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A very long conducting tube (hollow cylinder) has inner radius \(a\) and outer radius \(b .\) It carries charge per unit length \(+\alpha,\) where \(\alpha\) is a positive constant with units of \(\mathrm{C} / \mathrm{m} .\) A line of charge lies along the axis of the tube. The line of charge has charge per unit length \(+\alpha\). (a) Calculate the electric field in terms of \(\alpha\) and the distance \(r\) from the axis of the tube for (i) \(rb .\) Show your results in a graph of \(E\) as a function of \(r\). (b) What is the charge per unit length on (i) the inner surface of the tube and (ii) the outer surface of the tube?

Short Answer

Expert verified
The electric fields for different regions are: for \(rb\) it is \(E=\alpha b/\pi r \epsilon_0\). The charge per unit length on the inner surface of the tube is \(\sigma_1=-\alpha a/ 2b\) and on the outer surface is \(\sigma_2=0\).

Step by step solution

01

Apply Gauss's Law for r

The electric field inside the line of charge, where \(r
02

Apply Gauss's Law for a

For the area inside the conducting tube but outside the line of charge, \(a
03

Apply Gauss's Law for r>b

In the area outside the conducting tube, \(r>b\), Gauss's law encompasses the charge from the line of charge and the conducting tube. The enclosed charge is \(q=2\alpha b\). Solve Gauss's law equation to obtain: \(E=2\alpha b/2\pi r \epsilon_0 =\alpha b/ \pi r \epsilon_0\)
04

Calculate the Charge per Unit Length

For the inner tube surface (i), the field right outside is \(E_1=\alpha a/2\pi \epsilon_0\), and for the outer tube surface (ii), the field right inside is \(E_2=0\). The charge per unit length for either surface, according to Gauss's law, is given by the formula \(=E\epsilon_0/2\pi r\). Thus, the charge per unit length on the inner surface is \(\sigma_1=-E_1\epsilon_0/2\pi a\) and on the outer surface is \(\sigma_2=-E_2\epsilon_0/2\pi b = 0\)
05

Graph the Electric Field Function of r

Plot the function of electric field by considering x-axis as distance \(r\) and y-axis as electric field \(E\). The representation will be a piecewise function because of the different zones described (rb). The graph will show a linear function until rb.

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Key Concepts

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

Electric Field
The concept of the electric field is central to understanding electrostatic interactions. Electric fields are generated by electric charges and can be felt by other charges even in the absence of direct contact. They describe the force per unit charge that a test charge would experience in a given point in space. This abstract notion lets us calculate the forces in systems where charges are distributed over different shapes and sizes, like in the exercise with a charged conducting cylinder.

Using Gauss's Law, which relates the electric field to the charge distribution within a certain volume, we can calculate the electric field in different regions around the cylinder. Inside the central cavity where no charge is present, the field is zero. Immediately outside the central charge-carrying wire, the electric field increases linearly with distance due to symmetry and charge distribution. Within the cylinder walls, the field remains constant—interesting because this suggests that the cylinder shields its cavity from external fields. Outside the cylinder, the field decreases with distance, inversely proportional to the radius, in typical point-charge fashion. It's a beautiful illustration of how field lines begin on positive charges and end on negative charges, or at infinity if no negative charge is available to terminate the field lines.
Conducting Cylinder
A conducting cylinder, like the one in our exercise, is a useful construct in physics because it embodies principles of symmetry and charge distribution. Conductors allow charges to move freely until they reach an equilibrium where the electric field inside becomes zero. At equilibrium, any excess charge resides on the surface of a conductor, not in its interior.

The cylinder in the problem is particularly fascinating because it has an inherent hollow structure, which creates distinct regions where the electric field behaves differently. It is essential to consider the inner and outer surfaces while solving for electrical effects. When additional charges are present inside, such as the line of charge along the cylinder's axis, their influence must be accounted for. This system showcases the effects of electrostatic shielding, where the electric field within the conducting material and within its inner cavity is zero, protecting the inside from outside electrostatic effects.
Charge Density
Charge density typically refers to the amount of charge per unit volume, area, or length. In the context of our exercise, we're concerned with 'linear' charge density because the charges are distributed along a line and over the surfaces of a cylinder. The charge density is uniform (constant) for both the line of charge and the conducting cylinder.

As we analyze the charge distribution, it's crucial to understand how the cylinder's charged surfaces influence the electric field. For instance, Gauss's Law helps us deduce that for a charged conducting surface, the electric field is directly related to the surface charge density. In our problem, the inner and outer charge densities are essential for calculating the electric field in the neighboring regions. Surprisingly, despite the presence of charge, the charge density within the conductor's material remains zero, underscoring the fact that free charge only resides on the conductor's surfaces.
Electrostatics
Electrostatics is the study of stationary electric charges or the fields they generate. Many principles of electrostatics are exemplified in the exercise problem, such as the behavior of charges in a conductor and the resulting electric fields. It is a field governed by Coulomb's Law, which describes the force between two point charges, and by Gauss's Law, which relates electric fields to the enclosed charges.

Gauss's Law, in particular, offers a powerful method to solve problems in electrostatics with high symmetry. It simplifies complex integrals into manageable calculations, ideal for systems like our conducting cylinder. It emphasizes the connection between a charge and its electric field in a visually and mathematically intuitive way, highlighting the influence of charge distribution on the observed electric phenomena. Electrostatics is foundational for understanding not only theoretical scenarios like our exercise but also real-world applications like electrical insulators and capacitors.

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Most popular questions from this chapter

A point charge \(q_{1}=4.00 \mathrm{nC}\) is located on the \(x\) -axis at \(x=2.00 \mathrm{~m},\) and a second point charge \(q_{2}=-6.00 \mathrm{nC}\) is on the \(y\) -axis at \(y=1.00 \mathrm{~m}\). What is the total electric flux due to the two point charges through a spherical surface centered at the origin and with ra\(\operatorname{dius}(\mathrm{a}) 0.500 \mathrm{~m},(\mathrm{~b}) 1.50 \mathrm{~m},(\mathrm{c}) 2.50 \mathrm{~m} ?\)

Electric Fields in an Atom. 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?

A long line carrying a uniform linear charge density \(+50.0 \mu \mathrm{C} / \mathrm{m}\) runs parallel to and \(10.0 \mathrm{~cm}\) from the surface of \(\mathrm{a}\) 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 solid insulating sphere has radius \(R\) and carries positive charge distributed throughout its volume. The charge distribution has spherical symmetry but varies with radial distance \(r\) from the center of the sphere. The volume charge density is \(\rho(r)=\rho_{0}(1-r / R)\) where \(\rho_{0}\) is a constant with units of \(\mathrm{C} / \mathrm{m}^{3}\). (a) Derive an expression for the electric field as a function of \(r\) for \(rR .\) (c) At what value of \(r\), in terms of \(R,\) does the electric field have its maximum value?

A very long insulating cylinder with radius \(R_{\text {cylinder }}\) has nonuniform positive charge density \(\rho=\left(1-r / R_{\text {cylinder }}\right) \rho_{0}\) where \(\rho_{0}\) is constant and \(r\) is measured radially from the axis of the cylinder. A particle with charge \(-Q\) and mass \(M\) orbits the cylinder at a constant distance \(R_{\text {orbit }}>R_{\text {cylinder }}\) (a) What is the linear charge density \(\lambda\) of the tube, in terms of \(R_{\text {cylinder }}\) and \(\rho_{0} ?\) (b) Determine the period of the motion in terms of \(R_{\text {orbit }}\) (Hint: Use Gauss's law to determine the electric field, and therefore the electric force felt by the particle, that acts centripetally.)

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