Elements of Electromagnetics
Elements of Electromagnetics
6th Edition
ISBN: 9780190213879
Author: Sadiku
Publisher: Oxford University Press
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Question
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Chapter 4, Problem 4P

(a)

To determine

The electric field E at (0,0,0).

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The first charge (Q1) is Q.

The second charge (Q2) is Q.

Location of the Q1 charge (r1) is (a,0,0).

Location of the Q2 charge (r2) is (a,0,0).

Location of the E electric field (r) is (0,0,0).

Calculation:

Calculate the location of the electric field (E) using the relation.

r=xi+yj+zk=(0)i+(0)j+(0)k=0

Calculate the location of the Q1 charge in vector form using the relation.

r1=x1i+y1j+z1k=(a)i+(0)j+(0)k=ai

Calculate the location of the Q2 charge in vector form using the relation.

r2=x2i+y2j+z2k=(a)i+(0)j+(0)k=ai

Calculate the electric field (E).using the relation.

  E=Q1(rr1)4πε0|rr1|3+Q2(rr2)4πε0|rr2|3

  E=Q[0(ai)]4πε0|0(ai)|3+(Q)[0(ai)]4πε0|0(ai)|3=Qai4πε0(a3)Qai4πε0(a3)=2Qai4πε0a3=Q2πε0a2ax

Thus, the electric field E at (0,0,0) is Q2πε0a2ax.

(b)

To determine

The electric field E at (0,a,0).

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

Location of the E electric field (r) is (0,a,0).

Calculation:

Calculate the location of the E electric field using the relation.

r=xi+yj+zk=(0)i+(a)j+(0)k=aj

Calculate the electric field (E).using the relation.

  E=Q1(rr1)4πε0|rr1|3+Q2(rr2)4πε0|rr2|3

  E=Q[aj(ai)]4πε0|aj(ai)|3+(Q)[aj(ai)]4πε0|aj(ai)|3=Qa(ji)4πε0[a3(12+(1)2)3]Qa(j+i)4πε0[a3(12+(1)2)3]=Qa(ji)4πε0a322Qa(j+i)4πε0a322=Qaj4πε0a322Qai4πε0a322Qaj4πε0a322Qai4πε0a322

  E=2Qai4πε0a322=Q42πε0a2ax

Thus, the electric field E at (0,a,0) is Q42πε0a2ax.

(c)

To determine

The electric field E at (a,0,a).

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

Location of the E electric field (r) is (a,0,a).

Calculation:

Calculate the location of the E electric field using the relation.

r=xi+yj+zk=(a)i+(0)j+(a)k=ai+ak

Calculate the electric field (E).using the relation.

    E=Q1(rr1)4πε0|rr1|3+Q2(rr2)4πε0|rr2|3

  E=Q[(ai+ak)(ai)]4πε0|(ai+ak)(ai)|3+(Q)[(ai+ak)(ai)]4πε0|(ai+ak)(ai)|3=Qa(k)4πε0(ak)3Qa(2i+k)4πε0[a(2i+k)]3=Qak4πε0a3Qak4πε0a3(22+12)32Qai4πε0a3(22+12)3=Qak4πε0a3Qak20πε0a352Qai20πε0a35

  E=Qak4πε0a3(1155)Qai10πε0a35=Q105πε0a2[i+5.1k]=Q105πε0a2[ax+5.1az]

Thus, the electric field E at (a,0,a) is Q105πε0a2[ax+5.1az].

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Chapter 4 Solutions

Elements of Electromagnetics

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