Physics - Capacitor Question with Solution | TestHub

Total energy of the LC circuit is $U=\frac{Q_{max}^2}{2C}$

Energy stored in the electric field is $U_E=\frac{q^2}{2C}$

Energy stored in the magnetic field is $U_B=U-U_E$

Given that $U_E=U_B$, we have $U_E=\frac{U}{2}$

$\frac{q^2}{2C}=\frac{1}{2}\left(\frac{Q_{max}^2}{2C}\right)$

$q=\frac{Q_{max}}{\sqrt{2}}$

Substituting $Q_{max}=12Q$, we get:

$q=\frac{12Q}{\sqrt{2}}=6\sqrt{2}Q$

Answer: $q=6\sqrt{2}Q$