Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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Chapter 4, Problem 57P
To determine

Whether the electric flux density is a genuine electric flux density and the magnitude of flux crossing the given plane.

Expert Solution & Answer
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Explanation of Solution

Given:

The electric flux density (D) is (2ρsinϕaρcosϕ2ρaϕ)C/m2.

The limit of the radial axis (ρ) is ρ=1.

The limit of the angular axis (ϕ) is 0ϕπ/4.

The limit of axial axis (z) is 0<z<1.

Calculation:

Calculate the curl of field (×D) using the relation.

  ×D=1ρ|aρρaϕazρϕzDρρDϕDz|

  ×D=1ρ|aρρaϕazρϕz2ρsinϕρ(cosϕ2ρ)0|=1ρ[aρ(ϕ(0)z(cosϕ2))ρaϕ(ρ(0)z(2ρsinϕ))+az(ρ(cosϕ2)ϕ(2ρsinϕ))]=1ρ[aρ(00)ρaϕ(00)+az(02ρcosϕ)]=2ρcosϕρaz

  ×D=2cosϕaz0

The curl of the field is not equal to zero. Thus, the field is not genuine.

Calculate the volume charge density (ρv) using the relation.

  ρv=D

  ρv=1ρρ(ρDρ)+1ρϕ(Dϕ)+z(Dz)=1ρρ(2ρ2sinϕC/m2)+1ρϕ(cosϕ2ρC/m2)+z(0C/m2)=1ρ(4ρsinϕC/m3)+1ρ((sinϕ)2ρC/m3)+0=4sinϕ+sinϕ2ρ2C/m3

Calculate the flux (Ψ) using the relation.

  Ψ=vρvdv

  Ψ=010π401(4sinϕ+sinϕ2ρ2C/m3)ρdρdϕdz=010π4(4ρsinϕ+sinϕ2ρC/m2)dρdϕ[z]01=[10]010π4(4ρsinϕ+sinϕ2ρC/m2)dρdϕ=010π4(4ρsinϕ+sinϕ2ρC/m2)dρdϕ

  Ψ=01[4ρ(cosϕ)+cosϕ2ρC/m]0π4dρ=01[4ρ(cos(π4))+cos(π4)2ρ(4ρ(cos(0))+cos(0)2ρ)C/m]dρ=01[4ρ(22)+222ρ(4ρ(1)+12ρ)C/m]dρ=01[22ρ24ρ+4ρ+12ρC/m]dρ

  Ψ=[22(ρ22)24(lnρ)+4(ρ22)+12(lnρ)C]01={[22(122)24(ln1)+4(122)+12(ln1)C][22(022)24(ln0)+4(022)+12(ln0)C]=[(224(0)+2+12(0))(00+0+0)]C=22C

  Ψ=21.4142C=0.58578C

Thus, the given electric flux density (D) is not genuine_ and the magnitude of the flux crossing the given surface is 0.58578C_.

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Chapter 4 Solutions

Elements Of Electromagnetics

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