A very long conducting tube (hollow cylinder) has inner radius $A$ and outer radius $b$ . It carries charge per unit length $+α$ , where $α$ is a positive constant with units of C/m. A line of charge lies along the axis of the tube. The line of charge has charge per unit length. Calculate the electric field in terms of $α$ and the distance $r$ from the axis of the tube for (i) $r < a$ ; (ii) $a < r < b$ ; (iii) $r > b$ . Show your results in a graph of $E$ as a function of $R$ .

Verified step by step guidance

Step 1: Understand the problem setup. We have a hollow conducting cylinder with inner radius A and outer radius B, carrying a charge per unit length +α. Additionally, there is a line of charge along the axis of the tube with the same charge per unit length +α. We need to find the electric field at different regions: inside the inner radius, between the inner and outer radius, and outside the outer radius.

Step 2: Apply Gauss's Law to find the electric field for r < A. Since the line of charge is along the axis, consider a cylindrical Gaussian surface of radius r and length L inside the inner radius. The enclosed charge is the line charge, which is αL. Using Gauss's Law, $ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} $, where $ Q_{enc} = \alpha L $. The electric field $ E $ is radial and constant over the surface, so $ E(2\pi rL) = \frac{\alpha L}{\varepsilon_0} $. Solve for $ E $ to get $ E = \frac{\alpha}{2\pi \varepsilon_0 r} $.

Step 3: Determine the electric field for A < r < B. In this region, the Gaussian surface encloses both the line charge and the charge on the inner surface of the cylinder. The inner surface of the cylinder has a charge per unit length of -α to ensure the cylinder is neutral overall. Thus, the enclosed charge is zero, leading to $ E = 0 $ in this region.

Step 4: Calculate the electric field for r > B. Here, the Gaussian surface encloses the line charge and the entire cylinder, which has a net charge of +α per unit length. The enclosed charge is $ 2\alpha L $. Using Gauss's Law, $ E(2\pi rL) = \frac{2\alpha L}{\varepsilon_0} $. Solve for $ E $ to get $ E = \frac{2\alpha}{2\pi \varepsilon_0 r} = \frac{\alpha}{\pi \varepsilon_0 r} $.

Step 5: Graph the electric field E as a function of r. For r < A, the electric field decreases as $ \frac{1}{r} $. For A < r < B, the electric field is zero. For r > B, the electric field decreases as $ \frac{1}{r} $ but with a different coefficient. This graph will show discontinuities at r = A and r = B, reflecting the changes in enclosed charge.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. It is essential for calculating electric fields in symmetric charge distributions, such as cylindrical geometries. The law states that the total electric flux is equal to the enclosed charge divided by the permittivity of free space, allowing us to determine the electric field at various distances from the axis.

Electric Field in Cylindrical Coordinates

In cylindrical coordinates, the electric field due to a line of charge or a charged cylinder is radially symmetric. The field depends on the radial distance from the axis and is calculated using Gauss's Law. For a line of charge, the field decreases with distance, while for a charged cylinder, it varies based on the region considered (inside, between, or outside the cylinder).

Superposition Principle

The superposition principle states that the total electric field due to multiple charges is the vector sum of the fields due to each charge individually. In this problem, the field at any point is the sum of the field due to the line of charge and the field due to the charged tube. This principle is crucial for determining the net electric field in regions where multiple charge distributions are present.

Recommended video:

Guided course

03:32

Superposition of Sinusoidal Wave Functions

Related Practice

Textbook Question

An infinitely long cylindrical conductor has radius $r$ and uniform surface charge density $σ$ . In terms of $σ$ , what is the magnitude of the electric field produced by the charged cylinder at a distance $r > R$ from its axis? Then, express the result in terms of $λ$ and show that the electric field outside the cylinder is the same as if all the charge were on the axis.

Textbook Question

A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area $$\sigma$=5.00$\times$10^{-6}$ C/m². A small sphere of mass $m=8.00$\times$10^{-6}$ kg and charge $q$ is placed $3.00$ cm above the sheet of charge and then released from rest. What is $q$ if the sphere is released $1.50$ cm above the sheet?

Textbook Question

A very long conducting tube (hollow cylinder) has inner radius $A$ and outer radius $b$ . It carries charge per unit length $+ α$ , where $α$ is a positive constant with units of C/m. A line of charge lies along the axis of the tube. The line of charge has charge per unit length. What is the charge per unit length on (i) the inner surface of the tube and (ii) the outer surface of the tube?

views

Textbook Question

An infinitely long cylindrical conductor has radius $r$ and uniform surface charge density $σ$ . In terms of $σ$ and $R$ , what is the charge per unit length $λ$ for the cylinder?

views