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 $t\ge 0$,

• A. the charge on the capacitor is $Q=Q_0\cos \left(\frac{\pi }{2}+\frac{1}{\sqrt{LC}}\right)$.
• B. the charge on the capacitor is $Q_0\cos \left(\frac{\pi }{2}-\frac{1}{\sqrt{LC}}\right)$.
• C. the charge on the capacitor is $Q=-LC\frac{d^{2}Q}{d{t}^2}$
• D. the charge on the capacitor is $Q=-\frac{1}{\sqrt{LC}}\frac{d^{2}Q}{d{t}^2}$

Answer 1

Answer: (C)

We have the change $Q$ :
$Q=Q_0\cos \omega t$
$\frac{d^{2}Q}{dt2}=-Q_0{\omega }^2\cos \omega t$
$=-{\omega }^{2}Q=-\frac{Q}{LC}$
Therefore, the total change on the capacitor is
$Q=-LC\frac{d^{2}Q}{d{t}^2}$