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…

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)...

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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 =

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