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Problem: 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).

Problem: 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).

Inquiry into Physics
Inquiry into Physics
8th Edition
ISBN:9781337515863
Author:Ostdiek
Publisher:Ostdiek
Chapter7: Electricity
Section: Chapter Questions
Problem 8Q
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Problem: 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).
3 ACT 4. 4th A drive C Clan P Not M Sign Que A drive m mCc UP UP CO myG 4. Acce Al Bf A ust.blackboard.com/ultra/courses/_168281_1/cl/outline os M Gmail YouTube Maps y Image * Question Completion Status: QUESTION 1 Problem 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) Solution To find the electric field inside at r distance from the wire we will use the Gauss's law which is expressed as · dA = genc / epsilono We will choose a symmetric Gaussian surface, which is the surface a cylinder excluding its ends, then evaluate the dot product to obtain E A= genc / epsilon (Equation 1) Case 1: Inside the wire Since, r falls inside the wire, then all the enclosed charge must be: denc = 8piBR3/2 %3D 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 endosed, thus denc = %3D 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 = Click Save and Submit to save and submit. Click Save All Answers to save all ansuers. MacBook Air esc 000 000 EA F2 F3 F4
Transcribed Image Text:3 ACT 4. 4th A drive C Clan P Not M Sign Que A drive m mCc UP UP CO myG 4. Acce Al Bf A ust.blackboard.com/ultra/courses/_168281_1/cl/outline os M Gmail YouTube Maps y Image * Question Completion Status: QUESTION 1 Problem 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) Solution To find the electric field inside at r distance from the wire we will use the Gauss's law which is expressed as · dA = genc / epsilono We will choose a symmetric Gaussian surface, which is the surface a cylinder excluding its ends, then evaluate the dot product to obtain E A= genc / epsilon (Equation 1) Case 1: Inside the wire Since, r falls inside the wire, then all the enclosed charge must be: denc = 8piBR3/2 %3D 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 endosed, thus denc = %3D 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 = Click Save and Submit to save and submit. Click Save All Answers to save all ansuers. MacBook Air esc 000 000 EA F2 F3 F4
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