Explanation Given: The radius of inner conductor in coaxial cable is a and outer conductor goes from b to c . A charge of + Q is distributed along one meter length of inner conductor and charge of − Q is distributed along one meter length of outer conductor. Concept used: Gauss’s law states that the net electric flux that passes through the enclosed surface is equal to the total charge enclosed by the surface. Q enc = ∮ D ⋅ d S ......(1) Here, D is the electric flux density. Calculation: Consider that Gaussian surface is of length L . By the symmetry, the electric flux density is a function of ρ only and the components D ϕ and D z cancels by symmetry. There are only four regions especially: ρ ≤ a , a ≤ ρ ≤ b , b ≤ ρ ≤ c and ρ > c . Consider cylinders that are centered through z-axis, height h and radii as shown in cross-sectional view of coaxial cable in Figure 1. Through the top and bottom surfaces there is no flux. The flux through the Gaussian surface is, Q enc = ∮ D ⋅ d S = ∫ D ρ a ρ ⋅ ρ d ϕ d z a ρ = D ρ ρ ∫ 0 2 π d ϕ ∫ 0 L d z = 2 πρ L D ρ The flux is valid for all Gaussian surfaces. The charge for the region ρ < a is shown below. Q enc1 = ∫ ρ v d V = ρ v ∫ 0 ρ ρ d ρ ∫ 0 2 π d ϕ ∫ 0 L d z = 2 π L ρ v ⋅ ρ 2 2 | 0 ρ = πρ v L ρ 2 The volume charge density ρ v is, ρ v = Q V = Q π a 2 l = Q π a 2 Substitute Q π a 2 for ρ v in the expression for Q enc1 . Q enc1 = π ( Q π a 2 ) L ρ 2 = Q L ρ 2 a 2 Now, the flux density for the region ρ < a is, Q enc1 = 2 πρ L D ρ1 Q L ρ 2 a 2 = 2 πρ L D ρ1 D ρ1 = Q ρ 2 π a 2 Therefore, the electric flux density for the region ρ < a is Q ρ 2 π a 2 . The charge for the region a ≤ ρ ≤ b is shown below. Q enc2 = ∫ ρ v d V = ρ v ∫ 0 a ρ d ρ ∫ 0 2 π d ϕ ∫ 0 L d z = 2 π L ρ v ⋅ a 2 2 | 0 a = πρ v L a 2 The volume charge density ρ v is, ρ v = Q V = Q π a 2 l = Q π a 2 Substitute Q π a 2 for ρ v in the expression for Q enc2 . Q enc2 = π ( Q π a 2 ) L a 2 = Q L a 2 a 2 = Q L Now, the flux density for the region a ≤ ρ ≤ b is, Q enc2 = 2 πρ L D ρ2 Q L = 2 πρ L D ρ2 D ρ2 = Q 2 πρ Therefore, the electric flux density for the region a ≤ ρ ≤ b is Q 2 πρ

Fundamentals of Electromagnetics with Engineering Applications

1st Edition

ISBN: 9780470105757

Author: Stuart M. Wentworth

Publisher: Wiley, John & Sons, Incorporated

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Chapter 2, Problem 2.37P

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The equations of electric flux density for all ρ .

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Chapter 2, Problem 2.36P

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Chapter 2 Solutions

Fundamentals of Electromagnetics with Engineering Applications

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The equations of electric flux density for all ρ .

Solution Summary: The author calculates the electric flux density for all by Gauss's law, which states that the net electrical flux passes through the enclosed surface is equal to the total charge enclosed by the surface.

The equations of electric flux density for all ρ .

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Chapter 2 Solutions