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:

  1. Comparing the $L-C$ oscillations with normal SHM, we get

$$ \begin{array}{rlrl} & \frac{d^{2} Q}{d t^{2}} & =-\omega^{2} Q \\ & \text { Here, } & \omega^{2} & =\frac{1}{L C} \\ \therefore & Q & =-L C \frac{d^{2} Q}{d t^{2}} \end{array} $$



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