Question

A fully charged capacitor $C$ with initial charge $Q_{0}$ is connected to an ideal inductor of self-inductance $L$ at $t=0$ . The time at which energy is stored equally between the electric and the magnetic fields is

• A. $\frac{\pi }{4}\sqrt{L C}$
• B. $2\pi \sqrt{L C}$
• C. $\sqrt{L C}$
• D. $\pi \sqrt{L C}$

Answer 1

Answer: (A)

The charge on the capacitor at any instant of time is
$Q=Q_{0}cos\left(\omega t\right)$
$\Rightarrow i=\frac{d Q}{d t}=-\omega Q_{0}sin\left(\omega t\right)$
$\frac{1}{2}Li^{2}=\frac{Q^{2}}{2 C}$
$\left(\Rightarrow \frac{1}{2} L \, \left[\omega Q_{0} sin \left(\omega t\right)\right]\right)^{2}=\frac{\left(\left[Q_{0} cos \left(\omega t\right)\right]\right)^{2}}{2 C}$
$\omega =\frac{1}{\sqrt{L C}}$
$\Rightarrow tan\left(\omega t\right)=1$
$\omega t= tan^{- 1}\left(1\right)=\frac{\pi }{4}$
$t=\frac{\pi }{4 \omega }=\frac{\pi }{4}\sqrt{L C}$