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 48P

(a)

To determine

The energy expended in transferring 10nC charge from A(5,30°,0°) to B(5,90°,0°).

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The electric field intensity (E) is 20rsinθar+10rcosθaθV/m.

The first point (A) is (5,30°,0°).

The second point (B) is (5,90°,0°).

The magnitude of the charge (Q) is 10nC.

Calculation:

The energy expended to transfer the charge is equal to the work done to move the charge.

Calculate the work done (W) using the relation.

  W=QLEdl

  W=(10nC)A(5,30°,0°)B(5,90°,0°)(20rsinθar+10rcosθaθV/m)(drar+rdθaθ+rsinθdϕaϕ)W=(10nC)θ=3090[10r2cosθdθ]r=5,ϕ=0°W=(10nC)(10×52)[sinθ]30°90°W=1250nJ

Thus, the energy expended to transfer 10nC charge from A(5,30°,0°) to B(5,90°,0°) is 1250nJ_.

(b)

To determine

The energy expended in transferring 10nC charge from A(5,30°,0°) to C(10,30°,0°).

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The electric field intensity (E) is 12ρzcosϕaρ6ρzaϕ+6ρ2cosϕaz.

The first point (A) is (5,30°,0°)

The second point (C) is (10,30°,0°).

The magnitude of the charge (Q) is 10nC.

Calculation:

Calculate the work done (W) using the relation.

  W=QLEdl

  W=(10nC)A(5,30°,0°)B(10,30°,0°)(20rsinθar+10rcosθaθV/m)(drar+rdθaθ+rsinθdϕaϕ)W=(10nC)r=510[20rsinθdr]θ=30°,ϕ=0°W=(10nC)(20×sin(30°))[r22]510W=(10nC)(20×sin(30°))[1022522]

  W=3750nJ

Thus, the energy expended to transfer 10nC charge from A(5,30°,0°) to C(10,30°,0°) is 3750nJ_.

(c)

To determine

The energy expended in transferring 10nC charge from A(5,30°,0°) to D(5,30°,60°).

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The electric field intensity (E) is 12ρzcosϕaρ6ρzaϕ+6ρ2cosϕaz.

The first point (A) is (5,30°,0°) and the second point (D) is (5,30°,60°).

Calculation:

Calculate the work done (W) using the relation.

  W=QLEdl

  W=(10nC)A(5,30°,0°)B(5,30°,60°)(20rsinθar+10rcosθaθV/m)(drar+rdθaθ+rsinθdϕaϕ)W=(10nC)ϕ=0°60°[0dϕ]r=5,θ=30°W=0nJ

Thus, the energy expended to transfer 10nC charge from A(5,30°,0°) to D(5,30°,60°) is 0nJ_.

(d)

To determine

The energy expended in transferring 10nC charge from A(5,30°,0°) to E(10,90°,60°).

(d)

Expert Solution
Check Mark

Explanation of Solution

Given:

The electric field intensity (E) is 12ρzcosϕaρ6ρzaϕ+6ρ2cosϕaz.

The first point (A) is (5,30°,0°)

The second point (E) is (10,90°,60°).

The magnitude of the charge (Q) is 10nC.

Calculation:

Since the electrostatic force is conservative, the path of integration is immaterial.

Suppose the charge follow the path to from point A(5,30°,0°) to B(5,90°,0°) and then B(5,90°,0°) to B(10,90°,0°) and then B(10,90°,0°) to E(10,90°,60°).

Calculate the work done (W) by using the relation.

  W=QLEdl

  W=(10nC)A(5,30°,0°)E(10,90°,60°)(20rsinθar+10rcosθaθV/m)(drar+rdθaθ+rsinθdϕaϕ)W=(10nC)θ=30°90°r=510ϕ=0°60°(20rsinθar+10rcosθaθV/m)(drar+rdθaθ+rsinθdϕaϕ)W=(10nC)[[θ=30°90°[10r2cosθdθ]r=5,ϕ=0°]+[r=5r=10[20rsinθdr]θ=90°,ϕ=0°]+[ϕ=0°60°0dϕ]]W=(10nC)[(10×52)[sinθ]30°90°+20sin(90°)[r22]510+0]

  W=(10nC)[(10×52)(10.5)+20sin(90°)[1022522]]W=8750nJ

Thus, the energy expended to transfer 10nC charge from A(5,30°,0°) to E(10,90°,60°) is 8750nJ_.

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Chapter 4 Solutions

Elements Of Electromagnetics

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