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Chapter 1: Problem 83

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鈥攖he 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鈥攊n 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

(a) A point charge \(q\) is located at an arbitrary position inside a sphere (just an imaginary sphere in space) with radius \(R\). Show that the average electric field over the surface of the sphere is zero. Hint: Use an argument involving Newton's third law, along with what you know about spherical shells. (b) If the point charge \(q\) is instead located outside the sphere, a distance \(r\) from the center, show that the average electric field over the surface of the sphere has magnitude \(q / 4 \pi \epsilon_{0} r\). (c) Return to the case where the point charge \(q\) is located inside the sphere of radius \(R\). Let the distance from the center be \(r\). Use the above results to show that the average electric field over the entire volume of the sphere of radius \(R\) has magnitude qr \(/ 4 \pi \epsilon_{0} R^{3}\) and points toward the center (if \(q\) is positive).

Uniform field strength * We know from the example in Section \(1.11\) that the electric field inside a solid sphere with uniform charge density is proportional to \(r\). Assume instead that the charge density is not uniform, but depends only on \(r\). What should this dependence be so that the magnitude of the field at points inside the sphere is independent of \(r\) (except right at the center, where it isn't well defined)? What, should the dependence be in the analogous case where we have a cylinder instead of a sphere?

Field from a semicircle * A thin plastic rod bent into a semicircle of radius \(R\) has a charge \(Q\) distributed uniformly over its length. Find the electric field at the center of the semicircle.

Oscillating in a ring A ring with radius \(R\) has uniform positive charge density \(\lambda\). A particle with positive charge \(q\) and mass \(m\) is initially located at the center of the ring and is then given a tiny kick. If it is constrained to move in the plane of the ring, show that it undergoes simple harmonic motion (for small oscillations), and find the frequency. Hint: Find the potential energy of the particle when it is at a (small) radius, \(r\), by integrating over the ring, and then take the negative derivative to find the force. You will need to use the law of cosines and also the Taylor series \(1 / \sqrt{1+\epsilon} \approx 1-\epsilon / 2+3 \epsilon^{2} / 8\)

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.
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