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Chapter 24: Problem 29

Two Concentric Cylinders Two long, charged, concentric cylinders have radii of \(3.0\) and \(6.0 \mathrm{~cm}\). Assume the outer cylinder is hollow. The charge per unit length is \(5.0 \times 10^{-6} \mathrm{C} / \mathrm{m}\) on the inner cylinder and \(-7.0 \times 10^{-6} \mathrm{C} / \mathrm{m}\) on the outer cylinder. Find the electric field at (a) \(r=4.0 \mathrm{~cm}\) and (b) \(r=8.0 \mathrm{~cm}\), where \(r\) is the radial distance from the common central axis.

Short Answer

Expert verified
(a) The electric field at 4.0 cm is approximately 224,000 N/C away from the inner cylinder. (b) The electric field at 8.0 cm is approximately -45,000 N/C, pointing toward the system.

Step by step solution

01

Identify the Location of the Point

We need to find the electric field at two points: (a) at a distance of 4.0 cm and (b) at a distance of 8.0 cm from the central axis. Note that 4.0 cm lies between the inner and outer cylinders, while 8.0 cm is outside both cylinders.
02

Use Gauss's Law for Cylindrical Symmetry

Gauss's Law for a cylindrical symmetry states that the electric field E at a distance r from a line of charge with charge per unit length is given by:\[ E = \frac{\boldsymbol{\rho}_l}{2\pi \varepsilon_0 r} \]where \( \boldsymbol{\rho}_l \) is the charge per unit length, \( \boldsymbol{\rho}_0 \) is the permittivity of free space.
03

Calculate the Electric Field at r = 4.0 cm

At r = 4.0 cm, which is between the cylinders, only the inner cylinder's charge contributes to the electric field. Thus, we have:\[ E = \frac{5.0 \times 10^{-6} \text{ C/m}}{2\pi (8.85 \times 10^{-12} \text{ F/m}) \times 0.04 \text{ m}} \]\[ E = \frac{5.0 \times 10^{-6}}{2\pi (8.85 \times 10^{-12}) \times 0.04} \approx 224,000 \text{ N/C} \]
04

Calculate the Electric Field at r = 8.0 cm

At r = 8.0 cm, which is outside both cylinders, we consider the net charge per unit length. The charges on the inner and outer cylinders are +5.0 x 10^-6 C/m and -7.0 x 10^-6 C/m, respectively. The net charge per unit length is:\[ \boldsymbol{\rho}_l = 5.0 \times 10^{-6} - 7.0 \times 10^{-6} = -2.0 \times 10^{-6} \text{ C/m} \]Using Gauss's law:\[ E = \frac{-2.0 \times 10^{-6} \text{ C/m}}{2\pi (8.85 \times 10^{-12} \text{ F/m}) \times 0.08 \text{ m}} \]\[ E = \frac{-2.0 \times 10^{-6}}{2 \pi (8.85 \times 10^{-12}) \times 0.08} \approx -45,000 \text{ N/C} \]
05

Analyze the Direction of the Electric Field

The direction of the electric field at each point should be determined by the sign of the net charge producing the field. For r=4.0 cm, it points away from the positively charged inner cylinder. For r=8.0 cm, it points toward the overall negative charge of the system.

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Key Concepts

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

Gauss's Law
Gauss's Law is an essential concept in electromagnetism that tells us how electric fields relate to the distribution of electric charges. It states that the total electric flux through a closed surface is directly proportional to the charge enclosed within that surface. Mathematically, it is represented as \( abla \bullet \boldsymbol{E} = \frac{\rho}{\boldsymbol{\rho}_0} \), where \( \boldsymbol{\rho}_0 \) is the permittivity of free space and \( \rho \) is the charge density.
For our problem, we're dealing with cylindrical symmetry, where the surface we are considering is a cylindrical Gaussian surface. This means the electric field is uniform over the cylinder's surface. The use of Gauss's Law makes it easier to calculate the electric field in complex geometrical setups, such as two concentric cylinders.
Understanding Gauss's Law helps us see how the field at a point is influenced by the charges inside our chosen surface, simplifying the calculation of the electric field in the problem.
Cylindrical Symmetry in Electromagnetism
Cylindrical symmetry is crucial in this problem because it simplifies how we calculate electric fields around cylindrical objects. When we say a configuration has cylindrical symmetry, we mean its properties are unchanged if we rotate it around a central axis. This symmetry implies that the electric field in such a configuration is radial and only depends on the distance from the axis of the cylinders.
To use Gauss's Law effectively in these scenarios, we choose a cylindrical Gaussian surface co-axial with the charged cylinders. Since the field is radial, the electric field's magnitude only depends on the radial distance r from the axis.
This symmetry helps in breaking down complex field interactions into simpler calculations, as the electric field is constant over the surface of our cylindrical Gaussian surface. The problem illustrates this by making the calculation of the electric field straightforward at different radial distances, considering only the enclosed charges.
Electric Field Direction
The direction of the electric field is just as important as its magnitude. The electric field (\boldsymbol{E}) at a point around a charged object points away from positive charges and toward negative charges.
In the given exercise, for the point at \( r = 4.0 \text{ cm} \), which lies between the inner and outer cylinders, only the charge of the inner cylinders influences the field. Because the inner cylinder has a positive charge, the electric field points radially outward.
For the point at \( r = 8.0 \text{ cm} \), outside both cylinders, we look at the net charge enclosed. With the combined charges of the inner and outer cylinders resulting in a net negative charge, the field direction here points inward, towards the negatively charged outer cylinder.
Clearly differentiating and understanding the direction of the field is vital in correctly applying Gauss's Law and solving similar problems.

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Most popular questions from this chapter

Uniform Volume Charge Density Figure 24-29a shows a spherical shell of charge with uniform volume charge density \(\rho .\) Plot \(E\) due to the shell for distances \(r\) from the center of the shell ranging from zero to \(30 \mathrm{~cm}\). Assume that \(\rho=1.0 \times 10^{-6} \mathrm{C} / \mathrm{m}^{3}, a=10 \mathrm{~cm}\), and \(b=20 \mathrm{~cm}\).

Solid Nonconducting Sphere A solid nonconducting sphere of radius \(R\) has a nonuniform charge distribution of volume charge density \(\rho=\rho_{s} r / R\), where \(\rho_{s}\) is a constant and \(r\) is the distance from the center of the sphere. Show (a) that the total charge on the sphere is \(Q=\pi \rho_{s} R^{3}\) and \((b)\) that $$ |\vec{E}|=k \frac{|Q|}{R^{4}} r^{2} $$ gives the magnitude of the electric field inside the sphere.

Geiger Counter Figure \(24-33\) shows a Geiger counter, a device used to detect ionizing radiation (radiation that causes ionization of atoms). The counter consists of a thin, positively charged central wire surrounded by a concentric, circular, conducting cylinder with an equal negative charge. Thus, a strong radial electric field is set up inside the cylinder. The cylinder contains a low-pressure inert gas. When a particle of radiation enters the device through the cylinder wall, it ionizes a few of the gas atoms The resulting free electrons (labelled e) are drawn to the positive wire. However, the electric field is so intense that, between collisions with other gas atoms, the free electrons gain energy sufficient to ionize these atoms also. More free electrons are thereby created, and the process is repeated until the electrons reach the wire. The resulting "avalanche" of electrons is collected by the wire generating a signal that is used to record the passage of the original particle of radiation. Suppose that the radius of the central wire is \(25 \mu \mathrm{m}\), the radius of the cylinder \(1.4 \mathrm{~cm}\), and the length of the tube \(16 \mathrm{~cm}\). If the electric field component \(E_{r}\) at the cylinder's inner wall is \(+2.9 \times\) \(10^{4} \mathrm{~N} / \mathrm{C}\), what is the total positive charge on the central wire?

Large Metal Plates Two large metal plates of area \(1.0 \mathrm{~m}^{2}\) face each other. They are \(5.0 \mathrm{~cm}\) apart and have equal but opposite charges on their inner surfaces. If the magnitude \(|\vec{E}|\) of the electric field between the plates is \(55 \mathrm{~N} / \mathrm{C}\), what is the amount of charge on each plate? Neglect edge effects.

Four Charges You have four point charges, \(2 q, q,-q\), and \(-2 q .\) If possible describe how you would place a closed surface that encloses at least the charge \(2 q\) (and perhaps other charges) and through which the net electric flux is (a) 0 (b) \(+3 q / \varepsilon_{0}\) and (c) \(-2 q / \varepsilon_{0}\) -
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