1. Total energy in an ideal $LC$ circuit

In an oscillating $LC$ circuit, the total energy remains constant.

When the capacitor has maximum charge $Q$, the current is zero, so all the energy is electric:

$U_{\text{total}} = \frac{Q^2}{2C}$

2. Energy at any instant

If the instantaneous charge on the capacitor is $q$, then the electric energy is

$U_E = \frac{q^2}{2C}$

The magnetic energy in the inductor is

$U_B = U_{\text{total}} - U_E = \frac{Q^2}{2C} - \frac{q^2}{2C}$

3. Condition for equal sharing of energy

Given that energy is equally stored in electric and magnetic fields,

$U_E = U_B = \frac{U_{\text{total}}}{2}$

So,

$\frac{q^2}{2C} = \frac{1}{2}\cdot \frac{Q^2}{2C}$

Multiply both sides by $2C$:

$q^2 = \frac{Q^2}{2}$

Hence,

$q = \frac{Q}{\sqrt{2}}$

(we take the magnitude of charge as asked in the options)

4. Option check

• A: $\dfrac{Q}{2}$ ❌
• B: $\dfrac{Q}{\sqrt{3}}$ ❌
• C: $\dfrac{Q}{\sqrt{2}}$ ✅
• D: $Q$ ❌

Therefore, the correct answer is Option C.