An inductor of inductance 2.0 mH is connected across a charged capacitor of capacitance 5.0 uF and t

Q: Why is the current zero when the charge is maximum?

In an $L-C$ circuit, energy oscillates between the capacitor (electric field) and the inductor (magnetic field). When the charge is maximum, all energy is in the capacitor, meaning the magnetic energy (and thus the current) must be zero.

Q: How did we get the equation for $\frac{dI}{dt}$?

By differentiating the current equation $I(t) = -Q_0 \omega \sin(\omega t)$ with respect to time, we get $\frac{dI}{dt} = -Q_0 \omega^2 \cos(\omega t)$.

Q: Can we solve part (d) without energy conservation?

Yes! You can use the trigonometric identity $\sin^2(\omega t) + \cos^2(\omega t) = 1$. Since $I \propto \sin(\omega t)$ and $Q \propto \cos(\omega t)$, knowing one allows you to find the other.

Q: Why do we use the absolute value for $\frac{dI}{dt}$?

The question asks for the magnitude of the rate of change of current, which represents how fast the current is changing regardless of whether it's increasing or decreasing.

Q: What does the natural frequency $\omega$ represent physically?

It represents the rate at which the energy sloshes back and forth between the inductor and the capacitor, measured in radians per second.

Question

An inductor of inductance $2.0	ext{ mH}$ is connected across a charged capacitor of capacitance $5.0\,\mu	ext{F}$ and the resulting $L-C$ circuit is set oscillating at its natural frequency. Let $Q$ denote the instantaneous charge on the capacitor and $I$, the current in the circuit. It is found that the maximum value of $Q$ is $200\,\mu	ext{C}$.
(a) When $Q = 100\,\mu	ext{C}$, what is the value of $\left|\frac{dI}{dt}ight|$?
(b) When $Q = 200\,\mu	ext{C}$, what is the value of $I$?
(c) Find the maximum value of $I$.
(d) When $I$ is equal to one-half its maximum value, what is the value of $|Q|$?

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine an $L-C$ circuit where an inductor and a capacitor are connected together. The capacitor is fully charged initially, with a maximum charge $Q_0$ of $200\,\mu	ext{C}$. As soon as the circuit is closed, the charge begins to oscillate back and forth between the capacitor and the inductor. This is a classic example of simple harmonic motion in electromagnetism.

First, we need to find the natural frequency of this oscillation. The formula for the angular frequency $\omega$ is given by:

$$ \omega = \frac{1}{\sqrt{LC}} $$

Let's substitute the given values. The inductance $L$ is $2.0	ext{ mH}$ ($2.0 	imes 10^{-3}	ext{ H}$) and the capacitance $C$ is $5.0\,\mu	ext{F}$ ($5.0 	imes 10^{-6}	ext{ F}$).

$$ \omega = \frac{1}{\sqrt{(2.0 	imes 10^{-3})(5.0 	imes 10^{-6})}} = 10^4	ext{ rad/s} $$

### The Master Equations

Now let's set up our equations of motion. The charge $Q$ as a function of time can be written as:

$$ Q(t) = Q_0 \cos(\omega t) $$

The current $I$ is the rate of change of charge, $\frac{dQ}{dt}$, which upon differentiation gives:

$$ I(t) = \frac{dQ}{dt} = -Q_0 \omega \sin(\omega t) $$

And the rate of change of current, $\frac{dI}{dt}$, will be:

$$ \frac{dI}{dt} = -Q_0 \omega^2 \cos(\omega t) $$

### Solving the Sub-parts

**Part (a):** When the charge is $100\,\mu	ext{C}$, which is exactly half of the maximum charge $Q_0$, the value of $\cos(\omega t)$ must be $\frac{1}{2}$. We need to find the magnitude of $\frac{dI}{dt}$. Substituting the values into our equation:

$$ \left|\frac{dI}{dt}ight| = Q_0 \omega^2 |\cos(\omega t)| = (2.0 	imes 10^{-4})(10^4)^2 \left(\frac{1}{2}ight) = 10^4	ext{ A/s} $$

**Part (b):** We are asked for the current when the charge is at its maximum, $200\,\mu	ext{C}$. When the charge is maximum, $\cos(\omega t) = 1$, which means $\sin(\omega t) = 0$. Therefore, the current $I$ becomes zero. Physically, at this exact moment, all the energy is stored entirely within the capacitor's electric field.

**Part (c):** We need to find the maximum current. From the current equation, it's clear that the maximum current $I_{\max}$ occurs when $\sin(\omega t) = 1$. So $I_{\max}$ is simply $Q_0 \omega$.

$$ I_{\max} = (2.0 	imes 10^{-4})(10^4) = 2.0	ext{ A} $$

**Part (d):** When the current is half of its maximum, which is $1.0	ext{ A}$, what is the charge? Here we will use the principle of energy conservation. The total energy, $\frac{1}{2} L I_{\max}^2$, equals the sum of the instantaneous energies in the inductor and capacitor:

$$ \frac{1}{2} L I_{\max}^2 = \frac{1}{2} L I^2 + \frac{1}{2} \frac{Q^2}{C} $$

Rearranging and solving for $Q$:

$$ Q = \sqrt{LC(I_{\max}^2 - I^2)} $$

Substituting the values gives us:

$$ Q = \sqrt{(10^{-8})(2.0^2 - 1.0^2)} = \sqrt{3 	imes 10^{-8}} = \sqrt{3} 	imes 10^{-4}	ext{ C} \approx 1.732 	imes 10^{-4}	ext{ C} $$

This question beautifully tests the concepts of $L-C$ oscillations and energy conservation. Always be careful with unit conversions to avoid silly mistakes!

Visualized Solution

$Visual Anchor: L-C Circuit$

$Natural Frequency ω$

$Equations of Motion$

$Part (a): \frac{d I}{d t} when Q = 100 μ C$

$Part (b): I when Q = 200 μ C$

$Part (c): Maximum Current I_{m a x}$

$Part (d): ∣ Q ∣ when I = \frac{I _{m a x}}{2}$

$Conclusion$

The Sigma Insight: Alternating Current (AC) and Voltage

Analyzing the Setup

The Master Equations

Solving the Sub-parts

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visual Anchor: L-C Circuit

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Natural Frequency ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equations of Motion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Part (a): ​dtdI​​ when Q=100μC

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Part (b): I when Q=200μC

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Part (c): Maximum Current Imax​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Part (d): ∣Q∣ when I=2Imax​​

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