An electric charge Q is distributed uniformly along a thick and enormously long conducting wire with radius R and length L. Using Gauss's law, what is the electric field at distance r perpendicular to the wire? (Consider the cases inside and outside the wire) We will choose a symmetric Gaussian surface, which is the surface a cylinder excluding its ends, then evaluate the dotproduct to obtainA =(Equation 1)Case 1: Inside the wireSince, r falls inside the wire, then all the enclosed charge must be:Genc =On the other hand, the Gaussian surface inside the wire is given byA =Using Equation 1, the electric field in simplified form isE =Case 2: Outside the wireSince, r falls outside the wire, then, all the charge must be enclosed, thusdenc=On the other hand, the Gaussian surface outside the wire is given byA =Using Equation 1, the electric field in simplified form isE =

Equation 1 is, ∮E·dA=qencε0EA=qencε0

Answered: We will choose a symmetric Gaussian…

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter25: Gauss’s Law

Section: Chapter Questions

Problem 68PQ: Examine the summary on page 780. Why are conductors and charged sources with linear symmetry,...

See similar textbooks

Related questions

Concept explainers

Question

An electric charge Q is distributed uniformly along a thick and enormously long conducting wire with radius R and length L. Using Gauss's law, what is the electric field at distance r perpendicular to the wire? (Consider the cases inside and outside the wire)

We will choose a symmetric Gaussian surface, which is the surface a cylinder excluding its ends, then evaluate the dot
product to obtain
A =
(Equation 1)
Case 1: Inside the wire
Since, r falls inside the wire, then all the enclosed charge must be:
Genc =
On the other hand, the Gaussian surface inside the wire is given by
A =
Using Equation 1, the electric field in simplified form is
E =
Case 2: Outside the wire
Since, r falls outside the wire, then, all the charge must be enclosed, thus
denc=
On the other hand, the Gaussian surface outside the wire is given by
A =
Using Equation 1, the electric field in simplified form is
E =

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

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