Question

In L-C oscillation the maximum charge on the capacitor can be Q. If at any instant, electric energy and magnetic energy associated with the current is equal, then the charge on the capacitor at that instant is :

• A. $\frac{Q}{\sqrt{2}}$
• B. $\frac{Q}{2}$
• C. $\frac{3Q}{\sqrt{2}}$
• D. $\frac{3Q}{2}$

Answer 1

The correct option is A $\frac{Q}{\sqrt{2}}$

Total energy $U=\frac{1}{2}\frac{Q^2}{C}$

Let at any instant the charge on capacitor be $Q$

$\therefore$ Electrical energy = Magnetic energy $=\frac{1}{2}\frac{Q^2}{C}$

So, new total energy $=\frac{1}{2}\frac{Q^{\prime 2}}{C}+\frac{1}{2}\frac{Q^{\prime 2}}{C}=\frac{Q^{\prime 2}}{C}$

According to question

$\frac{1}{2}\frac{Q^2}{C}=\frac{Q^{\prime 2}}{C}$

$\Rightarrow Q^{\prime }=\frac{Q}{\sqrt{2}}$