In an oscillating L-C circuit, the maximum charge on the capacitor is Q. The charge on the capacitor

Q: Why is the total energy equal to the maximum energy in the capacitor?

At the instant the capacitor holds its maximum charge $Q$, the current in the circuit is zero. Since magnetic energy depends on current ($U_B = \frac{1}{2}Li^2$), it is zero. Thus, all the system's energy is stored in the electric field of the capacitor.

Q: Can we solve this using the equations of Simple Harmonic Motion (SHM)?

Yes, absolutely! You can write $q = Q \cos(\omega t)$ and $i = -Q\omega \sin(\omega t)$. By equating $U_E$ and $U_B$, you can find the specific time $t$, and then plug it back into the charge equation. However, the energy conservation method is much faster and less prone to algebraic errors.

Q: What happens to the energy when it's not equally divided?

The energy continuously transforms back and forth between the electric field and the magnetic field. When one decreases, the other increases by the exact same amount, ensuring their sum always equals the constant total energy.

Q: Why did we replace $U_B$ with $U_E$ in the conservation equation?

The question asks us to find the instantaneous charge $q$. Since $q$ is a variable in the formula for electric energy ($U_E = \frac{q^2}{2C}$), it is mathematically strategic to express the entire equation in terms of $U_E$ to isolate and solve for $q$.

Q: Is the total energy always perfectly constant in real life?

In an ideal $LC$ circuit with zero resistance, yes. However, in real-world circuits, the connecting wires and the inductor have some resistance. This causes the energy to gradually dissipate as heat (Joule heating), leading to damped oscillations.

Question

In an oscillating $L-C$ circuit, the maximum charge on the capacitor is $Q$. The charge on the capacitor when the energy is stored equally between the electric and magnetic fields is

Options:

(A) $\frac{Q}{2}$

(B) $\frac{Q}{\sqrt{3}}$

(C) $\frac{Q}{\sqrt{2}}$

(D) $Q$

Sigma Percentile · Accepted Answer

An $LC$ circuit is one of the most poetic systems in physics. It behaves exactly like a mechanical pendulum, but instead of kinetic and potential energy trading places, the energy sloshes back and forth between the **electric field** of the capacitor and the **magnetic field** of the inductor.

### Analyzing the Setup

Imagine the moment the circuit is switched on. Let's say the capacitor is fully charged with a maximum charge, $Q$. At this precise instant, the current in the circuit is zero. Because there is no current flowing through the inductor, there is no magnetic field.

Therefore, all the energy in the system is purely electrical. We can write the total energy of our system as the maximum energy stored in the capacitor:

$$U_{	ext{total}} = U_{E_{	ext{max}}} = \frac{Q^2}{2C}$$

### The Master Equation

As time ticks forward, the capacitor begins to discharge, pushing current through the inductor. The electric energy decreases, and the magnetic energy grows. The question asks us to investigate a very specific, magical instant: the exact moment when the energy is shared equally between the electric and magnetic fields.

Mathematically, this condition is written as:

$$U_E = U_B$$

By the fundamental law of conservation of energy, the sum of the electric energy and magnetic energy at any instant must always equal the total maximum energy we started with.

$$U_E + U_B = U_{	ext{total}}$$

Since the energies are perfectly equal at this instant, we can cleverly replace the magnetic energy ($U_B$) with the electric energy ($U_E$) in our conservation equation. This simplifies our path immensely:

$$U_E + U_E = U_{	ext{total}}$$

$$2U_E = U_{	ext{total}}$$

### Final Calculation

Now, let's substitute the formulas for these energies. The instantaneous electric energy is $\frac{q^2}{2C}$, and the total energy is $\frac{Q^2}{2C}$. Plugging these in gives us:

$$2 \left( \frac{q^2}{2C} ight) = \frac{Q^2}{2C}$$

The denominator, $2C$, cancels out beautifully from both sides of the equation. We are left with a very clean algebraic relation:

$$2q^2 = Q^2$$

Taking the square root of both sides, we find the instantaneous charge $q$:

**$$q = \frac{Q}{\sqrt{2}}$$**

This elegant result tells us that when the energy is split 50-50, the charge on the capacitor is $1/\sqrt{2}$ times its maximum value. If you were asked to find the current at this instant, you could use the exact same logic, substituting $U_E$ with $U_B$ to find that $i = I_{	ext{max}}/\sqrt{2}$. Always keep exploring!

Animated Solution for Physics - Electromagnetic Induction: In an oscillating $L - C$ circuit, the maximum charge on the capacitor is $Q$ . The charge on the capacitor when the energy is stored equally between the electric and magnetic fields is

Select Answer:

Visualized Solution

$L C$ Circuit Oscillation

Maximum Energy

The Given Condition

Energy Conservation

Substituting the Condition

Expanding the Energies

Final Calculation

The Way Forward

The Sigma Insight: Alternating Current (AC) and Voltage

Analyzing the Setup

The Master Equation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LC Circuit Oscillation

Maximum Energy

The Given Condition

Energy Conservation

Substituting the Condition

Expanding the Energies

Final Calculation

The Way Forward

The Sigma Insight: Alternating Current (AC) and Voltage

Analyzing the Setup

The Master Equation

Final Calculation

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In a series resonant L-C-R circuit, the voltage across R is 100 V and R=1 kΩ with C=2μF. The resonant frequency ω is 200 rad/s. At resonance, the voltage across L is

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$L C$ Circuit Oscillation

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