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Ch 22: Gauss' Law
Chapter 22, Problem 24

Charge q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00 cm from the center of the sphere, the electric field due to the charge distribution has magnitude E = 940 N/C. What are (a) the volume charge density for the sphere?

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1
Calculate the volume of the sphere using the formula for the volume of a sphere, \( V = \frac{4}{3} \pi R^3 \), where \( R \) is the radius of the sphere.
Use Gauss's Law, \( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \), where \( Q_{\text{enc}} \) is the charge enclosed by the Gaussian surface and \( \epsilon_0 \) is the permittivity of free space. Choose a Gaussian surface that is a sphere with radius \( r \) (8.00 cm in this case) which encloses the entire charge of the sphere.
Since the electric field \( E \) is given at the distance \( r \), and the area of the Gaussian surface is \( A = 4\pi r^2 \), the left side of Gauss's Law becomes \( E \cdot 4\pi r^2 \).
Solve Gauss's Law for \( Q_{\text{enc}} \) using the expression from step 3 and the known value of \( \epsilon_0 \).
Calculate the volume charge density, \( \rho \), using the formula \( \rho = \frac{Q_{\text{enc}}}{V} \), where \( V \) is the volume of the sphere calculated in step 1.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Field

The electric field is a vector field that represents the force exerted by an electric charge on other charges in its vicinity. It is defined as the force per unit charge and is measured in newtons per coulomb (N/C). For a uniformly charged sphere, the electric field outside the sphere behaves as if all the charge were concentrated at the center, allowing for straightforward calculations using Gauss's law.
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Volume Charge Density

Volume charge density, denoted by ρ, is a measure of the amount of electric charge per unit volume within a given region of space. It is expressed in coulombs per cubic meter (C/m³). For a uniformly charged insulating sphere, the volume charge density can be calculated by dividing the total charge by the volume of the sphere, which is given by the formula V = (4/3)πR³.
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Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. Mathematically, it states that the total electric flux is equal to the enclosed charge divided by the permittivity of free space. This law is particularly useful for calculating electric fields in symmetric charge distributions, such as spheres, as it simplifies the process of finding the electric field at a distance from the charge.
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Related Practice
Textbook Question
A flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60° to a uniform electric field of magnitude 14 N/C. (c) For what angle φ between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest? Explain your answers.
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Textbook Question
A hemispherical surface with radius r in a region of uniform electric field E→ has its axis aligned parallel to the direction of the field. Calculate the flux through the surface.

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Textbook Question
You measure an electric field of 1.25×106 N/C at a distance of 0.150 m from a point charge. There is no other source of electric field in the region other than this point charge. (a) What is the electric flux through the surface of a sphere that has this charge at its center and that has radius 0.150 m? (b) What is the magnitude of this charge?
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Textbook Question
Charge q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00 cm from the center of the sphere, the electric field due to the charge distribution has magnitude E = 940 N/C. What are (b) the electric field at a distance of 2.00 cm from the sphere's center?
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Textbook Question
A conductor with an inner cavity, like that shown in Fig. 22.23c, carries a total charge of +5.00 nC. The charge within the cavity, insulated from the conductor, is −6.00 nC. How much charge is on (a) the inner surface of the conductor and (b) the outer surface of the conductor?
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Textbook Question
A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area σ = 5.00×10−6 C/m2. (a) A small sphere of mass m = 8.00×10−6 kg and charge q is placed 3.00 cm above the sheet of charge and then released from rest. (a) If the sphere is to remain motionless when it is released, what must be the value of q?
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