A fully charged capacitor C with initial charge q_0 is connected to a coil of self inductance L at t

Q: Why do we use cosine for charge instead of sine?

Because at $t=0$, the capacitor is fully charged. The cosine function starts at its maximum value ($\cos(0)=1$), which perfectly matches our initial condition.

Q: Where did the negative sign in the current go?

When we equate the energies, the current term is squared ($\frac{1}{2}Li^2$). Squaring a negative value makes it positive, so the negative sign vanishes.

Q: Why is $\omega = \frac{1}{\sqrt{LC}}$?

This is the natural angular frequency of an LC circuit, derived from the differential equation of the circuit: $\frac{d^2q}{dt^2} + \frac{1}{LC}q = 0$.

Q: Why do we take $\tan(\omega t) = 1$ and not $-1$?

The equation $\tan^2(\omega t) = 1$ gives $\tan(\omega t) = \pm 1$. We want the *first* time the energies are equal, which corresponds to the smallest positive angle, $\frac{\pi}{4}$, where $\tan$ is positive.

Q: Does the total energy of the system change?

No, in an ideal LC circuit with no resistance, the total energy remains constant. It just continuously transforms back and forth between the electric field of the capacitor and the magnetic field of the inductor.

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

(B) $2\pi\sqrt{LC}$

(C) $\sqrt{LC}$

(D) $\pi\sqrt{LC}$

Sigma Percentile · Accepted Answer

### The Dance of Energy in an LC Circuit

Imagine a fully charged capacitor connected to an inductor. The moment we close the circuit, the charge doesn't just disappear; it starts to oscillate back and forth, much like a pendulum swinging.

Since it starts with maximum charge, we can describe this charge using a cosine function:

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

Now, how fast does this oscillation happen? That's determined by the angular frequency, $\omega$, which is simply $\frac{1}{\sqrt{LC}}$. This is the natural rhythm of our LC circuit.

### The Flow of Current

As the charge moves, it creates a current. Current is just the rate of change of charge, so we differentiate our charge equation with respect to time.

The derivative of cosine is negative sine, giving us the current:

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

The energy in this system constantly shifts between two forms. The capacitor stores electric energy, $U_E$, which is $\frac{q^2}{2C}$. The inductor stores magnetic energy, $U_B$, which is $\frac{1}{2}Li^2$.

### Finding the Perfect Balance

The question asks for the exact moment when these two energies are perfectly equal. So, let's set $U_E$ equal to $U_B$. This is our master equation for the problem:

$$\frac{q^2}{2C} = \frac{1}{2}Li^2$$

Let's substitute our expressions for charge and current into this equation. Don't get intimidated by the squares; just carefully plug in the cosine term for $q$ and the sine term for $i$:

$$\frac{(q_0 \cos(\omega t))^2}{2C} = \frac{1}{2}L(-q_0 \omega \sin(\omega t))^2$$

Squaring both sides, the negative sign on the current vanishes. We get:

$$\frac{q_0^2 \cos^2(\omega t)}{2C} = \frac{1}{2}L q_0^2 \omega^2 \sin^2(\omega t)$$

### The Magic of Substitution

Here is a catch. We have an $\omega^2$ on the right. Remember our angular frequency? $\omega^2$ is simply $\frac{1}{LC}$. Let's substitute that in. Notice how beautifully the $L$ cancels out!

$$\frac{1}{2}L q_0^2 \left(\frac{1}{LC}ight) \sin^2(\omega t) = \frac{q_0^2 \sin^2(\omega t)}{2C}$$

Now, look at the equation. The $\frac{q_0^2}{2C}$ terms cancel out completely from both sides! We are left with a very simple trigonometric equation:

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

Which means:

$$	an^2(\omega t) = 1$$

### The Final Countdown

Since we want the first time this happens, we take $	an(\omega t) = 1$. This gives:

$$\omega t = \frac{\pi}{4}$$

Finally, dividing by $\omega$, which is multiplying by $\sqrt{LC}$, we get our answer:

$$t = \frac{\pi}{4\omega} = \frac{\pi}{4}\sqrt{LC}$$

**This is the exact moment the energy is perfectly balanced between the electric and magnetic fields!**

Animated Solution for Physics - Electromagnetic Induction: 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

Select Answer:

Visualized Solution

LC Circuit Oscillation

Angular Frequency $ω$

Current $i (t)$ in the Circuit

Energy $U_{E}$ and $U_{B}$

Equating $U_{E} = U_{B}$

Substituting Expressions

Simplifying the Equation

Using $ω^{2} = \frac{1}{L C}$

Solving for Time $t$

Final Calculation

The Way Forward

The Sigma Insight: Alternating Current (AC) and Voltage

The Dance of Energy in an LC Circuit

The Flow of Current

Finding the Perfect Balance

The Magic of Substitution

The Final Countdown

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