Electro Magnetic Induction And Alternating Currents Question 411
Question: A fully charged capacitor C with initial charge $ q_{0} $ is connected to a coil of self-inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic fields is
Options:
A) $ \pi \sqrt{LC} $
B) $ \frac{\pi }{4}\sqrt{LC} $
C) $ 2\pi \sqrt{LC} $
D) $ \sqrt{LC} $
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Answer:
Correct Answer: B
Solution:
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Charge on the capacitor at any time ‘f’,
$ q=q_{0}\cos \omega t $ …(i) At time when energy stored equally in electric and magnetic field, at this time Energy of a capacitor $ =\frac{1}{2} $
Total energy $ \frac{1}{2}\frac{q^{2}}{C}=\frac{1}{2}( \frac{1}{2}\frac{q_{0}^{2}}{C} )\Rightarrow q=\frac{q_{0}}{\sqrt{2}} $
From equation (i)
$ \frac{q_{0}}{\sqrt{2}}=q_{0}\cos \omega t $ $ \cos \omega t=\frac{1}{\sqrt{2}}\Rightarrow \omega t={{\cos }^{-1}}( \frac{1}{\sqrt{2}} )=\frac{\pi }{4} $ $ t=\frac{\pi }{4\omega }=\frac{\pi }{4}\sqrt{LC}( \therefore \omega =\frac{1}{\sqrt{LC}} ) $
