1. Set up the LC oscillation

A charged capacitor $C$ with initial charge $q_0$ is connected to an inductor $L$. The system performs LC oscillations.

The charge on the capacitor at time $t$ is

$q(t)=q_0\cos \omega t$

where

$\omega = \frac{1}{\sqrt{LC}}$

The current is

$i(t)=\frac{dq}{dt}=-\omega q_0 \sin \omega t$

2. Write electric and magnetic energies

• Energy stored in the capacitor:

$U_E = \frac{q^2}{2C} = \frac{q_0^2}{2C}\cos^2 \omega t$

• Energy stored in the inductor:

$U_B = \frac{1}{2}Li^2$

Substitute $i = -\omega q_0 \sin \omega t$:

$U_B = \frac{1}{2}L\omega^2 q_0^2 \sin^2 \omega t$

Since

$\omega^2 = \frac{1}{LC}$

we get

$L\omega^2 = \frac{1}{C}$

Hence,

$U_B = \frac{q_0^2}{2C}\sin^2 \omega t$

3. Condition for equal sharing of energy

For equal energy in electric and magnetic fields,

$U_E = U_B$

So,

$\frac{q_0^2}{2C}\cos^2 \omega t = \frac{q_0^2}{2C}\sin^2 \omega t$

$\cos^2 \omega t = \sin^2 \omega t$

This gives

$\tan^2 \omega t = 1$

The first positive time is

$\omega t = \frac{\pi}{4}$

Therefore,

$t = \frac{\pi}{4\omega} = \frac{\pi}{4}\sqrt{LC}$

4. Match with options

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

So the correct option is A.

5. Comparison with stored answer

Stored correct answer: A

Derived answer: A

They match.