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 On the other hand, the Gaussian surface outside the wire is given by Using Equation 1, the electric field in simplified form is E-
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 On the other hand, the Gaussian surface outside the wire is given by Using Equation 1, the electric field in simplified form is E-
Physics for Scientists and Engineers
10th Edition
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter23: Continuous Charge Distributions And Gauss's Law
Section: Chapter Questions
Problem 23P: Figure P23.23 represents the top view of a cubic gaussian surface in a uniform electric field E...
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Question
Please fill in the blanks for both questions.
![QUESTION 3
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
fE - dÃ=
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
qenc =
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%2F23bafcf0-6961-469b-aaf0-450fb2c2d54f%2F5df9b266-934c-4d6b-8137-b24e3259da0f%2F60xo0kh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:QUESTION 3
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
fE - dÃ=
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
qenc =
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 =
![QUESTION 4
Consider a cube with edge length L immersed in a uniform electric field E along the x-direction as shown in the figure below. Calculate the electric flux through the closed
surface.
L
We need to solve for the electric flux through each face of the cube and take the sum to calculate the net electric flux:
Onet =01 + 02 + 03 + @xy + 0xz+ ®yz
The flux through surface 1 is: 01 = 0.
%3D
The flux through surface 2 is: 02 =
The flux through surface 3 is: 03 =
The flux through the surface on the xy plane is:
ху
The flux through the surface on the xz plane is: Oxz*
The flux through the surface on the yz plane is:
Dyz'
Taking the algebraic sum, the net electric flux through the cube is: Onet
111](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F23bafcf0-6961-469b-aaf0-450fb2c2d54f%2F5df9b266-934c-4d6b-8137-b24e3259da0f%2F28cs5xy_processed.jpeg&w=3840&q=75)
Transcribed Image Text:QUESTION 4
Consider a cube with edge length L immersed in a uniform electric field E along the x-direction as shown in the figure below. Calculate the electric flux through the closed
surface.
L
We need to solve for the electric flux through each face of the cube and take the sum to calculate the net electric flux:
Onet =01 + 02 + 03 + @xy + 0xz+ ®yz
The flux through surface 1 is: 01 = 0.
%3D
The flux through surface 2 is: 02 =
The flux through surface 3 is: 03 =
The flux through the surface on the xy plane is:
ху
The flux through the surface on the xz plane is: Oxz*
The flux through the surface on the yz plane is:
Dyz'
Taking the algebraic sum, the net electric flux through the cube is: Onet
111
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