Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 22: Problem 84

A uniform sphere has a radius \(R\) and a total charge \(+Q,\) uniformly distributed throughout its volume. It is surrounded by a thick spherical shell carrying a total charge \(-Q,\) also uniformly distributed, and having an outer radius of \(2 R\). What is the electric field as a function of \(R ?\)

Short Answer

Expert verified
Question: Determine the electric field as a function of the radius R for a uniformly charged sphere with total charge +Q and radius R, surrounded by a uniformly charged spherical shell with outer radius 2R and total charge -Q. Answer: The electric field as a function of R is given by: $$ E(r) = \frac{Qr}{4\pi\epsilon_0 R^3} $$

Step by step solution

01

Determine the electric field of a uniformly charged sphere

First, let's find the electric field of a uniformly charged sphere. To do this, we use Gauss' Law: $$ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0} $$ For a sphere of radius r (with r <= R) centered at the origin, the enclosed charge is $$ Q_{inclosed} = \frac{Q}{\frac{4}{3}\pi R^3} \times \frac{4}{3}\pi r^3 $$ Then, the electric field is same for each point on the spherical surface enclosing the uniformly charged sphere, so: $$ E \times 4\pi r^2 = \frac{Q}{\frac{4}{3}\pi R^3} \times \frac{4}{3}\pi r^3 \div \epsilon_0 $$ Solving for E, we get: $$ E_{sphere}(r) = \frac{Qr}{4\pi\epsilon_0 R^3} $$
02

Determine the electric field of a uniformly charged spherical shell

Now, let's find the electric field of a uniformly charged spherical shell. To do this, we again use Gauss' Law, but this time we'll use a Gaussian surface that is a sphere centered at the origin, with 2R >= r >= R. Since there is no charge enclosed by this Gaussian surface, we have: $$ \oint \vec{E} \cdot d\vec{A} = 0 $$ Then, we can say that the electric field due to the shell at the Gaussian surface is canceled by the electric field due to the inner uniformly charged sphere: $$ E_{shell}(r) = -E_{sphere}(r) $$
03

Combine the electric fields

Now, we combine the electric fields from step 1 and step 2 to obtain the total electric field as a function of R: For r <= R, the electric field is only due to the charged sphere, so: $$ E(r) = \frac{Qr}{4\pi\epsilon_0 R^3} $$ For R <= r <= 2R, the electric field is due to both the sphere and the shell, so, as we derived, the electric field is: $$ E(r) = -E_{shell}(r) = -\left(-\frac{Qr}{4\pi\epsilon_0 R^3}\right) = \frac{Qr}{4\pi\epsilon_0 R^3} $$ Thus, the electric field as a function of R is: $$ E(r) = \frac{Qr}{4\pi\epsilon_0 R^3} $$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two charges, \(+e\) and \(-e,\) are a distance of \(0.68 \mathrm{nm}\) apart in an electric field, \(E,\) that has a magnitude of \(4.4 \mathrm{kN} / \mathrm{C}\) and is directed at an angle of \(45^{\circ}\) with the dipole axis. Calculate the dipole moment and thus the torque on the dipole in the electric field.

A sphere centered at the origin has a volume charge distribution of \(120 \mathrm{nC} / \mathrm{cm}^{3}\) and a radius of \(12 \mathrm{~cm}\). The sphere is centered inside a conducting spherical shell with an inner radius of \(30.0 \mathrm{~cm}\) and an outer radius of \(50.0 \mathrm{~cm}\). The charge on the spherical shell is \(-2.0 \mathrm{mC}\). What is the magnitude and direction of the electric field at each of the following distances from the origin? a) at \(r=10.0 \mathrm{~cm}\) c) at \(r=40.0 \mathrm{~cm}\) b) at \(r=20.0 \mathrm{~cm}\) d) at \(r=80.0 \mathrm{~cm}\)

A cube has an edge length of \(1.00 \mathrm{~m} .\) An electric field acting on the cube from outside has a constant magnitude of \(150 \mathrm{~N} / \mathrm{C}\) and its direction is also constant but unspecified (not necessarily along any edges of the cube). What is the total charge within the cube?

Research suggests that the electric fields in some thunderstorm clouds can be on the order of \(10.0 \mathrm{kN} / \mathrm{C}\). Calculate the magnitude of the electric force acting on a particle with two excess electrons in the presence of a \(10.0-\mathrm{kN} / \mathrm{C}\) field.

The electric flux through a spherical Gaussian surface of radius \(R\) centered on a charge \(Q\) is \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right) .\) What is the electric flux through a cubic Gaussian surface of side \(R\) centered on the same charge \(Q ?\) a) less than \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right)\) b) more than \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right)\) c) equal to \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right)\) d) cannot be determined from the information given
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Classical Mechanics

Read Explanation

Linear Momentum

Read Explanation

Circular Motion and Gravitation

Read Explanation

Waves Physics

Read Explanation

Thermodynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free