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Chapter 24: Problem 21

Nonconducting Spherical Shell In Fig. \(24-29 b\), a nonconducting spherical shell, of inner radius \(a\) and outer radius \(b\), has a positive volume charge density \(\rho=A / r\) (within its thickness), where \(A\) is a constant and \(r\) is the distance from the center of the shell. In addition, a positive point charge \(q\) is located at that center. What value should \(A\) have if the electric field in the shell \((a \leq r \leq b)\) is to be uniform? (Hint: The constant \(A\) depends on \(a\) but not on \(b .\) )

Short Answer

Expert verified
A should be \dfrac{q}{2 \pi \textbf{\textbackslash \textdegree}_0 a^2}.

Step by step solution

01

- Understand the Problem

We need to find the constant value of the parameter A such that the electric field inside a nonconducting spherical shell with given charge density remains uniform.
02

- Express the Charge Density

The volume charge density is given by \(\rho = \dfrac{A}{r}\).
03

- Apply Gauss's Law

Gauss's law states that \(abla \bullet \textbf{E} = \frac{\rho}{\textbf{\textbackslash \textdegree}_0} \). To ensure the electric field is uniform, consider a spherical Gaussian surface of radius \(r\).
04

- Calculate the Charge Enclosed

The total charge enclosed within the shell from radius \(a\) to \(r\) is \[ Q_{enc} = \int_a^r \left( \textbf\frac{A}{r'} \times 4\pi r'^2 \right)dr' = 4 \pi A \int_a^r \frac{r'^2}{r'}dr' = 4 \pi A \int_a^r r'dr' = 2 \pi A (r^2 - a^2) \].
05

- Apply the Result to Gauss's Law

Using Gauss's law for a spherical Gaussian surface with uniform electric field \(E\), the electric flux is \[ E \cdot 4 \pi r^2 = \dfrac{Q_{enc}}{\textbf{\textbackslash \textdegree}_0} = \dfrac{q + 2 \pi A (r^2 - a^2)}{\textbf{\textbackslash \textdegree}_0} \].
06

Step 6- Solve for Uniform Electric Field

For the electric field \(E\) to be uniform, the expression on the right must be independent of \(r\), therefore: \[ E 4 \pi \textbf{\textbackslash \textdegree}_0 r^2 = q + 2 \pi A (r^2 - a^2) \rightarrow 2 \pi A \textbf{\textbackslash \textdegree}_0 r^2 = A \rightarrow A = \dfrac{q}{2 \pi \textbf{\textbackslash \textdegree}_0 a^2} \].

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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 key principle in electromagnetism. It relates the electric flux through a closed surface to the charge enclosed by that surface. Mathematically, Gauss's Law is given by the formula \[\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\epsilon_0}\].
Here, \( \oint \mathbf{E} \cdot d\mathbf{A} \) is the electric flux through a closed surface, \(Q_{enc} \) is the total charge enclosed within that surface, and \( \epsilon_0 \) is the permittivity of free space.
To use Gauss's Law effectively, choose a Gaussian surface that mirrors the symmetry of the charge distribution. For example, when dealing with a spherical charge distribution, a spherical Gaussian surface simplifies calculations.
By evaluating the electric field on this surface, we can relate it to the enclosed charge and solve for unknowns like electric field strength or charge density.
Electric Field
The electric field represents the force per unit charge exerted on a positive test charge at any point in space. It is denoted by the symbol \(\mathbf{E}\) and measured in newtons per coulomb (N/C).
For a spherical shell, the electric field inside the shell (between the inner radius \(a\) and outer radius \(b\)) is governed by the charge distribution within the shell and the point charge at the center. According to Gauss's Law, for a spherical surface, the electric field at a distance \(r\) is given by: \[\mathbf{E} = \frac{1}{4\pi\epsilon_0} \frac{Q_{enc}}{r^2}\].
This formula shows that the electric field depends on the total enclosed charge \(Q_{enc}\) and the distance \(r\) from the center.
In the case of a uniform electric field, the enclosed charge must be arranged such that the resulting field does not vary with \(r\), simplifying our calculations.
Charge Density
Charge density \(\rho\) describes how charge is distributed in space. It is defined as the amount of charge per unit volume, area, or length, depending on the context. Volume charge density is given by \[\rho = \frac{A}{r}\], where \(A\) is a constant and \(r\) is the distance from the center of the shell in this exercise.
In nonconducting spherical shells, the volume charge density can vary with \(r\) leading to a different charge distribution throughout the shell's thickness.
To find the value of \(A\), given that the electric field must remain uniform within the shell, we can integrate over the volume of the shell to find the total enclosed charge and relate it to the electric field using Gauss's Law. This change in charge density ensures a balance between the point charge at the center and the charge within the shell for a uniform electric field.

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Most popular questions from this chapter

Square Metal Plate A square metal plate of edge length \(8.0 \mathrm{~cm}\) and negligible thickness has a total charge of \(6.0 \times 10^{-6} \mathrm{C}\) (a) Estimate the magnitude \(E\) of the electric field just off the center of the plate (at, say, a distance of \(0.50 \mathrm{~mm}\) ) by assuming that the charge is spread uniformly over the two faces of the plate. (b) Estimate \(E\) at a distance of \(30 \mathrm{~m}\) (large relative to the plate size) by assuming that the plate is a point charge.

Geiger Counter Figure \(24-33\) shows a Geiger counter, a device used to detect ionizing radiation (radiation that causes ionization of atoms). The counter consists of a thin, positively charged central wire surrounded by a concentric, circular, conducting cylinder with an equal negative charge. Thus, a strong radial electric field is set up inside the cylinder. The cylinder contains a low-pressure inert gas. When a particle of radiation enters the device through the cylinder wall, it ionizes a few of the gas atoms The resulting free electrons (labelled e) are drawn to the positive wire. However, the electric field is so intense that, between collisions with other gas atoms, the free electrons gain energy sufficient to ionize these atoms also. More free electrons are thereby created, and the process is repeated until the electrons reach the wire. The resulting "avalanche" of electrons is collected by the wire generating a signal that is used to record the passage of the original particle of radiation. Suppose that the radius of the central wire is \(25 \mu \mathrm{m}\), the radius of the cylinder \(1.4 \mathrm{~cm}\), and the length of the tube \(16 \mathrm{~cm}\). If the electric field component \(E_{r}\) at the cylinder's inner wall is \(+2.9 \times\) \(10^{4} \mathrm{~N} / \mathrm{C}\), what is the total positive charge on the central wire?

Interpreting Gauss Gauss' law states $$ \oint_{A} \vec{E} \cdot d \vec{A}=q_{A} / \varepsilon_{0} $$ where \(A\) is a surface and \(q_{A}\) is a charge. (a) Which of the following statements are true about the surface \(A\) appearing in Gauss' law for the equation to hold? You may list any number of these statements including all or none. i. The surface \(A\) must be a closed surface (must cover a volume). ii. The surface \(A\) must contain all the charges in the problem. iii. The surface \(A\) must be a highly symmetrical surface like a sphere or a cylinder. iv. The surface \(A\) must be a conductor. v. The surface \(A\) is purely imaginary. vi. The normals to the surface \(A\) must all be in the same direction as the electric field on the surface. (b) Which of the following statements are true about the charge \(q_{A}\) appearing in Gauss' law? You may list any number of these statements including all or none. i. The charge \(q_{A}\) must be all the charge lying on the Gaussian surface. ii. The charge \(q_{A}\) must be the charge lying within the Gaussian surface. iii. The charge \(q_{A}\) must be all the charge in the problem. iv. The charge \(q_{A}\) flows onto the Gaussian surface once the surface is established. v. The electric field \(E\) in the integral on the left of Gauss' law is due only to the charge \(q_{A}\). vi. The electric field \(E\) in the integral on the left on Gauss' law is due to all charges in the problem.

An Electron Is Shot An electron is shot directly toward the center of a large metal plate that has excess negative charge with surface charge density \(-2.0 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) If the initial kinetic energy of the electron is \(1.60 \times 10^{-17} \mathrm{~J}\) and if the electron is to stop (owing to electrostatic repulsion from the plate) just as it reaches the plate, how far from the plate must it be shot?

Conducting Sphere A conducting sphere of radius \(10 \mathrm{~cm}\) has an unknown charge. If the electric field \(15 \mathrm{~cm}\) from the center of the sphere has the magnitude \(3.0 \times 10^{3} \mathrm{~N} / \mathrm{C}\) and is directed radially inward, what is the net charge on the sphere?
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