Electromagnetic Induction and Alternating Current 4 Question 14
####14. If the total charge stored in the $L C$ circuit is $Q_{0}$, then for $t \geq 0$,
$(2006,6 \mathrm{M})$
(a) the charge on the capacitor is $Q=Q_{0} \cos \frac{\pi}{2}+\frac{t}{\sqrt{L C}}$
(b) the charge on the capacitor is $Q=Q_{0} \cos \frac{\pi}{2}-\frac{t}{\sqrt{L C}}$
(c) the charge on the capacitor is $Q=-L C \frac{d^{2} Q}{d t^{2}}$
(d) the charge on the capacitor is $Q=-\frac{1}{\sqrt{L C}} \frac{d^{2} Q}{d t^{2}}$
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Answer:
Correct Answer: 14. (c)
Solution:
- Comparing the $L-C$ oscillations with normal SHM, we get
$$ \begin{array}{rlrl} & \frac{d^{2} Q}{d t^{2}} & =-\omega^{2} Q \\ & \text { Here, } & \omega^{2} & =\frac{1}{L C} \\ \therefore & Q & =-L C \frac{d^{2} Q}{d t^{2}} \end{array} $$
