A fully charged capacitor C with initial char | vedclass

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(D) The total energy in an $LC$ circuit is constant and is given by $U = U_E + U_B$,where $U_E$ is the electric energy and $U_B$ is the magnetic energ

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$\frac{\pi}{4} \sqrt{LC}$

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(D) The total energy in an $LC$ circuit is constant and is given by $U = U_E + U_B$,where $U_E$ is the electric energy and $U_B$ is the magnetic energy.At $t = 0$,the charge on the capacitor is $q_0$,so the total energy is $U = \frac{q_0^2}{2C}$.We want to find the time $t$ when $U_E = U_B = \frac{U}{2}$.Since $U_E = \frac{q^2}{2C}$,we have $\frac{q^2}{2C} = \frac{1}{2} \frac{q_0^2}{2C}$,which implies $q^2 = \frac{q_0^2}{2}$,or $q = \frac{q_0}{\sqrt{2}}$.The charge on the capacitor at time $t$ is given by $q(t) = q_0 \cos(\omega t)$,where $\omega = \frac{1}{\sqrt{LC}}$.Setting $q(t) = \frac{q_0}{\sqrt{2}}$,we get $\cos(\omega t) = \frac{1}{\sqrt{2}}$.This implies $\omega t = \frac{\pi}{4}$.Substituting $\omega = \frac{1}{\sqrt{LC}}$,we get $t = \frac{\pi}{4} \sqrt{LC}$.

$A$ 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 fields is

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