Question

The capacitor of capacitance $C$ can be charged (with the help of a resistance R) by a voltage source $V$, by closing switch $S_{1}$ while keeping switch $S_{2}$ open. The capacitor can be connected in series with an inductor $L$ by closing switch $S_{2}$ and opening $S_{1}$.
If the total charge stored in the LC circuit is $Q_{0}$, then for $t \geq 0$,

• A. the charge on the capacitor is $Q =Q_{0} \cos \left(\frac{\pi}{2}+\frac{1}{\sqrt{L C}}\right)$.
• B. the charge on the capacitor is $Q_{0} \cos \left(\frac{\pi}{2}-\frac{1}{\sqrt{L C}}\right)$.
• C. the charge on the capacitor is $Q =- LC \frac{d^{2} Q}{d t^{2}}$
• D. the charge on the capacitor is $Q =-\frac{1}{\sqrt{L C}} \frac{d^{2} Q}{d t^{2}}$

Answer 1

Answer: (C)

We have the change $Q$ :
$ Q =Q_{0} \cos \omega t $
$\frac{d^{2} Q}{d t 2}=-Q_{0} \omega^{2} \cos \omega t$
$=-\omega^{2} Q=-\frac{Q}{L C}$
Therefore, the total change on the capacitor is
$Q =-L C \frac{d^{2} Q}{d t^{2}}$