/*! 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 21 A nonconducting spherical shell,... [FREE SOLUTION] | 91Ó°ÊÓ

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A nonconducting spherical shell, with an inner radius of \(4.0\) \(\mathrm{cm}\) and an outer radius of \(6.0 \mathrm{~cm}\), has charge spread nonuniformly through its volume between its inner and outer surfaces. The volume charge density \(\rho\) is the charge per unit volume, with the unit coulomb per cubic meter. For this shell \(\rho=b / r\), where \(r\) is the distance in meters from the center of the shell and \(b=3.0 \mu \mathrm{C} / \mathrm{m}^{2} .\) What is the net charge in the shell?

Short Answer

Expert verified
The net charge in the shell is approximately \(7.54 \times 10^{-8} \mathrm{C}\).

Step by step solution

01

Set up the integral for total charge

The total charge \( Q \) is given by integrating the charge density \( \rho \) over the volume of the shell. The volume element in spherical coordinates is \( dV = 4 \pi r^2 dr \). The charge density is \( \rho = \frac{b}{r} \). Therefore, the total charge is \[ Q = \int_{R_1}^{R_2} \rho \, dV = \int_{R_1}^{R_2} \frac{b}{r} \cdot 4\pi r^2 \, dr \], where \( R_1 = 0.04 \) m and \( R_2 = 0.06 \) m are the inner and outer radii.
02

Simplify the integrand

Substitute the given charge density \( \rho = \frac{b}{r} \) into the integrand: \[ Q = 4\pi b \int_{R_1}^{R_2} r \, dr \]. Now, it reduces to a simpler form as \( b \) and \( 4\pi \) are constants.
03

Evaluate the integral

Integrate the expression \( \int r \, dr \), which results in \( \frac{r^2}{2} \). Therefore, the integral becomes: \[ Q = 4\pi b \left[ \frac{r^2}{2} \right]_{R_1}^{R_2} = 4\pi b \left( \frac{R_2^2}{2} - \frac{R_1^2}{2} \right) \].
04

Substitute the limits of integration

Insert the values \( R_1 = 0.04 \) and \( R_2 = 0.06 \) into the expression: \[ Q = 4\pi \cdot 3.0 \times 10^{-6} \left( \frac{0.06^2}{2} - \frac{0.04^2}{2} \right) \].
05

Calculate the result

First, calculate \( 0.06^2 - 0.04^2 = 0.0036 - 0.0016 = 0.0020 \). Then, calculate \( 4\pi \cdot 3.0 \times 10^{-6} \cdot 0.0010 = 12\pi \times 10^{-6} \cdot 0.0010 = 3.77 \times 10^{-8} \). Finally, the total charge is \( Q = 7.54 \times 10^{-8} \).

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Key Concepts

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

Volume Charge Density
Volume charge density, represented as \( \rho \), is a fundamental concept when studying electrostatics. It describes how charge is distributed throughout a given volume. The unit of volume charge density is coulombs per cubic meter (C/m³). Understanding this concept is crucial for solving problems involving charge distributions that are not uniform, such as in materials where charge varies with position.
In the given exercise, the volume charge density is nonuniform and is expressed as \( \rho = \frac{b}{r} \), where \( b \) is a constant denoted in microcoulombs per square meter, and \( r \) is the radial distance from the center of the sphere in meters.
This formula indicates that the charge density diminishes as you move outward from the center, inversely proportional to \( r \). Thus, knowing how to work with and integrate this density can help find the total charge in a specified region of space, such as the spherical shell in our exercise.
Spherical Coordinates
Spherical coordinates are often used in physics and engineering, especially when dealing with problems exhibiting spherical symmetry. This system is handy because it simplifies the description of points in three-dimensional space using three parameters:
  • \( r \): the radial distance from the origin.
  • \( \theta \): the polar angle measured from the z-axis.
  • \( \phi \): the azimuthal angle in the x-y plane.

For a volume like a spherical shell, integrating in spherical coordinates streamlines the process. The volume element in this system is represented as \( dV = r^2 \sin \theta \, dr \, d\theta \, d\phi \). However, in our problem, due to symmetry and focusing on volume, the expression simplifies to \( dV = 4\pi r^2 \, dr \).
This simplification accounts for the full angle rotations \( \theta \) and \( \phi \), reflecting the symmetry in all directions around the sphere. The exercise focusses on calculating integral over the radial distance, highlighting how useful spherical coordinates can be for evaluating integrals in problems concerning spherical shells.
Nonconducting Spherical Shell
A nonconducting spherical shell is a hollow object with a specified inner and outer radius. It is made from a material that does not allow current to flow through it, meaning charge distributes only over its volume, if present, and remains static.
In this particular problem, the shell is described with an inner radius of 4 cm and an outer radius of 6 cm. Upon taking spherical symmetry into account, the shell doesn't permit any electrical conduction but can still house stationary charges.
The challenges presented with such nonconducting shells often involve understanding how charge is distributed across its volume, especially when the volume charge density isn't uniform. This non-conductive property allows charges to maintain a specific spatial distribution, dependent on variables like distance from the center, thus making computations like finding the net charge integral important.
By carefully integrating the charge density across the shell's volume, accounting for both inner and outer surfaces, we can derive quantities like total net charge, providing insight into the electrostatic behavior of the shell.

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Most popular questions from this chapter

If a cat repeatedly rubs against your cotton slacks on a dry day, the charge transfer between the cat hair and the cotton can leave you with an excess charge of \(-2.00 \mu \mathrm{C}\). (a) How many electrons are transferred between you and the cat? You will gradually discharge via the floor, but if instead of waiting, you immediately reach toward a faucet, a painful spark can suddenly appear as your fingers near the faucet. (b) In that spark, do electrons flow from you to the faucet or vice versa? (c) Just before the spark appears, do you induce positive or negative charge in the faucet? (d) If, instead, the cat reaches a paw toward the faucet, which way do electrons flow in the resulting spark? (e) If you stroke a cat with a bare hand on a dry day, you should take care not to bring your fingers near the cat's nose or you will hurt it with a spark. Considering that cat hair is an insulator, explain how the spark can appear.

Three identical conducting spheres form an equilateral triangle of side length \(d=20.0 \mathrm{~cm}\). The sphere radii are much smaller than \(d\), and the sphere charges are \(q_{A}=-2.00 \mathrm{nC}, q_{B}=-4.00\) \(\mathrm{nC}\), and \(q_{C}=+8.00 \mathrm{nC}\). (a) What is the magnitude of the electrostatic force between spheres \(A\) and \(C\) ? The following steps are then taken: \(A\) and \(B\) are connected by a thin wire and then discon-nected; \(B\) is grounded by the wire, and the wire is then removed; \(B\) and \(C\) are connected by the wire and then disconnected. What now are the magnitudes of the electrostatic force (b) between spheres \(A\) and \(C\) and (c) between spheres \(B\) and \(C ?\)

Two small, positively charged spheres have a combined charge of \(5.0 \times 10^{-5} \mathrm{C}\). If each sphere is repelled from the other by an electrostatic force of \(1.0 \mathrm{~N}\) when the spheres are \(2.0 \mathrm{~m}\) apart, what is the charge on the sphere with the smaller charge?

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of \(0.108 \mathrm{~N}\) when their center-to-center separation is \(50.0 \mathrm{~cm} .\) The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of \(0.0360 \mathrm{~N}\). Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other?

Calculate the number of coulombs of positive charge in 250 \(\mathrm{cm}^{3}\) of (neutral) water. (Hint: A hydrogen atom contains one proton; an oxygen atom contains eight protons.)

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