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)
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)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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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)
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
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:
denc
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 =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4dc607df-687a-4fcc-ae41-8e9f4a1de482%2Ff767212f-3e57-4563-920a-a9a383e2154f%2Fu56qapd_processed.png&w=3840&q=75)
Transcribed Image Text: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
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:
denc
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
%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 =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4dc607df-687a-4fcc-ae41-8e9f4a1de482%2Ff767212f-3e57-4563-920a-a9a383e2154f%2F1b0rgpp_processed.png&w=3840&q=75)
Transcribed Image Text:Case 2: Outside the wire
Since, r falls outside the wire, then, all the charge must be enclosed, 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 =
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