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 DE - dÀ= qenc / epsilon0 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 = Q / epsilon0 (Equation 1) Case 1: Inside the wire Since, r falls inside the wire, then all the enclosed charge must be: denc = 0 On the other hand, the Gaussian surface inside the wire is given by A = 2pirL Using Equation 1, the electric field in simplified form is E = 0 / 2piRLepsilono Case 2: Outside the wire Since, r falls outside the wire, then, all the charge must be enclosed, thus denc = Q On the other hand, the Gaussian surface outside the wire is given by A = 2pirL Using Equation 1, the electric field in simplified form is E = Q / 2piRLepsilon0
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 DE - dÀ= qenc / epsilon0 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 = Q / epsilon0 (Equation 1) Case 1: Inside the wire Since, r falls inside the wire, then all the enclosed charge must be: denc = 0 On the other hand, the Gaussian surface inside the wire is given by A = 2pirL Using Equation 1, the electric field in simplified form is E = 0 / 2piRLepsilono Case 2: Outside the wire Since, r falls outside the wire, then, all the charge must be enclosed, thus denc = Q On the other hand, the Gaussian surface outside the wire is given by A = 2pirL Using Equation 1, the electric field in simplified form is E = Q / 2piRLepsilon0
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Transcribed Image Text: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= qenc
/ epsilon0
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 = Q
/ epsilon0
(Equation 1)
Case 1: Inside the wire
Since, r falls inside the wire, then all the enclosed charge must be:
denc = 0
On the other hand, the Gaussian surface inside the wire is given by
A = 2pirL
Using Equation 1, the electric field in simplified form is
E = 0
( 2piRLepsilon0
Case 2: Outside the wire
Since, r falls outside the wire, then, all the charge must be enclosed, thus
denc
Q
On the other hand, the Gaussian surface outside the wire is given by
A = 2pirl
Using Equation 1, the electric field in simplified form is
E = Q
/ 2piRLepsilon0
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