Homework Help is Here – Start Your Trial Now!
Science
Physics
GAUSS'S LAW TO CHARGE INSULATORS: Gauss's law is useful in determining electric fields when the charge distribution is characterized by a high degree of symmetry. The following examples demonstrate ways of choosing the gaussian surface over which the surface integral can be simplified and the electric field determined. In choosing the surface, we should always take advantage of the symmetry of the charge distribution so that we can remove E from the integral and solve for it. The goal in this type of calculation is to determine a surface that satisfies one or more of the following conditions: 1. The value of the electric field can be argued by symmetry to be constant over the surface. 2. The dot product can be expressed as a simple algebraic product E dA because E and dA are parallel. 3. The dot product is zero because E and dA are perpendiculars. 4. The field can be argued to be zero over the surface.

GAUSS'S LAW TO CHARGE INSULATORS: Gauss's law is useful in determining electric fields when the charge distribution is characterized by a high degree of symmetry. The following examples demonstrate ways of choosing the gaussian surface over which the surface integral can be simplified and the electric field determined. In choosing the surface, we should always take advantage of the symmetry of the charge distribution so that we can remove E from the integral and solve for it. The goal in this type of calculation is to determine a surface that satisfies one or more of the following conditions: 1. The value of the electric field can be argued by symmetry to be constant over the surface. 2. The dot product can be expressed as a simple algebraic product E dA because E and dA are parallel. 3. The dot product is zero because E and dA are perpendiculars. 4. The field can be argued to be zero over the surface.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
See similar textbooks
icon
Related questions
Question
Urgent! Please give me the correct answer. Show complete solution
GAUSS'S LAW TO CHARGE INSULATORS: Gauss's law is useful in determining electric fields when the charge distribution is characterized by a high degree of symmetry. The following examples demonstrate ways of choosing the gaussian surface over which the surface integral can be simplified and the electric field determined. In choosing the surface, we should always take advantage of the symmetry of the charge distribution so that we can remove E from the integral and solve for it. The goal in this type of calculation is to determine a surface that satisfies one or more of the following conditions: 1. The value of the electric field can be argued by symmetry to be constant over the surface. 2. The dot product can be expressed as a simple algebraic product E dA because E and dA are parallel. 3. The dot product is zero because E and dA are perpendiculars. 4. The field can be argued to be zero over the surface.
Transcribed Image Text:GAUSS'S LAW TO CHARGE INSULATORS: Gauss's law is useful in determining electric fields when the charge distribution is characterized by a high degree of symmetry. The following examples demonstrate ways of choosing the gaussian surface over which the surface integral can be simplified and the electric field determined. In choosing the surface, we should always take advantage of the symmetry of the charge distribution so that we can remove E from the integral and solve for it. The goal in this type of calculation is to determine a surface that satisfies one or more of the following conditions: 1. The value of the electric field can be argued by symmetry to be constant over the surface. 2. The dot product can be expressed as a simple algebraic product E dA because E and dA are parallel. 3. The dot product is zero because E and dA are perpendiculars. 4. The field can be argued to be zero over the surface.
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Relativistic speed and time
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON
SEE MORE TEXTBOOKS