/*! 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 59 mmh A solid nonconducting sphere... [FREE SOLUTION] | 91Ó°ÊÓ

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mmh A solid nonconducting sphere has a positive charge \(q\) spread uniformly throughout its volume. The charge density or charge per unit volume, therefore, is \(\frac{q}{\frac{4}{3} \pi R^{3}} .\) Use Gauss' law to show that the electric field at a point within the sphere at a radius \(r\) has a magnitude of \(\frac{q r}{4 \pi \epsilon_{0} R^{3}}\) (Hint: For a Gaussian surface, use a sphere of radius r centered within the solid sphere of radius. Note that the net charge within any volume is the charge density times the volume.)

Short Answer

Expert verified
The electric field within the sphere at radius \( r \) is \( \frac{q r}{4 \pi \epsilon_0 R^3} \).

Step by step solution

01

Understand Charge Density

The charge density \( \rho \) for the sphere is given by the formula \( \rho = \frac{q}{\frac{4}{3} \pi R^3} \). This indicates the amount of charge per unit volume within the sphere, where \( q \) is the total charge and \( R \) is the radius of the sphere.
02

Define Gaussian Surface

To apply Gauss' law, consider a Gaussian surface inside the sphere at a radius \( r \), where \( r < R \). This surface is an imaginary sphere concentric with the solid sphere and has radius \( r \).
03

Calculate Charge Enclosed by Gaussian Surface

The charge \( q' \) enclosed by the Gaussian surface can be calculated using the charge density and the volume of the smaller sphere:\[q' = \rho \times \frac{4}{3} \pi r^3 = \left( \frac{q}{\frac{4}{3} \pi R^3} \right) \times \frac{4}{3} \pi r^3 = \frac{q r^3}{R^3}\]
04

Apply Gauss' Law

Gauss' law states that the electric flux \( \Phi_E \) through a closed surface is \( \Phi_E = \frac{q'}{\epsilon_0} \), where \( \epsilon_0 \) is the permittivity of free space. For a spherical surface, \( \Phi_E = E \cdot 4\pi r^2 \). Equating these gives:\[E \cdot 4\pi r^2 = \frac{q r^3}{\epsilon_0 R^3}\]
05

Solve for Electric Field

Rearrange the equation to solve for \( E \):\[E = \frac{q r}{4 \pi \epsilon_0 R^3}\]Thus, the magnitude of the electric field at a point within the sphere at radius \( r \) is \( \frac{q r}{4 \pi \epsilon_0 R^3} \), which confirms the given result.

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Key Concepts

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

Electric Field
The electric field is a crucial concept in understanding electrical forces. It represents the influence that a charged object exerts in the space around it. This field extends outward from the charged object and affects other charges that enter its realm. In essence, the electric field can be thought of as the 'force field' that emanates from an electric charge. This field is symbolized by the letter \( E \) and is measured in units of newtons per coulomb (N/C). To determine the strength and direction of an electric field created by a charge, one must utilize Coulomb's law. The equation derived from Gauss' law, as used in this problem, goes a step further. It provides a simple method for calculating the electric field for symmetrical charge distributions. Here, the nonconducting sphere has a uniform charge distribution.The Gaussian surface—a hypothetical sphere inside the charged body—allows us to apply Gauss' law effectively. The surface encloses some charge which helps us obtain the electric field inside the nonconducting sphere.
Charge Density
Charge density is a measure that describes how much charge is spread over a certain volume. It provides critical insight into the distribution of electric charge within a given object. For instance, if we place charge \( q \) into a nonconducting sphere with radius \( R \), the charge density \( \rho \) quantifies this by detailing the charge for each unit of volume.In mathematical terms, for a sphere, the charge density \( \rho \) is calculated as \( \rho = \frac{q}{\frac{4}{3} \pi R^3} \). - \( q \) represents the total charge contained within the sphere.- \( \displaystyle \frac{4}{3} \pi R^3 \) is the formula for calculating the volume of the sphere.When solving for the electric field inside a nonconducting sphere, this density simplifies the process. By knowing how the charge is distributed, we can determine the fraction that exists within the Gaussian surface at any smaller radius \( r \), aiding in our application of Gauss' Law.
Nonconducting Sphere
A nonconducting sphere is a type of insulator characterized by having no free-flowing charge within its volume. Unlike conductive materials, where electrons can move freely, nonconducting materials keep their charge stationary. This fundamental characteristic affects how electric fields are generated and calculated within such objects. In the exercise scenario, the sphere is uniformly charged throughout its entire structure. This means the charge doesn't move, spreading evenly and consistently across the volume. This uniformity simplifies mathematical calculations because every point within the sphere has the same charge behavior, making the electric field predictable. By placing a Gaussian sphere inside a nonconducting sphere, Gauss' law becomes a powerful tool. The imaginary Gaussian surface helps in calculating the electric field at any point within the sphere's radius. The approach leverages the uniform charge density to find out how much charge the inner Gaussian sphere encloses, thus allowing calculation of the field inside the sphere. In summary, nonconducting spheres retain charge in a fixed position, drastically influencing how these electric fields are understood and computed based on their inherent properties.

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Most popular questions from this chapter

A charge \(+q\) is located at the origin, while an identical charge is located on the \(x\) axis at \(x=+0.50 \mathrm{m}\). A third charge of \(+2 q\) is located on the \(x\) axis at such a place that the net electrostatic force on the charge at the origin doubles, its direction remaining unchanged. Where should the third charge be located?

Two objects carry initial charges that are \(q_{1}\) and \(q_{2}\), respectively, where \(\left|q_{2}\right|>\left|q_{1}\right| .\) They are located \(0.200 \mathrm{m}\) apart and behave like point charges. They attract each other with a force that has a magnitude of \(1.20 \mathrm{N} .\) The objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to the magnitude of the initial attractive force. What are the magnitudes of the initial charges on the objects? Section 18.6 The Electric Field Section 18.7 Electric Field Lines Section 18.8 The Electric Field Inside a Conductor: Shielding

Two parallel plate capacitors have circular plates. The magnitude of the charge on these plates is the same. However, the electric field between the plates of the first capacitor is \(2.2 \times 10^{5} \mathrm{N} / \mathrm{C},\) whereas the field within the second capacitor is \(3.8 \times 10^{5} \mathrm{N} / \mathrm{C} .\) Determine the ratio \(r_{2} / r_{1}\) of the plate radius for the second capacitor to the plate radius for the first capacitor.

A spring with an unstrained length of \(0.074 \mathrm{m}\) and a spring constant of \(2.4 \mathrm{N} / \mathrm{m}\) hangs vertically downward from the ceiling. A uniform electric field directed vertically upward fills the region containing the spring. A sphere with a mass of \(5.1 \times 10^{-3} \mathrm{kg}\) and a net charge of \(+6.6 \mu \mathrm{C}\) is attached to the lower end of the spring. The spring is released slowly, until it reaches equilibrium. The equilibrium length of the spring is \(0.059 \mathrm{m} .\) What is the magnitude of the external electric field?

A small object, which has a charge \(q=7.5 \mu \mathrm{C}\) and mass \(m=9.0 \times 10^{-5} \mathrm{kg},\) is placed in a constant electric field. Starting from rest, the object accelerates to a speed of \(2.0 \times 10^{3} \mathrm{m} / \mathrm{s}\) in a time of \(0.96 \mathrm{s}\). Determine the magnitude of the electric field.

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