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Step-by-Step Solution
Step 1: State the energies in the LC circuit
• The energy stored in the magnetic field of the inductor is
$ \displaystyle E_\text{mag} = \tfrac{1}{2} L\,i^2,\!$
• The energy stored in the electric field of the capacitor is
$ \displaystyle E_\text{elec} = \tfrac{1}{2}\,\tfrac{q^2}{C}.\!$
Step 2: Condition for equal energy sharing
At the instant when the energy is equally shared between the magnetic and electric fields, we have:
$ \displaystyle \tfrac{1}{2} L\,i^2 \;=\; \tfrac{1}{2}\,\tfrac{q^2}{C}.\!$
Step 3: Express charge and current in terms of time
For an LC circuit, the charge on the capacitor at time $t$ is:
$ \displaystyle q(t) = q_0 \cos \bigl(\omega t\bigr), \quad \text{where} \;\omega = \tfrac{1}{\sqrt{LC}}.\!$
The current is the time derivative of charge:
$ \displaystyle i(t) = \frac{dq}{dt} \;=\;-\,q_0\,\omega\,\sin\bigl(\omega t\bigr).\!$
Step 4: Substitute into the equality condition
From Step 2, we need
$ \displaystyle L\,i^2 \;=\;\frac{q^2}{C}.\!$
Substituting
$ \displaystyle i(t) = -\,q_0 \,\omega \,\sin\bigl(\omega t\bigr)\!$
and
$ \displaystyle q(t) = q_0 \cos\bigl(\omega t\bigr)\!$
gives
\[
L\,\bigl[q_0^2\,\omega^2\,\sin^2(\omega t)\bigr] \;=\;\frac{q_0^2\,\cos^2(\omega t)}{C}.
\]
Step 5: Simplify and solve for time
Canceling out the common factors $q_0^2$ and recalling that
$\omega^2 = \tfrac{1}{LC},$
we get:
\[
L \,\bigl(\tfrac{1}{LC}\bigr)\,\sin^2(\omega t) \;=\;\frac{\cos^2(\omega t)}{C}.
\]
Thus,
\[
\tfrac{1}{C}\,\sin^2(\omega t) \;=\;\tfrac{1}{C}\,\cos^2(\omega t),
\]
which implies
\[
\sin^2(\omega t) \;=\;\cos^2(\omega t)\quad \Longrightarrow \quad \tan^2(\omega t) = 1.
\]
Hence,
\[
\tan(\omega t) = 1 \quad\Longrightarrow\quad \omega t = \frac{\pi}{4},\;\frac{3\pi}{4},\dots
\]
The earliest positive time occurs when
$\omega t = \tfrac{\pi}{4}.$
Since
$\omega = \tfrac{1}{\sqrt{LC}},$
we have
\[
t = \frac{\pi}{4\,\omega} = \frac{\pi}{4}\,\sqrt{LC}.
\]
Final Answer
$ \displaystyle \boxed{ \, t \;=\;\frac{\pi}{4}\,\sqrt{LC}\, }\!
