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Chapter 22: Problem 38

(II) A very long solid nonconducting cylinder of radius \(R_{1}\) is uniformly charged with a charge density \(\rho_{\mathrm{E}}\). It is surrounded by a concentric cylindrical tube of inner radius \(R_{2}\) and outer radius \(R_{3}\) as shown in Fig. \(22-36,\) and it too carries a uniform charge density \(\rho_{\mathrm{E}}\). Determine the electric field as a function of the distance \(R\) from the center of the cylinders for \((a) 0R_{3}\). \((e)\) If \(\rho_{\mathrm{E}}=15 \mu \mathrm{C} / \mathrm{m}^{3}\) and \(R_{1}=\frac{1}{2} R_{2}=\frac{1}{3} R_{3}=5.0 \mathrm{~cm}\) plot \(E\) as a function of \(R\) from \(R=0\) to \(R=20.0 \mathrm{~cm} .\) Assume the cylinders are very long compared to \(R_{3}\)

Short Answer

Expert verified
For each region: (a) \( E = \frac{\rho_{\mathrm{E}}R}{2\epsilon_0} \) for \( 0R_3 \).

Step by step solution

01

Understand the Problem

We are tasked with finding the electric field at different ranges for a charged solid cylinder and its surrounding tube. The charge density \( \rho_{\mathrm{E}} \) is identical for both materials. We need to apply Gauss's Law to solve each part.
02

Apply Gauss's Law

Gauss's Law states \( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \). To find the electric field, we must integrate the electric field across a Gaussian surface, usually a cylindrical surface coaxial with the charged cylinder.
03

Calculate Electric Field for \(0

Here, the Gaussian surface is within the solid cylinder. The enclosed charge is \( Q_{\text{enc}} = \rho_{\mathrm{E}} \pi R^2 L \). Substitute into Gauss's Law: \( E(2\pi RL) = \frac{\rho_{\mathrm{E}} \pi R^2 L}{\epsilon_0} \). Simplify to find \( E = \frac{\rho_{\mathrm{E}}R}{2\epsilon_0} \).
04

Calculate Electric Field for \(R_1

In this range, the Gaussian surface is outside the solid cylinder but inside the hollow cylinder. Enclosed charge is from the solid cylinder only: \( Q_{\text{enc}} = \rho_{\mathrm{E}} \pi R_1^2 L \). Thus, \( E(2\pi RL) = \frac{\rho_{\mathrm{E}} \pi R_1^2 L}{\epsilon_0} \), leading to \( E = \frac{\rho_{\mathrm{E}}R_1^2}{2\epsilon_0 R} \).
05

Calculate Electric Field for \(R_2

Now, the Gaussian surface is inside the cylindrical shell. The enclosed charge includes the solid cylinder and the part of the shell up to \( R \): \( Q_{\text{enc}} = \rho_{\mathrm{E}}(\pi R_1^2 L + (\pi R^2 - \pi R_2^2)L) \). Thus, \( E = \frac{\rho_{\mathrm{E}}(R_1^2 + R^2 - R_2^2)}{2\epsilon_0 R} \).
06

Calculate Electric Field for \(R>R_3\)

Beyond the outer radius, the Gaussian surface encapsulates both the solid cylinder and the full cylindrical shell. The total charge is \( Q_{\text{enc}} = \rho_{\mathrm{E}}(\pi R_1^2 L + \pi(R_3^2 - R_2^2)L) \). Electric field \( E = \frac{\rho_{\mathrm{E}} (R_1^2 + R_3^2 - R_2^2)}{2\epsilon_0 R} \).
07

Understand the Parameters

Given \( \rho_{\mathrm{E}}=15 \mu \mathrm{C} / \mathrm{m}^3 \) and \( R_1 = 5\text{ cm}, R_2 = 10\text{ cm}, R_3 = 15 \text{ cm} \). This means \( R_1 = \frac{1}{2} R_2 = \frac{1}{3} R_3 \). Set up the conditions for solving and plotting the electric field profile.
08

Plot the Electric Field

Using the expressions for \( E \) found in steps 3-6, plot \( E(R) \) from \( R = 0 \) to \( R = 20 \text{ cm} \). Calculate the values of \( E \) for each region and compile into a single graph of \( E \) versus \( R \).

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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 a pivotal principle in electromagnetism that helps us calculate electric fields generated by symmetrical charge distributions easily. The law states: \[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \]This means that the electric flux through a closed surface is proportional to the charge enclosed by that surface, divided by the permittivity of free space, \(\epsilon_0\). When applying Gauss's Law, the choice of the Gaussian surface is crucial. It should be simple and reflect the symmetry of the charge distribution. For cylindrical charge distributions, like in this exercise, a coaxial cylindrical Gaussian surface simplifies calculations. This symmetry makes the electric field constant over the surface, facilitating easy integration.To solve the problem, for each region defined by the radii \(R_1\), \(R_2\), and \(R_3\), we encapsulate the relevant portion of the charge distribution.Utilizing Gauss's Law clarifies how the electric field behaves around and within different regions of charged objects, particularly in cylindrical geometries. This is not only essential for theoretical understanding but also for practical problem-solving in physics.
Cylindrical Charge Distribution
A cylindrical charge distribution is when charges are distributed evenly along a cylinder, either being solid or forming a shell. This exercise involves a solid cylinder with a uniform charge density, \( \rho_{\text{E}} \), and a coaxial cylindrical tube, also uniformly charged. For cylindrical objects, the unique symmetry makes electric field calculations more manageable. When dealing with infinite or very long cylinders, as in this case, we can assume that the electric field points radially outward from the axis due to symmetry.For regions inside and outside the cylinder, the electric field behaves differently:
  • Inside the solid cylinder (\(0 The field is directly proportional to the distance from the axis, as we accumulate more enclosed charge with distance.
  • Between the solid and hollow cylinders (\(R_1 The field decreases with distance, as the internal charge is fixed, and the Gaussian surface increases.
  • Inside the hollow cylinder (\(R_2 The charge affects the field both from the solid cylinder and the cylindrical shell's accumulating charge.
  • Outside all cylinders (\(R>R_3\)): The electric field is determined by the total enclosed charge from both cylinders, diminishing with distance.
Understanding cylindrical charge distributions allows for precise electric field calculations as each region can be evaluated independently.
Charge Density
Charge density, symbolized as \( \rho_{\text{E}} \) in this context, is a measure of how much charge is present per unit volume within a material. In this exercise, it's given in microcoulombs per cubic meter (\( \mu \mathrm{C}/\mathrm{m}^3 \)). This is an essential factor in determining the electric field using Gauss's Law. When a solid cylinder and an outer tube are both uniformly charged, the density helps establish how much charge is enclosed by a Gaussian surface.In mathematical terms, for a given volume \( V \), the total charge \( Q \) can be calculated as:\[ Q = \rho_{\text{E}} \times V \]For cylindrical geometries:- **Solid Cylinder:** The volume enclosed equals \( \pi R^2 L \), leading to charge \( Q = \rho_{\text{E}} \pi R^2 L \).- **Shell Segment:** The charge for a region within the hollow cylinder is calculated by the volume it occupies between two radii (\(R_2\) to \(R_3\)), resulting in \( Q = \rho_{\text{E}} (\pi R_3^2 L - \pi R_2^2 L) \).Understanding charge density allows one to navigate the relations between volume, charge, and the subsequent electric field. In practical applications, it influences how fields are manipulated in conductive environments, making it a critical aspect of designing electronic devices and systems.

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Most popular questions from this chapter

(II) A flat square sheet of thin aluminum foil, \(25 \mathrm{~cm}\) on a side, carries a uniformly distributed \(275 \mathrm{nC}\) charge. What, approximately, is the electric field (a) \(1.0 \mathrm{~cm}\) above the center of the sheet and (b) \(15 \mathrm{~m}\) above the center of the sheet?

(II) A solid metal sphere of radius \(3.00 \mathrm{~m}\) carries a total charge of \(-5.50 \mu \mathrm{C}\). What is the magnitude of the electric field at a distance from the sphere's center of (a) \(0.250 \mathrm{~m}\), (b) \(2.90 \mathrm{~m},\) (c) \(3.10 \mathrm{~m},\) and \((d) 8.00 \mathrm{~m} ?\) How would the answers differ if the sphere were \((e)\) a thin shell, or \((f)\) a solid nonconductor uniformly charged throughout?

(II) A ring of charge with uniform charge density is completely enclosed in a hollow donut shape. An exact copy of the ring is completely enclosed in a hollow sphere. What is the ratio of the flux out of the donut shape to that out of the sphere?

A conducting spherical shell (Fig. 22-49) has inner radius \(=10.0 \mathrm{~cm},\) outer radius \(=15.0 \mathrm{~cm},\) and has a \(+3.0 \mu \mathrm{C}\) point charge at the center. A charge of \(-3.0 \mu \mathrm{C}\) is put on the conductor. ( \(a\) ) Where on the conductor does the \(-3.0 \mu \mathrm{C}\) end up? \((b)\) What is the electric field both inside and outside the shell?

(III) Charge is distributed within a solid sphere of radius \(r_{0}\) in such a way that the charge density is a function of the radial position within the sphere of the form: \(\rho_{\mathrm{E}}(r)=\rho_{0}\left(r / r_{0}\right) .\) If the total charge within the sphere is \(Q\) (and positive), what is the electric field everywhere within the sphere in terms of \(Q, r_{0},\) and the radial position \(r ?\)
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