An electric charge Q is distributed uniformly along a thick and enormously long conducting wire with radius R and length L. UsingGauss's law, what is the electric field at distance r perpendicular to the wire? (Consider the cases inside and outside the wire)SolutionTo find the electric field inside at r distance from the wire we will use the Gauss's law which is expressed asWe will choose a symmetric Gaussian surface, which is the surface a cylinder excluding its ends, then evaluate the dot product to obtainA =(Equation 1)Case 1: Inside the wireSince, r falls inside the wire, then all the enclosed charge must be:GencOn 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 =

Gauss's law, what is the electric field at distance r perpendicular to the wire? (Consider the cases inside and Solution To find the electric field inside at r distance from the wire we will use the Gauss's law which is expressed as We will choose a symmetric Gaussian surface, which is the surface a cylinder excluding its ends, then evalu A- Case 1: Inside the wire Since. r falls inside the wire, then all the enclosed charge must be: Penc 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=

Gauss's law, what is the electric field at distance r perpendicular to the wire? (Consider the cases inside and Solution To find the electric field inside at r distance from the wire we will use the Gauss's law which is expressed as We will choose a symmetric Gaussian surface, which is the surface a cylinder excluding its ends, then evalu A- Case 1: Inside the wire Since. r falls inside the wire, then all the enclosed charge must be: Penc 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=

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter6: Gauss's Law

Section: Chapter Questions

Problem 61P: An uncharged spherical conductor S of radius R has two spherical cavities A and B of radii a and b,...

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