inductance $L$ at $t=0$. The time at which the energy is stored equally between the electric and the magnetic field is :
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Verified Answer
The correct answer is:
$\frac{\pi}{4} \sqrt{L C}$
$\frac{\pi}{4} \sqrt{L C}$
Charge oscillates simple harmonic motion $q=q_0 \sin \omega t, U=\frac{1}{2} \frac{q^2}{C}$
$$
\begin{aligned}
& q=\frac{q_0}{\sqrt{2}} \Rightarrow \omega t=\frac{\pi}{4} \\
& \Rightarrow t=\frac{T}{8}=\frac{2 \pi}{8} \sqrt{L C}=\frac{\pi}{4} \sqrt{L C}
\end{aligned}
$$
$$
\begin{aligned}
& q=\frac{q_0}{\sqrt{2}} \Rightarrow \omega t=\frac{\pi}{4} \\
& \Rightarrow t=\frac{T}{8}=\frac{2 \pi}{8} \sqrt{L C}=\frac{\pi}{4} \sqrt{L C}
\end{aligned}
$$
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