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L-C Oscillations. A capacitor with capacitance \(6.00 \times\) \(10^{-5} \mathrm{F}\) is charged by connecting it to a \(12.0-\mathrm{V}\) battery. The capacitor is disconnected from the battery and connected across an inductor with \(L=1.50 \mathrm{H}\) (a) What are the angular frequency \(\omega\) of the electrical oscillations and the period of these oscillations (the time for one oscillation \() ?(\text { b) What is the initial charge on the }\) capacitor? (c) How much energy is initially stored in the capacitor? (d) What is the charge on the capacitor 0.0230 \(\mathrm{s}\) after the connection to the inductor is made? Interpret the sign of your answer.(e) At the time given in part (d), what is the current in the inductor? Interpret the sign of your answer. (f) At the time given in part (d), how much electrical energy is stored in the capacitor and how much is stored in the inductor?

Step by step solution

01

Determine Angular Frequency

The angular frequency of oscillations in an LC circuit is given by \( \omega = \frac{1}{\sqrt{LC}} \). Given \( L = 1.50 \text{ H} \) and \( C = 6.00 \times 10^{-5} \text{ F} \), we can plug these values into the formula:\[ \omega = \frac{1}{\sqrt{1.50 \times 6.00 \times 10^{-5}}} = 333 \text{ rad/s} \]

02

Calculate the Period of Oscillations

The period \( T \) of the oscillations is the reciprocal of the frequency. We know that the frequency \( f \) is \( \frac{\omega}{2\pi} \), so the period is given by:\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{333} \approx 0.0189 \text{ s} \]

03

Find Initial Charge on the Capacitor

The initial charge \( Q \) on the capacitor can be found using the relationship \( Q = CV \). Using the given \( C = 6.00 \times 10^{-5} \text{ F} \) and \( V = 12.0 \text{ V} \), we have:\[ Q = 6.00 \times 10^{-5} \times 12.0 = 7.20 \times 10^{-4} \text{ C} \]

04

Calculate Initial Energy in the Capacitor

The initial energy \( U \) stored in the capacitor is given by \( \frac{1}{2}CV^2 \). Substituting the given values:\[ U = \frac{1}{2} \times 6.00 \times 10^{-5} \times 12.0^2 = 4.32 \times 10^{-3} \text{ J} \]

05

Compute Charge on Capacitor at 0.0230 s

The charge on the capacitor at time \( t \), \( Q(t) \), in an LC circuit can be expressed as \( Q(t) = Q_0 \cos(\omega t) \), where \( Q_0 \) is the initial charge.At \( t = 0.0230 \text{ s} \), with \( \omega = 333 \) and \( Q_0 = 7.20 \times 10^{-4} \):\[ Q(0.0230) = 7.20 \times 10^{-4} \cos(333 \times 0.0230) \approx -4.54 \times 10^{-4} \text{ C} \]

06

Determine Current in the Inductor at 0.0230 s

The current \( I(t) \) in the inductor is given by \( I(t) = -\omega Q_0 \sin(\omega t) \).Substituting the known values at \( t = 0.0230 \text{ s} \):\[ I(0.0230) = -333 \times 7.20 \times 10^{-4} \sin(333 \times 0.0230) \approx 0.222 \text{ A} \]

07

Calculate Energy Stored in Capacitor and Inductor at 0.0230 s

The energy stored in the capacitor at time \( t \) is \( U_C = \frac{1}{2} C [Q(t)]^2 \), and in the inductor, it is \( U_L = \frac{1}{2} L I^2 \).Using \( Q(0.0230) = -4.54 \times 10^{-4} \):\[ U_C = \frac{1}{2} \times 6.00 \times 10^{-5} \times (-4.54 \times 10^{-4})^2 \approx 6.17 \times 10^{-5} \text{ J} \]Using \( I(0.0230) = 0.222 \text{ A} \):\[ U_L = \frac{1}{2} \times 1.50 \times (0.222)^2 \approx 3.70 \times 10^{-3} \text{ J} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Frequency Calculation

In an LC circuit, calculating the angular frequency is essential. Angular frequency, denoted as \( \omega \), signifies how rapidly the oscillations occur within these circuits. The formula employed to determine \( \omega \) is: \( \omega = \frac{1}{\sqrt{LC}} \). Here, \( L \) is the inductance in henrys (H), and \( C \) is the capacitance in farads (F).

Plug in the values to find the angular frequency for any circuit configuration. For the example given, with \( L = 1.50 \text{ H} \) and \( C = 6.00 \times 10^{-5} \text{ F} \), the calculation becomes \( \omega = \frac{1}{\sqrt{1.50 \times 6.00 \times 10^{-5}}} \), resulting in a \( \omega \) of approximately \( 333 \text{ rad/s} \).

This numeric insight into angular frequency helps in fully grasping how fast a specific LC circuit undergoes oscillation cycles.

Period of Oscillation

The period of oscillation, denoted as \( T \), indicates the duration needed to complete one full cycle of an electrical oscillation in an LC circuit. It is the reciprocal of the frequency, which is derived from angular frequency.

To find the period, use the formula \( T = \frac{2\pi}{\omega} \). A useful approach involves first calculating the frequency, \( f = \frac{\omega}{2\pi} \), then inverting the result to find \( T \). For the current scenario, given \( \omega = 333 \text{ rad/s} \), the period of oscillation translates to \( T = \frac{2\pi}{333} \approx 0.0189 \text{ seconds} \).

This value signifies the time it takes for one cycle of oscillation, helping predict the circuit's behavior over time.

Energy Stored in Capacitor

The initial energy stored in a charged capacitor is critical for understanding its potential to do work. Energy in a capacitor is stored as an electric field, resulting from separated charges across its plates.

The equation \( U = \frac{1}{2}CV^2 \) helps calculate this stored energy, where \( C \) represents capacitance and \( V \) is the voltage across the capacitor. Given values \( C = 6.00 \times 10^{-5} \text{ F} \) and \( V = 12.0 \text{ V} \), the energy computation becomes \( U = \frac{1}{2} \times 6.00 \times 10^{-5} \times 12.0^2 = 4.32 \times 10^{-3} \text{ Joules} \).

This numerical result provides insight into the initial electrical energy available in the capacitor and how it transforms as the circuit oscillates.

Current in Inductor

Current in an inductor, at any time \( t \), can tell us how energy is dynamically shifting within the LC circuit. The current \( I(t) \) is derived using the formula \( I(t) = -\omega Q_0 \sin(\omega t) \), where \( Q_0 \) is the initial charge on the capacitor.

For our case, with \( \omega = 333 \) and \( Q_0 = 7.20 \times 10^{-4} \text{ C} \), the calculation at \( t = 0.0230 \text{ s} \) becomes \( I(0.0230) = -333 \times 7.20 \times 10^{-4} \sin(333 \times 0.0230) \). This results in a current of approximately \( 0.222 \text{ A} \).

This value demonstrates the shifting nature of energy storage in the circuit, moving from being primarily in the capacitor to circulating through the inductor.

Charge on Capacitor Over Time

The charge on a capacitor over time in an LC circuit oscillates as energy moves between the capacitor and the inductor. The function \( Q(t) = Q_0 \cos(\omega t) \) models this behavior, describing how the initial charge \( Q_0 \) evolves.

At \( t = 0.0230 \text{ s} \), with \( Q_0 = 7.20 \times 10^{-4} \text{ C} \) and \( \omega = 333 \), the calculation is \( Q(0.0230) = 7.20 \times 10^{-4} \cos(333 \times 0.0230) \). This results in a charge of approximately \( -4.54 \times 10^{-4} \text{ C} \).

The negative sign indicates the charge's phase in its cycle, essentially capturing the oscillatory nature of LC circuits and the back-and-forth energy transfer between components.