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) Use the attached pictures as basis for the solution

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Chapter6: Gauss's Law
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Problem 86AP: Two non-conducting spheres of radii R1 and R2 are uniformly charged with charge densities p1 and p2...
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An electric charge Q is distributed uniformly along a thick and enormously long conducting wire with radius 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)

Use the attached pictures as basis for the solution

 

PE dA=
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
A =
Using Equation 1, the electric field in simplified form is
Transcribed Image Text:PE dA= 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 A = Using Equation 1, the electric field in simplified form is
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
A =
Using Equation 1, the electric field in simplified form is
Transcribed Image Text: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 A = Using Equation 1, the electric field in simplified form is
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