/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 A dielectric of permittivity \(3... [FREE SOLUTION] | 91Ó°ÊÓ

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A dielectric of permittivity \(3.5 \times 10^{-11} \mathrm{~F} / \mathrm{m}\) completely fills the volume between two capacitor plates. For \(t>0\) the electric flux through the dielectric is \(\left(8.0 \times 10^{3} \mathrm{~V} \cdot \mathrm{m} / \mathrm{s}^{3}\right) t^{3} .\) The dielectric is ideal and nonmagnetic; the conduction current in the dielectric is zero. At what time does the displacement current in the dielectric equal \(21 \mu \mathrm{A} ?\)

Short Answer

Expert verified
The displacement current in the dielectric equals 21 µA at t = 0.7 s.

Step by step solution

01

Understand the displacement current

Displacement current is a concept in electromagnetism where a varying electric field gives rise to a form of current despite the absence of free charges. It is represented by the equation \(I_d = \epsilon \frac{d\Phi_E}{dt}\), where \(I_d\) is the displacement current, \(\epsilon\) is the permittivity of the material, and \(\frac{d\Phi_E}{dt}\) is the rate of change of the electric flux.
02

Relate given values to the equation

From the problem, we know the value of displacement current \(I_d = 21 \mu A\), permittivity of the dielectric \(\epsilon = 3.5 \times 10^{-11} F/m\), and the electric flux as a function of time \(\Phi_E = (8.0 \times 10^{3} V \cdot m/s^{3}) t^{3}\). All these values can be substituted into the equation of the displacement current.
03

Substitute the values and solve for time

On substituting these values into the equation \(I_d = \epsilon \frac{d\Phi_E}{dt}\), we have \(21 \times 10^{-6} A = 3.5 \times 10^{-11} F/m \times \frac{d}{dt}(8.0 \times 10^{3} V \cdot m/s^{3} t^{3})\). Solving this equation, the derivative of the flux with respect to time is \(\frac{d\Phi_E}{dt} = 24,000 t^{2}\). Substituting back into the equation, we will have \(21 \times 10^{-6} A = 3.5 \times 10^{-11} F/m \times 24,000 t^{2}\). Solving for \(t\), we get \(t = 0.7s\)

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Key Concepts

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

Permittivity
Permittivity is a fundamental property of materials that quantifies their ability to permit electric field lines to pass through them. More formally, it is a measure of the electric susceptibility of a dielectric material. The permittivity of a material is represented by the symbol \( \epsilon \) and is expressed in the units of farads per meter (F/m). In the context of capacitors, the permittivity of the dielectric material between the plates determines the capacitance of the capacitor, the amount of electric charge it can store for a given electric potential.
In the given exercise, the dielectric has a permittivity of \( 3.5 \times 10^{-11} \rm{{F/m}} \) which indicates its capacity to store electric energy. The higher the permittivity, the more electric flux the material can support, leading to a higher capacitance. Students often struggle to visualize permittivity but can think of it as a measure of how 'dense' the electric field can be within a material.
Electric Flux
Electric flux is a key term in electromagnetism that represents the number of electric field lines passing through a given area. Think of it as a measure of the 'flow' of the electric field through a surface. It's often denoted by \( \Phi_E \) and influences many aspects of electromagnetic phenomena, such as the generation of electric fields around charged objects.
In our exercise, the electric flux varies with time according to the function \( \Phi_E = (8.0 \times 10^{3} \rm{V} \cdot \rm{m/s^{3}}) t^{3} \), with \( t \) being the time. The cubic relationship with time indicates that the electric flux – and thus the strength of the electric field – increases rapidly as time progresses. Visualizing electric flux can be challenging for students; one can imagine it as the number of field lines 'flowing' through a virtual surface, which changes over time in this particular scenario.
Electromagnetism
Electromagnetism is one of the four fundamental forces that govern the universe and it encompasses the interaction between electrically charged particles. Within this field of study, we explore concepts such as electric fields, magnetic fields, and their interdependence. Displacement current, introduced by James Clerk Maxwell, is a cornerstone concept in electromagnetism that allows us to understand how changing electric fields can create magnetic fields, even without the physical movement of charge that constitutes a traditional electric current.

Understanding Displacement Current

The exercise presents a practical application of displacement current, which occurs within the dielectric of a capacitor. Even in the absence of free charge movement, the displacement current is a manifestation of a time-varying electric field within the dielectric, critical for electromagnetic wave propagation. Clear comprehension of this concept helps bridge one's understanding between electricity and magnetism in electromagnetic waves and is essential for grasping advanced topics in physics and engineering.
Capacitor Dielectric
The capacitor dielectric is the insulating material placed between the plates of a capacitor. Its primary role is to increase the capacitor's ability to store electrical energy. This is accomplished by reducing the electric field within the capacitor, which allows the capacitor to hold more charge at a given voltage, thereby increasing its capacitance. The effectiveness of a dielectric material is determined by its permittivity, as discussed earlier.

Role in Electric Fields

A dielectric material in a capacitor polarizes when exposed to an electric field, decreasing the field's strength inside the material compared to a vacuum. This polarization effect increases the capacitor's charge storage capacity without requiring additional voltage. For students, understanding dielectrics can be facilitated by looking at how these materials influence the behavior of capacitors and the efficiency of energy storage. In real-world applications, dielectrics are crucial not only for capacitors but also for a wide range of electronic components, where they improve performance, stability, and energy efficiency.

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Most popular questions from this chapter

A small, closely wound coil has \(N\) turns, area \(A\), and resistance \(R\). The coil is initially in a uniform magnetic field that has magnitude \(B\) and a direction perpendicular to the plane of the loop. The coil is then rapidly pulled out of the field so that the flux through the coil is reduced to zero in time \(\Delta t\). (a) What are the magnitude of the average \(\operatorname{emf} \mathcal{E}_{\text {av }}\) and average current \(I_{\mathrm{av}}\) induced in the coil? (b) The total charge \(Q\) that flows through the coil is given by \(Q=I_{\mathrm{av}} \Delta t .\) Derive an expression for \(Q\) in terms of \(N, A, B,\) and \(R .\) Note that \(Q\) does not depend on \(\Delta t .\) (c) What is \(Q\) if \(N=150\) turns, \(A=4.50 \mathrm{~cm}^{2}, R=30.0 \Omega,\) and \(B=0.200 \mathrm{~T} ?\)

The battery for a certain cell phone is rated at \(3.70 \mathrm{~V}\). According to the manufacturer it can produce \(3.15 \times 10^{4} \mathrm{~J}\) of electrical energy, enough for \(5.25 \mathrm{~h}\) of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.

Compact Fluorescent Bulbs. Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but they last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a \(100 \mathrm{~W}\) incandescent bulb uses only \(23 \mathrm{~W}\) of power. The compact bulb lasts 10,000 hours, on the average, and costs \(\$ 11.00,\) whereas the incandescent bulb costs only \(\$ 0.75\), but lasts just 750 hours. The study assumed that electricity costs \(\$ 0.080\) per kilowatt-hour and that the bulbs are on for \(4.0 \mathrm{~h}\) per day. (a) What is the total cost (including the price of the bulbs) to run each bulb for 3.0 years? (b) How much do you save over 3.0 years if you use a compact fluorescent bulb instead of an incandescent bulb? (c) What is the resistance of a "100 W" fluorescent bulb? (Remember, it actually uses only \(23 \mathrm{~W}\) of power and operates across \(120 \mathrm{~V} .\) )

The armature of a small generator consists of a flat, square coil with 120 turns and sides with a length of \(1.60 \mathrm{~cm} .\) The coil rotates in a magnetic field of \(0.0750 \mathrm{~T}\). What is the angular speed of the coil if the maximum emf produced is \(24.0 \mathrm{mV} ?\)

A typical small flashlight contains two batteries, each having an emf of \(1.5 \mathrm{~V},\) connected in series with a bulb having resistance \(17 \Omega .\) (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for \(5.0 \mathrm{~h}\), what is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.

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