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 in the form of electric field in the capacitor and the magnetic field in the inductor is:

• A. \(\pi\sqrt{LC}\)
• B. \(\dfrac{\pi}{4}\sqrt{LC}\)
• C. \(2\pi\sqrt{LC}\)
• D. \(\sqrt{LC}\)

Hint:

In an LC circuit: \[ U_E = U_B \quad \text{when} \quad \omega t = \frac{\pi}{4} \] This occurs at one-eighth of the total oscillation period.

Answer 1

The Correct Option is B

Step 1: Recall the nature of an LC circuit. When a charged capacitor is connected to an inductor, the system performs electromagnetic oscillations with angular frequency: \[ \omega = \frac{1}{\sqrt{LC}} \]
Step 2: Write expressions for energies. Electric energy stored in the capacitor at time \(t\): \[ U_E = \frac{q^2}{2C} \] Magnetic energy stored in the inductor: \[ U_B = \frac{1}{2}Li^2 \]
Step 3: Express charge and current as functions of time. For an LC circuit: \[ q = q_0 \cos(\omega t) \] \[ i = -q_0\omega \sin(\omega t) \]
Step 4: Substitute into energy expressions. Electric energy: \[ U_E = \frac{q_0^2}{2C}\cos^2(\omega t) \] Magnetic energy: \[ U_B = \frac{1}{2}L(q_0^2\omega^2)\sin^2(\omega t) \] Since \(\omega^2 = \frac{1}{LC}\), \[ U_B = \frac{q_0^2}{2C}\sin^2(\omega t) \]
Step 5: Condition for equal energies. \[ U_E = U_B \] \[ \cos^2(\omega t) = \sin^2(\omega t) \] \[ \tan^2(\omega t) = 1 \Rightarrow \omega t = \frac{\pi}{4} \]
Step 6: Find the required time. \[ t = \frac{\pi}{4\omega} = \frac{\pi}{4}\sqrt{LC} \]
Hence, the correct answer is \(\boxed{\dfrac{\pi}{4}\sqrt{LC}}\).