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 DE dÃ= Q / piR2L 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 = qenc / epsilono (Equation 1) Case 1: Inside the wire Since, r falls inside the wire, then all the enclosed charge must be: qenc = Qr2/R2 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= Qr / 2piR2Lepsilono 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 =
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 DE dÃ= Q / piR2L 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 = qenc / epsilono (Equation 1) Case 1: Inside the wire Since, r falls inside the wire, then all the enclosed charge must be: qenc = Qr2/R2 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= Qr / 2piR2Lepsilono 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 =
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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
DE - dÃ= Q
piR2L
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 =
qenc
/ epsilon0
(Equation 1)
Case 1: Inside the wire
Since, r falls inside the wire, then all the enclosed charge must be:
qenc = Qr2/R2
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 = Qr
/ 2piR2Lepsilono
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 =
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