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 52P

(a)

To determine

The electric field intensity and the volume charge density.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The potential distribution (V) is (2x2+4y2)V.

Calculation:

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

  E=V

  E=(xax+yay+zaz)((2x2+4y2)V)=((2x2+4y2V)xax+(2x2+4y2V)yay+(2x2+4y2V)zaz)=((4x+0)ax+(0+8y)ay+(0)az)V/m=4xax8yayV/m

Take the permeability of free space (ε0) as 8.854×1012F/m.

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

  ρv=ε0E

  ρv=(8.854×1012F/m)(xax+yay+zaz)(4xax8yayV/m)=(8.854×1012F/m)((4xV/m)x+(8yV/m)y+(0V/m)z)=(8.854×1012F/m)(4V/m28V/m2+0V/m2)=(12V/m2)(8.854×1012F/m)

  ρv=1.06248×1010C/m3

Thus, the electric field intensity is 4xax8yayV/m_ and the volume charge density is 1.06248×1010C/m3_.

(b)

To determine

The electric field intensity and the volume charge density.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The potential distribution (V) is (10ρ2sinϕ+6ρz)V.

Calculation:

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

  E=V

  E=(ρaρ+1ρϕaϕ+zaz)((10ρ2sinϕ+6ρz)V)=((10ρ2sinϕ+6ρzV)ρaρ+1ρ(10ρ2sinϕ+6ρzV)ϕaϕ+(10ρ2sinϕ+6ρzV)zaz)=((20ρsinϕ+6z)aρ+1ρ(10ρ2cosϕ+0)aϕ+(0+6ρ)az)V/m=(20ρsinϕ+6z)aρ10ρcosϕaϕ6ρazV/m

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

  ρv=ε0E

  ρv=ε0[(1ρρρaρ+1ρϕaϕ+zaz)((20ρsinϕ+6z)aρ10ρcosϕaϕ6ρazV/m)]=ε0(1ρ(ρ(20ρsinϕ+6z)V/m)ρ+1ρ(10ρcosϕV/m)ϕ+(6ρV/m)z)=ε0(1ρ(40ρsinϕ6zV/m2)+1ρ(10ρsinϕV/m2)+0)=ε0(40sinϕ6zρ+10sinϕV/m2)

  ρv=(30ε0sinϕ6zρε0)C/m3

Thus, the electric field intensity is (20ρsinϕ+6z)aρ10ρcosϕaϕ6ρazV/m_ and the volume charge density is (30ε0sinϕ6zρε0)C/m3_.

(c)

To determine

The electric field intensity and the volume charge density.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The potential distribution (V) is (5r2cosθsinϕ)V.

Calculation:

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

  E=V

  E=(rar+1rθaθ+1rsinθϕaϕ)((5r2cosθsinϕ)V)=((5r2cosθsinϕV)rar+1r(5r2cosθsinϕV)θaθ+1rsinθ(5r2cosθsinϕV)ϕaϕ)=((10rcosθsinϕ)ar+1r(5r2sinθsinϕ)aθ+1rsinθ(5r2cosθcosϕ)aϕ)V/m=10rcosθsinϕar+5rsinθsinϕaθ5rcotθcosϕaϕV/m

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

  ρv=ε0E

  ρv=ε0[(1r2rr2ar+1rsinθθsinθaθ+1rsinθϕaϕ)(10rcosθsinϕar+5rsinθsinϕaθ5rcotθcosϕaϕV/m)]=ε0(1r2(10r3cosθsinϕV/m)r+1rsinθ(5rsin2θsinϕV/m)θ+1rsinθ(5rcotθcosϕV/m)ϕ)=ε0(1r2(30r2cosθsinϕV/m2)+1rsinθ(5r(2sinθcosθ)sinϕV/m2)+1rsinθ(5rcotθsinϕV/m2))=ε0(30cosθsinϕ+10cosθsinϕ+5cosθsinϕsin2θV/m2)

  ρv=(20cosθsinϕε0+5cosθsinϕsin2θε0)C/m3

Thus, the electric field intensity is 10rcosθsinϕar+5rsinθsinϕaθ5rcotθcosϕaϕV/m_ and the volume charge density is (20cosθsinϕε0+5cosθsinϕsin2θε0)C/m3_.

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Chapter 4 Solutions

Elements Of Electromagnetics

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