/*! 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 39 The Coaxial Cable. A long coaxia... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 39

The Coaxial Cable. A long coaxial cable consists of an inner cylindrical conductor with radius \(a\) and an outer coaxial cylinder with inner radius \(b\) and outer radius \(c .\) The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length \lambda. Calculate the electric field (a) at any point between the cylinders a distance \(r\) from the axis and (b) at any point outside the outer cylinder. (c) Graph the magnitude of the electric field as a function of the distance \(r\) from the axis of the cable, from \(r=0\) to \(r=2 c\) . (d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.

Short Answer

Expert verified
(a) \(E = \frac{\lambda}{2\pi\varepsilon_0 r}\) for \(a < r < b\), (b) \(E = 0\) for \(r > c\), (c) graph shows \(E\) decreases as \(1/r\) between \(a < r < b\), and then \(E = 0\), (d) \(-\lambda\) on inner surface, \(+\lambda\) on outer surface.

Step by step solution

01

Understanding the System

Consider a coaxial cable setup with a cylindrical inner conductor of radius \(a\) carrying a positive charge per unit length \(\lambda\), and a cylindrical outer conductor with an inner radius \(b\) and outer radius \(c\). The outer cylinder is neutral overall and rests on insulating supports.
02

Applying Gauss鈥檚 Law for Region a < r < b

To find the electric field between the cylinders, we apply Gauss鈥檚 Law. Consider a cylindrical Gaussian surface of radius \(r\) (where \(a < r < b\)) and length \(L\), coaxial with the conductors. The charge enclosed by this surface is \(\lambda L\). According to Gauss's Law, the electric field \(E\) is given by: \[ E \, (2 \pi r L) = \lambda L / \varepsilon_0 \]Solving for \(E\), we get:\[ E = \frac{\lambda}{2\pi\varepsilon_0 r} \]
03

Analysis for Region r > c

Now consider the electric field at a point outside the outer cylinder (\(r > c\)). Here, the Gaussian surface will not include any charge since the outer layer is neutral as a whole. Thus, the electric field \(E\) at any point where \(r > c\) is \(0\).
04

Graphing the Electric Field

To graph the electric field as a function of distance \(r\) from \(r=0\) to \(r=2c\):- For \(0 < r < a\), inside the inner cylinder, \(E = 0\) since the inner conductor's charge is uniformly distributed on its surface.- From \(a < r < b\), \(E = \frac{\lambda}{2\pi\varepsilon_0 r}\).- For \(b < r < c\), the field remains the same until the end of the inner conductor since the outer conductor adds no additional charge.- Beyond \(c\) (\(r > c\)), \(E = 0\) as stated before.
05

Charge Distribution on the Outer Cylinder

For the outer cylinder to be neutral overall:- The inner surface of the outer cylinder will acquire a charge \(-\lambda\) per unit length to counterbalance the field from the inner conductor.- The outer surface of the outer cylinder will seem neutral since \(-\lambda\) from the inner conductor is balanced by \(+\lambda\) from the outer surface, which is dispersed such that it does not affect points outside the outer cylinder.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Electric Field in Coaxial Cables
In a coaxial cable, the electric field plays a crucial role in understanding how the charges are distributed between the conductors. The electric field is essentially the region around the charged objects where other charges would feel a force. Here, considering a coaxial cable with a charged inner conductor, we utilize Gauss's Law to calculate the electric field at various regions.
  • Between the cylinders (radius between inner and outer conductor), the electric field is non-zero. Using the cylindrical symmetry, the field can be calculated as: \[ E = \frac{\lambda}{2\pi\varepsilon_0 r} \]
  • Outside the outer cylinder, there is no net charge affecting the Gaussian surface; hence, the electric field is zero.
The electric field gives insight into where charges might move, which is critical for understanding the behavior of components like coaxial cables in practical applications.
Understanding Coaxial Cables
Coaxial cables are a central piece of technology when it comes to communication systems. They consist of two cylindrical conductors; an inner conductor that carries the signal and an outer conductor that acts as a shielding layer. The space between these layers is usually filled with insulating material.
  • The inner conductor has radius \(a\) and is charged with a uniform charge density \(\lambda\).
  • The outer conductor spans from inner radius \(b\) to outer radius \(c\), and it is typically neutrally charged.
  • Coaxial cables provide good electromagnetic interference (EMI) shielding due to their design, preventing external fields from interfering with the internal carried signals.
Understanding their architecture assists in assessing how electrical fields form and spread within these systems.
Charge Distribution in Coaxial Structures
Charge distribution within coaxial cables determines many characteristics of the cable performance. In the coaxial setup, charges on the inner conductor induce equal and opposite charges on the inner surface of the outer conductor. This charge behavior is crucial for the cable's functionality.
  • The inner conductor has a positive charge density \(\lambda\) which is balanced by an induced charge of \(-\lambda\) on the inner surface of the outer conductor.
  • The outermost layer of the outer cylinder remains neutral, having a balancing positive charge of \(\lambda\), ensuring no electric field extends beyond the cable.
  • This distribution is vital to ensure that the cable exerts no external electric fields, making it a highly efficient setup for signal integrity.
Charge management is integral to ensuring that the coaxial cable operates without interference from external charges or fields.
Role of Cylindrical Conductors
Cylindrical conductors are foundational in the setup of coaxial cables due to their unique properties. Their geometry allows for uniform distribution of electric fields and efficient containment of electromagnetic fields within the cable.
  • The inner cylindrical conductor carries the main signal or current, having a specific radius \(a\) and a positive charge density \(\lambda\).
  • The outer cylindrical shell further aids in shielding, having an inner radius \(b\) and outer radius \(c\).
  • The design supports maintaining a conductive path for return currents, which helps in diminishing signal loss.
Cylindrical conductors are chosen in coaxial cables due to these advantageous characteristics, providing reliability and efficiency in signal transmission and shielding.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A long line carrying a uniform linear charge density \(+50.0 \mu C / m\) runs parallel to and 10.0 cm from the surface of a large, flat plastic sheet that has a uniform surface charge density of \(-100 \mu \mathrm{C} / \mathrm{m}^{2}\) on one side. Find the location of all points where an \(\alpha\) particle would feel no force due to this arrangement of charged objects.

A negative charge \(-Q\) is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. (a) Is there any excess charge induced on the inner surface of the piece of metal? If so, find its sign and magnitude. (b) Is there any excess charge on the outside of the piece of metal? Why or why not? (c) Is there an electric field in the cavity? Explain. (d) Is there an electric field within the metal? Why or why not? Is there an electric field outside the piece of metal? Explain why or why not. (e) Would someone outside the solid measure an electric field due to the charge \(-Q ?\) Is it reasonable to say that the grounded conductor has shielded the region from the effects of the charge \(-Q ?\) In principle, could the same thing be done for gravity? Why or why not?

Thomson's Model of the Atom. In the early years of the 20 th century, a leading model of the structure of the atom was that of the English physicist J. J. Thomson (the discoverer of the electron). In Thomson's model, an atom consisted of a sphere of positively charged material in which were embedded negatively charged electrons, like chocolate chips in a ball of cookie dough. Consider such an atom consisting of one electron with mass \(m\) and charge \(-e,\) which may be regarded as a point charge, and a uniformly charged sphere of charge \(+e\) and radius \(R\) (a) Explain why the equilibrium position of the electron is at the center of the nucleus. (b) In Thomson's model, it was assumed that the positive material provided little or no resistance to the motion of the electron. If the electron is displaced from equilibrium by a distance less than \(R,\) show that the resulting motion of the electron will be simple harmonic, and calculate the frequency of oscillation. Hint: Review the definition of simple harmonic motion in Section \(14.2 .\) If it can be shown that the net force on the electron is of this form, then it follows that the motion is simple harmonic. Conversely, if the net force on the electron does not follow this form, the motion is not simple harmonic.) (c) By Thomson's time, it was known that excited atoms emit light waves of only certain frequencies. In his model, the frequency of emitted light is the same as the oscillation frequency of the electron or electrons in the atom. What would the radius of a Thomson-model atom have to be for it to produce red light of frequency \(4.57 \times 10^{14}\) Hz? Compare your answer to the radii of real atoms, which are of the order of \(10^{-10} \mathrm{m}\) (see Appendix For data about the electron). (d) If the electron were displaced from equilibrium by a distance greater than \(R,\) would the electron oscillate? Would its motion be simple harmonic? Explain your reasoning. (Historical note: In \(1910,\) the atomic nucleus was discovered, proving the Thomson model to be incorrect. An atom's positive charge is not spread over its volume as Thomson supposed, but is concentrated in the tiny nucleus of radius \(10^{-14}\) to \(10^{-15} \mathrm{m} . )\)

An electron is released from rest at a distance of 0.300 \(\mathrm{m}\) from a large insulating sheet of charge that has uniform surface charge density \(+2.90 \times 10^{-12} \mathrm{C} / \mathrm{m}^{2}\) . (a) How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 0.050 \(\mathrm{m}\) from the sheet? (b) What is the speed of the electron when it is 0.050 \(\mathrm{m}\) from the sheet?

The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately \(7.4 \times 10^{-15} \mathrm{m}\) (a) What is the electric field this nucleus produces just outside its surface? (b) What magnitude of electric field does it produce at the distance of the electrons, which is about \(1.0 \times 10^{-10} \mathrm{m} ?\) (c) The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Waves Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Modern Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Solid State Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.