Question

A fully charged capacitor 'C' with initial charge 'q0' is connected to a coil of self inductance 'L' at t = 0. The time at which the energy is stored equally between the electric field and the magnetic field is

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

Answer 1

The Correct Option is C

A fully charged capacitor \(C\) with an initial charge \(q_0\) is connected to a coil of self-inductance \(L\) at time \(t = 0\). We want to find the time \(t\) at which the energy stored in the electric field of the capacitor is equal to the energy stored in the magnetic field of the inductor.

In an LC circuit, the charge on the capacitor oscillates sinusoidally. The charge \(q(t)\) on the capacitor as a function of time is given by:

\(q(t) = q_0 \cos(\omega t)\)

Where \(\omega\) is the angular frequency of oscillation, and \(\omega = \frac{1}{\sqrt{LC}}\). 

The energy stored in the capacitor at time \(t\) is:

\(U_C = \frac{q(t)^2}{2C} = \frac{q_0^2 \cos^2(\omega t)}{2C}\)

The current \(i(t)\) in the inductor is the derivative of the charge with respect to time:

\(i(t) = \frac{dq(t)}{dt} = -q_0 \omega \sin(\omega t)\)

The energy stored in the inductor at time \(t\) is:

\(U_L = \frac{1}{2}Li(t)^2 = \frac{1}{2}L(-q_0 \omega \sin(\omega t))^2 = \frac{1}{2}Lq_0^2 \omega^2 \sin^2(\omega t)\)

Since \(\omega = \frac{1}{\sqrt{LC}}\), then \(\omega^2 = \frac{1}{LC}\), and we can rewrite the inductor energy as:

\(U_L = \frac{1}{2}Lq_0^2 \frac{1}{LC} \sin^2(\omega t) = \frac{q_0^2 \sin^2(\omega t)}{2C}\)

We want to find the time \(t\) when \(U_C = U_L\). Therefore:

\(\frac{q_0^2 \cos^2(\omega t)}{2C} = \frac{q_0^2 \sin^2(\omega t)}{2C}\)

This implies \(\cos^2(\omega t) = \sin^2(\omega t)\), which means \(\tan^2(\omega t) = 1\), or \(\tan(\omega t) = \pm 1\).

The first positive solution for \(\omega t\) is \(\omega t = \frac{\pi}{4}\). Therefore:

\(t = \frac{\pi}{4\omega} = \frac{\pi}{4 \times \frac{1}{\sqrt{LC}}} = \frac{\pi}{4} \sqrt{LC}\)

Therefore, the time at which the energy is stored equally between the electric field and the magnetic field is:

\(\frac{\pi}{4} \sqrt{LC}\)

Answer 2

In an LC circuit, the charge on the capacitor oscillates with time as:

\(q(t) = q_0 \cos(\omega t)\)

where \(\omega = \frac{1}{\sqrt{LC}}\) is the angular frequency.

The energy stored in the capacitor is given by:

\(U_C = \frac{q^2}{2C} = \frac{q_0^2 \cos^2(\omega t)}{2C}\)

The current in the inductor is given by:

\(i(t) = \frac{dq}{dt} = -q_0 \omega \sin(\omega t)\)

The energy stored in the inductor is given by:

\(U_L = \frac{1}{2} L i^2 = \frac{1}{2} L (q_0^2 \omega^2 \sin^2(\omega t)) = \frac{1}{2} L q_0^2 \frac{1}{LC} \sin^2(\omega t) = \frac{q_0^2 \sin^2(\omega t)}{2C}\)

We want to find the time \(t\) when the energy is equally stored between the electric field and the magnetic field, i.e., \(U_C = U_L\). So,

\(\frac{q_0^2 \cos^2(\omega t)}{2C} = \frac{q_0^2 \sin^2(\omega t)}{2C}\)

\(\cos^2(\omega t) = \sin^2(\omega t)\)

\(\tan^2(\omega t) = 1\)

\(\tan(\omega t) = \pm 1\)

The smallest positive solution is when \(\omega t = \frac{\pi}{4}\). Therefore,

\(t = \frac{\pi}{4\omega} = \frac{\pi}{4} \sqrt{LC}\)

The time at which the energy is stored equally between the electric field and the magnetic field is \(\frac{\pi}{4} \sqrt{LC}\).