Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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Question
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Chapter 4, Problem 17P

(a)

To determine

The electric field at P(5,1,4).

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Surface charge density (ρx) at x=2 is 10μC/m2.

Surface charge density (ρy) at y=3 is 20μC/m2.

Surface charge density (ρz) at z=5 is 30μC/m2.

Calculation:

Calculate the electric field (E5) at point x=5  using the relation.

  E5=ρx2ε0ax

  E5=10μC/m2×106C1μC2(10936πC2/Nm2)ax=(565486.6N/C×103kN/C1N/C)ax=565.486axkN/C

Calculate the electric field (E1) at point y=1  using the relation.

  E1=ρy2ε0ay

  E1=20μC/m2×106C1μC2(10936πC2/Nm2)ay=(1130.97×103N/C×103kN/C1N/C)ay=1130.97aykN/C

Calculate the electric field (E4) at point z=4 using the relation.

  E4=ρz2ε0az

  E4=30μC/m2×106C1μC2(10936πC2/Nm2)(az)=(1696.46×103N/C×103kN/C1N/C)az=1696.46azkN/C

Calculate the electric field (EP) at point P(5,1,4) using the relation.

  EP=E5+E1+E4

  EP=565.486axkN/C1130.97aykN/C1696.46azkN/C

Thus, the electric field at P(5,1,4) is (565.486ax1130.97ay1696.46az)kN/C.

(b)

To determine

The electric field at R(0,2,1).

(b)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

Calculate the electric field (E0) at point x=0  using the relation.

  E0=ρx2ε0(ax)

  E0=10μC/m2×106C1μC2(10936πC2/Nm2)(ax)=(565486.6N/C×103kN/C1N/C)ax=565.486axkN/C

Calculate the electric field (E2) at point y=2  using the relation.

  E2=ρy2ε0(ay)

  E2=20μC/m2×106C1μC2(10936πC2/Nm2)(ay)=(1130.97×103N/C×103kN/C1N/C)ay=1130.97aykN/C

Calculate the electric field (E1) at point z=1  using the relation.

  E1=ρz2ε0(az)

  E1=30μC/m2×106C1μC2(10936πC2/Nm2)(az)=(1696.46×103N/C×103kN/C1N/C)(az)=1696.46azkN/C

Calculate the electric field (ER) at point R(0,2,1) using the relation.

  ER=E5+E1+E4

  ER=565.486axkN/C1130.97aykN/C1696.46azkN/C

Thus, the electric field at R(0,2,1) is (565.486ax1130.97ay1696.46az)kN/C.

(c)

To determine

The electric field at Q(3,4,10).

(c)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

Calculate the electric field (E3) at point x=3  using the relation.

  E3=ρx2ε0ax

  E3=10μC/m2×106C1μC2(10936πC2/Nm2)ax=(565486.6N/C×103kN/C1N/C)ax=565.486axkN/C

Calculate the electric field (E4) at point y=4  using the relation.

  E4=ρy2ε0(ay)

  E4=20μC/m2×106C1μC2(10936πC2/Nm2)(ay)=(1130.97×103N/C×103kN/C1N/C)ay=1130.97aykN/C

Calculate the electric field (E10) at point z=10 using the relation.

  E10=ρz2ε0az

  E10=30μC/m2×106C1μC2(10936πC2/Nm2)az=(1696.46×103N/C×103kN/C1N/C)az=1696.46azkN/C

Calculate the electric field (EQ) at point (3,4,10) using the relation.

  EP=E3+E4+E10

  EP=565.486axkN/C+1130.97aykN/C+1696.46azkN/C

Thus, the electric field at Q(3,4,10) is (565.486ax+1130.97ay+1696.46az)kN/C.

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Chapter 4 Solutions

Elements Of Electromagnetics

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