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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 PE- dÀ = qenc / epsilono 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 / epsilono (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

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 PE- dÀ = qenc / epsilono 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 / epsilono (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

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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 PE - dÃ= qenc / epsilon0 %3D 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 = 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
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 PE - dÃ= qenc / epsilon0 %3D 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 = 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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