/*! 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 83 Potential energy of a cylinder *... [FREE SOLUTION] | 91Ó°ÊÓ

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Potential energy of a cylinder * * Problem \(1.24\) gives one way of calculating the energy per unit length stored in a solid cylinder with radius \(a\) and uniform volume charge density \(\rho .\) Calculate the energy here by using Eq. (1.53) to find the total energy per unit length stored in the electric field. Don't forget to include the field inside the cylinder. You will find that the energy is infinite, so instead calculate the energy relative to the configuration where all the charge is initially distributed uniformly over a hollow cylinder with large radius \(R\). (The field outside radius \(R\) is the same in both configurations, so it can be ignored when calculating the relative energy.) In terms of the total charge \(\lambda\) per unit length in the final cylinder, show that the energy per unit length can be written as \(\left(\lambda^{2} / 4 \pi \epsilon_{0}\right)(1 / 4+\ln (R / a))\).

Short Answer

Expert verified
The energy per unit length of the solid cylinder with a uniform volume charge density, when calculated relative to an initial state where the charge is uniformly distributed on a hollow cylinder, can be expressed as \(\left(\lambda^{2} / 4 \pi \epsilon_{0}\right)(1 / 4+\ln (R / a))\).

Step by step solution

01

Set up the electric field and energy equations

Start by noting that the electric field of a line charge is given by \(E = \lambda / (2 \pi \epsilon_{0} r)\), where \(\lambda\) is the charge per unit length, \(\epsilon_{0}\) is vacuum permittivity, and \(r\) is the distance from the line charge. The total energy density in the electric field can be written as \(u = \epsilon_{0} E^2 / 2\). Substitute the electric field into the energy density equation to find the energy per unit length.
02

Integrate over the volume of the cylinder

Integrate the energy over the volume of the cylinder. The integration should be done over \(r\) from \(a\) to \(R\) and over the total angle of \(2 \pi\). The volume element is \(rd \phi dz dr\), where \(d \phi\) and \(dz\) are the differential elements for the angle and length, respectively.
03

Add the energies

Add the energy of the electric field inside and outside the solid cylinder together. When calculating the energy of the outer cylinder, subtract the energy of the initial, hollow cylinder from the final configuration.
04

Simplify the result

After adding these energy densities together and performing the integration, simplify the result. By doing the necessary calculations, it can be shown that the energy per unit length of the cylinder can be written as \(\left(\lambda^{2} / 4 \pi \epsilon_{0}\right)(1 / 4+\ln (R / a))\).

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Key Concepts

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

Electric Field of a Line Charge
Understanding the electric field created by a line charge is fundamental to many problems in electrostatics. Imagine a wire or a uniformly charged cylinder that's very long—the distribution of charge along the length is uniform, and for calculations, it is often approximated as infinite. The key to this concept lies in the symmetry of the arrangement, which allows us to conclude that the electric field only points radially outward and its magnitude depends solely on the radial distance from the line charge, not on the angle or the length.

The electric field created by a line charge is described by the equation: \[ E = \frac{\lambda}{2 \pi \epsilon_{0} r} \],where \( E \) is the electric field, \( \lambda \) is the charge per unit length, \( \epsilon_{0} \) is the permittivity of free space, and \( r \) is the radial distance from the line charge. If the line charge is enclosed, such as within a cylinder, this field applies within the material up to the surface.
Energy Density in an Electric Field
The concept of energy density is pivotal when describing how much energy is stored in an electric field per unit volume. Energy density, denoted by \( u \), is given by the expression:\[u = \frac{\epsilon_{0} E^2}{2}\].This equation cleverly illustrates that the energy density is proportional to the square of the electric field's magnitude. The factor of \( \frac{1}{2} \) comes from the integral of electric field work done in assembling the charge distribution gradually from zero to its final configuration.

When we deal with a uniform volume charge distribution inside a cylinder, we are looking at integrating this energy density over the entire volume to find the total energy stored in the electric field. This integral accounts for how the energy density changes with the distance from the line charge, giving us insightful information on how the total energy is distributed along the cylinder's dimensions.
Integration in Electric Field Calculations
The process of integration is a mathematical tool that plays a crucial role in electric field calculations. Integration allows us to sum up infinitesimally small quantities to obtain a total value—in our case, the total energy stored in an electric field throughout a volume. For a cylinder, we start by integrating the energy density across a cross-sectional slice, then extend this to cover the cylinder's length.

The integral for the energy in the electric field of a charged cylinder might look like this:\[ U = \int_{Volume} u(r) \, dV = \int_{a}^{R} \int_{0}^{2\pi} \int_{0}^{L} \frac{\epsilon_{0} (\lambda / (2 \pi \epsilon_{0} r))^2}{2} \, r \, d\phi \, dz \, dr \].Here, \(U\) is the total energy, and we integrate over the cylindrical coordinates (radius \(r\), angle \(\phi\), and length \(z\)). The integral boundaries correspond to the dimensions of the cylinder from its inner radius \(a\) to its outer radius \(R\), and over its entire angle and length. The result of this integration gives us the energy stored in the electric field per unit length when considering the charge distribution's transition from a hollow cylinder to a uniformly charged solid one.

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Most popular questions from this chapter

Field at the end of a cylinder (a) Consider a half-infinite hollow cylindrical shell (that is, one that extends to infinity in one direction) with radius \(R\) and uniform surface charge density \(\sigma .\) What is the electric field at the midpoint of the end face? (b) Use your result to determine the field at the midpoint of a half-infinite solid cylinder with radius \(R\) and uniform volume charge density \(\rho\), which can be considered to be built up from many cylindrical shells.

Field from a spherical shell, right and wrong ** The electric field outside and an infinitesimal distance away from a uniformly charged spherical shell, with radius \(R\) and surface charge density \(\sigma\), is given by Eq. (1.42) as \(\sigma / \epsilon_{0}\). Derive this in the following way. (a) Slice the shell into rings (symmetrically located with respect to the point in question), and then integrate the field contributions from all the rings. You should obtain the incorrect result of \(\sigma / 2 \epsilon_{0}\) (b) Why isn't the result correct? Explain how to modify it to obtain the correct result of \(\sigma / \epsilon_{0} .\) Hint: You could very well have performed the above integral in an effort to obtain the electric field an infinitesimal distance inside the shell, where we know the field is zero. Does the above integration provide a good description of what's going on for points on the shell that are very close to the point in question?

Potential energy of a cylinder A cylindrical volume of radius \(a\) is filled with charge of uniform density \(\rho\). We want to know the potential energy per unit length of this cylinder of charge, that is, the work done per unit length in assembling it. Calculate this by building up the cylinder layer by layer, making use of the fact that the field outside a cylindrical distribution of charge is the same as if all the charge were located on the axis. You will find that the energy per unit length is infinite if the charges are brought in from infinity, so instead assume that they are initially distributed uniformly over a hollow cylinder with large radius \(R\). Write your answer in terms of the charge per unit length of the cylinder, which is \(\lambda=\rho \pi a^{2}\). (See Exercise \(1.83\) for a different method of solving this problem.)

Thundercloud You observe that the passage of a particular thundercloud overhead causes the vertical electric field strength in the atmosphere, measured at the ground, to rise to \(3000 \mathrm{~N} / \mathrm{C}\) (or \(\mathrm{V} / \mathrm{m})\). (a) How much charge does the thundercloud contain, in coulombs per square meter of horizontal area? Assume that the width of the cloud is large compared with the height above the ground. (b) Suppose there is enough water in the thundercloud in the form of \(1 \mathrm{~mm}\) diameter drops to make \(0.25 \mathrm{~cm}\) of rainfall, and that it is those drops that carry the charge. How large is the electric field strength at the surface of one of the drops?

Zero field \(?\) Four charges, \(q,-q, q\), and \(-q\), are located at equally spaced intervals on the \(x\) axis. Their \(x\) values are \(-3 a,-a, a\), and \(3 a\), respectively. Does there exist a point on the \(y\) axis for which the electric field is zero? If so, find the \(y\) value.

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