Calculate the values of V , E , and D and find the values of density ρ s for ρ values of 5 mm and 15 mm.

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Answer 1

Question

Chapter 6, Problem 14P

To determine

Calculate the values of V, E, and D and find the values of density ρs for ρ values of 5 mm and 15 mm.

Expert Solution & Answer

Answer to Problem 14P

The values of V, E, and D are 36.91 V_, 9.102aρ kV/m_, and 161aρ nC/m2_ respectively. The values of density ρs for ρ values of 5 mm and 15 mm are 322 nC/m2_ and −107.3 nC/m2_ respectively.

Explanation of Solution

Calculation:

Refer to FIGURE 6.31 in the textbook.

Write the expression for the voltage V.

V=Voln(ρa)ln(ba)

V=Voln(aρ)ln(ab) (1)

Write the expression for the electric field.

E=−∇V

Substitute Voln(aρ)ln(ab) for V.

E=−∇(Voln(aρ)ln(ab))=−Voρln(ba)aρ

E=Voρln(ab)aρ (2)

Write the expression for the displacement field vector.

D=εE

Substitute εoεr for ε and −Voρln(ba)aρ for E.

D=εoεr(−Voρln(ba)aρ)

D=−εoεrVoρln(ba)aρ (3)

Write the expression for the expression for density.

ρs=±εoεrVoρln(ab) (4)

Consider that Vo=100 V, b=5 mm, a=15 mm, and εr=2.

Substitute 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (1).

V=100ln(1510)ln(155)=36.91 V

Substitute 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (2).

E=100 V(10 mm)ln(15 mm5 mm)aρ=9.102aρ kV/m

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (3).

D=−(10−936π F⋅m−1)(2)100 V(10 mm)ln(5 mm15 mm)aρ=161aρ×10−9 C/m2 {∵1 C=1 F⋅1 V}=161aρ nC/m2

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 5 mm for ρ, and 5 mm for b in Equation (4) to find the density ρs when ρ=5 mm.

ρs=(10−936π F⋅m−1)(2)100 V(5 mm)ln(15 mm5 mm)=322×10−9 C/m2=322 nC/m2

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 5 mm for ρ, and 15 mm for b in Equation (4) to find the density ρs when ρ=15 mm.

ρs=−(10−936π F⋅m−1)(2)100 V(15 mm)ln(15 mm5 mm)=−107.3×10−9 C/m2=−107.3 nC/m2

Conclusion:

Thus, the values of V, E, and D are 36.91 V_, 9.102aρ kV/m_, and 161aρ nC/m2_ respectively. The values of density ρs for ρ values of 5 mm and 15 mm are 322 nC/m2_ and −107.3 nC/m2_ respectively.

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Answer 2

Question

Chapter 6, Problem 14P

To determine

Calculate the values of V, E, and D and find the values of density ρs for ρ values of 5 mm and 15 mm.

Expert Solution & Answer

Answer to Problem 14P

The values of V, E, and D are 36.91 V_, 9.102aρ kV/m_, and 161aρ nC/m2_ respectively. The values of density ρs for ρ values of 5 mm and 15 mm are 322 nC/m2_ and −107.3 nC/m2_ respectively.

Explanation of Solution

Calculation:

Refer to FIGURE 6.31 in the textbook.

Write the expression for the voltage V.

V=Voln(ρa)ln(ba)

V=Voln(aρ)ln(ab) (1)

Write the expression for the electric field.

E=−∇V

Substitute Voln(aρ)ln(ab) for V.

E=−∇(Voln(aρ)ln(ab))=−Voρln(ba)aρ

E=Voρln(ab)aρ (2)

Write the expression for the displacement field vector.

D=εE

Substitute εoεr for ε and −Voρln(ba)aρ for E.

D=εoεr(−Voρln(ba)aρ)

D=−εoεrVoρln(ba)aρ (3)

Write the expression for the expression for density.

ρs=±εoεrVoρln(ab) (4)

Consider that Vo=100 V, b=5 mm, a=15 mm, and εr=2.

Substitute 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (1).

V=100ln(1510)ln(155)=36.91 V

Substitute 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (2).

E=100 V(10 mm)ln(15 mm5 mm)aρ=9.102aρ kV/m

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (3).

D=−(10−936π F⋅m−1)(2)100 V(10 mm)ln(5 mm15 mm)aρ=161aρ×10−9 C/m2 {∵1 C=1 F⋅1 V}=161aρ nC/m2

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 5 mm for ρ, and 5 mm for b in Equation (4) to find the density ρs when ρ=5 mm.

ρs=(10−936π F⋅m−1)(2)100 V(5 mm)ln(15 mm5 mm)=322×10−9 C/m2=322 nC/m2

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 5 mm for ρ, and 15 mm for b in Equation (4) to find the density ρs when ρ=15 mm.

ρs=−(10−936π F⋅m−1)(2)100 V(15 mm)ln(15 mm5 mm)=−107.3×10−9 C/m2=−107.3 nC/m2

Conclusion:

Thus, the values of V, E, and D are 36.91 V_, 9.102aρ kV/m_, and 161aρ nC/m2_ respectively. The values of density ρs for ρ values of 5 mm and 15 mm are 322 nC/m2_ and −107.3 nC/m2_ respectively.

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Answer 3

Question

Chapter 6, Problem 14P

To determine

Calculate the values of V, E, and D and find the values of density ρs for ρ values of 5 mm and 15 mm.

Expert Solution & Answer

Answer to Problem 14P

The values of V, E, and D are 36.91 V_, 9.102aρ kV/m_, and 161aρ nC/m2_ respectively. The values of density ρs for ρ values of 5 mm and 15 mm are 322 nC/m2_ and −107.3 nC/m2_ respectively.

Explanation of Solution

Calculation:

Refer to FIGURE 6.31 in the textbook.

Write the expression for the voltage V.

V=Voln(ρa)ln(ba)

V=Voln(aρ)ln(ab) (1)

Write the expression for the electric field.

E=−∇V

Substitute Voln(aρ)ln(ab) for V.

E=−∇(Voln(aρ)ln(ab))=−Voρln(ba)aρ

E=Voρln(ab)aρ (2)

Write the expression for the displacement field vector.

D=εE

Substitute εoεr for ε and −Voρln(ba)aρ for E.

D=εoεr(−Voρln(ba)aρ)

D=−εoεrVoρln(ba)aρ (3)

Write the expression for the expression for density.

ρs=±εoεrVoρln(ab) (4)

Consider that Vo=100 V, b=5 mm, a=15 mm, and εr=2.

Substitute 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (1).

V=100ln(1510)ln(155)=36.91 V

Substitute 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (2).

E=100 V(10 mm)ln(15 mm5 mm)aρ=9.102aρ kV/m

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (3).

D=−(10−936π F⋅m−1)(2)100 V(10 mm)ln(5 mm15 mm)aρ=161aρ×10−9 C/m2 {∵1 C=1 F⋅1 V}=161aρ nC/m2

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 5 mm for ρ, and 5 mm for b in Equation (4) to find the density ρs when ρ=5 mm.

ρs=(10−936π F⋅m−1)(2)100 V(5 mm)ln(15 mm5 mm)=322×10−9 C/m2=322 nC/m2

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 5 mm for ρ, and 15 mm for b in Equation (4) to find the density ρs when ρ=15 mm.

ρs=−(10−936π F⋅m−1)(2)100 V(15 mm)ln(15 mm5 mm)=−107.3×10−9 C/m2=−107.3 nC/m2

Conclusion:

Thus, the values of V, E, and D are 36.91 V_, 9.102aρ kV/m_, and 161aρ nC/m2_ respectively. The values of density ρs for ρ values of 5 mm and 15 mm are 322 nC/m2_ and −107.3 nC/m2_ respectively.

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Answer 4

Question

Chapter 6, Problem 14P

To determine

Calculate the values of V, E, and D and find the values of density ρs for ρ values of 5 mm and 15 mm.

Expert Solution & Answer

Answer to Problem 14P

The values of V, E, and D are 36.91 V_, 9.102aρ kV/m_, and 161aρ nC/m2_ respectively. The values of density ρs for ρ values of 5 mm and 15 mm are 322 nC/m2_ and −107.3 nC/m2_ respectively.

Explanation of Solution

Calculation:

Refer to FIGURE 6.31 in the textbook.

Write the expression for the voltage V.

V=Voln(ρa)ln(ba)

V=Voln(aρ)ln(ab) (1)

Write the expression for the electric field.

E=−∇V

Substitute Voln(aρ)ln(ab) for V.

E=−∇(Voln(aρ)ln(ab))=−Voρln(ba)aρ

E=Voρln(ab)aρ (2)

Write the expression for the displacement field vector.

D=εE

Substitute εoεr for ε and −Voρln(ba)aρ for E.

D=εoεr(−Voρln(ba)aρ)

D=−εoεrVoρln(ba)aρ (3)

Write the expression for the expression for density.

ρs=±εoεrVoρln(ab) (4)

Consider that Vo=100 V, b=5 mm, a=15 mm, and εr=2.

Substitute 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (1).

V=100ln(1510)ln(155)=36.91 V

Substitute 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (2).

E=100 V(10 mm)ln(15 mm5 mm)aρ=9.102aρ kV/m

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (3).

D=−(10−936π F⋅m−1)(2)100 V(10 mm)ln(5 mm15 mm)aρ=161aρ×10−9 C/m2 {∵1 C=1 F⋅1 V}=161aρ nC/m2

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 5 mm for ρ, and 5 mm for b in Equation (4) to find the density ρs when ρ=5 mm.

ρs=(10−936π F⋅m−1)(2)100 V(5 mm)ln(15 mm5 mm)=322×10−9 C/m2=322 nC/m2

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 5 mm for ρ, and 15 mm for b in Equation (4) to find the density ρs when ρ=15 mm.

ρs=−(10−936π F⋅m−1)(2)100 V(15 mm)ln(15 mm5 mm)=−107.3×10−9 C/m2=−107.3 nC/m2

Conclusion:

Thus, the values of V, E, and D are 36.91 V_, 9.102aρ kV/m_, and 161aρ nC/m2_ respectively. The values of density ρs for ρ values of 5 mm and 15 mm are 322 nC/m2_ and −107.3 nC/m2_ respectively.

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Answer 5

Chapter 6, Problem 14P

To determine

Calculate the values of V, E, and D and find the values of density ρs for ρ values of 5 mm and 15 mm.

Expert Solution & Answer

Answer to Problem 14P

The values of V, E, and D are 36.91 V_, 9.102aρ kV/m_, and 161aρ nC/m2_ respectively. The values of density ρs for ρ values of 5 mm and 15 mm are 322 nC/m2_ and −107.3 nC/m2_ respectively.

Explanation of Solution

Calculation:

Refer to FIGURE 6.31 in the textbook.

Write the expression for the voltage V.

V=Voln(ρa)ln(ba)

V=Voln(aρ)ln(ab) (1)

Write the expression for the electric field.

E=−∇V

Substitute Voln(aρ)ln(ab) for V.

E=−∇(Voln(aρ)ln(ab))=−Voρln(ba)aρ

E=Voρln(ab)aρ (2)

Write the expression for the displacement field vector.

D=εE

Substitute εoεr for ε and −Voρln(ba)aρ for E.

D=εoεr(−Voρln(ba)aρ)

D=−εoεrVoρln(ba)aρ (3)

Write the expression for the expression for density.

ρs=±εoεrVoρln(ab) (4)

Consider that Vo=100 V, b=5 mm, a=15 mm, and εr=2.

Substitute 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (1).

V=100ln(1510)ln(155)=36.91 V

Substitute 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (2).

E=100 V(10 mm)ln(15 mm5 mm)aρ=9.102aρ kV/m

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (3).

D=−(10−936π F⋅m−1)(2)100 V(10 mm)ln(5 mm15 mm)aρ=161aρ×10−9 C/m2 {∵1 C=1 F⋅1 V}=161aρ nC/m2

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 5 mm for ρ, and 5 mm for b in Equation (4) to find the density ρs when ρ=5 mm.

ρs=(10−936π F⋅m−1)(2)100 V(5 mm)ln(15 mm5 mm)=322×10−9 C/m2=322 nC/m2

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 5 mm for ρ, and 15 mm for b in Equation (4) to find the density ρs when ρ=15 mm.

ρs=−(10−936π F⋅m−1)(2)100 V(15 mm)ln(15 mm5 mm)=−107.3×10−9 C/m2=−107.3 nC/m2

Conclusion:

Thus, the values of V, E, and D are 36.91 V_, 9.102aρ kV/m_, and 161aρ nC/m2_ respectively. The values of density ρs for ρ values of 5 mm and 15 mm are 322 nC/m2_ and −107.3 nC/m2_ respectively.

Answer 6

Chapter 6, Problem 14P

To determine

Calculate the values of V, E, and D and find the values of density ρs for ρ values of 5 mm and 15 mm.

Expert Solution & Answer

Answer to Problem 14P

The values of V, E, and D are 36.91 V_, 9.102aρ kV/m_, and 161aρ nC/m2_ respectively. The values of density ρs for ρ values of 5 mm and 15 mm are 322 nC/m2_ and −107.3 nC/m2_ respectively.

Explanation of Solution

Calculation:

Refer to FIGURE 6.31 in the textbook.

Write the expression for the voltage V.

V=Voln(ρa)ln(ba)

V=Voln(aρ)ln(ab) (1)

Write the expression for the electric field.

E=−∇V

Substitute Voln(aρ)ln(ab) for V.

E=−∇(Voln(aρ)ln(ab))=−Voρln(ba)aρ

E=Voρln(ab)aρ (2)

Write the expression for the displacement field vector.

D=εE

Substitute εoεr for ε and −Voρln(ba)aρ for E.

D=εoεr(−Voρln(ba)aρ)

D=−εoεrVoρln(ba)aρ (3)

Write the expression for the expression for density.

ρs=±εoεrVoρln(ab) (4)

Consider that Vo=100 V, b=5 mm, a=15 mm, and εr=2.

Substitute 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (1).

V=100ln(1510)ln(155)=36.91 V

Substitute 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (2).

E=100 V(10 mm)ln(15 mm5 mm)aρ=9.102aρ kV/m

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (3).

D=−(10−936π F⋅m−1)(2)100 V(10 mm)ln(5 mm15 mm)aρ=161aρ×10−9 C/m2 {∵1 C=1 F⋅1 V}=161aρ nC/m2

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 5 mm for ρ, and 5 mm for b in Equation (4) to find the density ρs when ρ=5 mm.

ρs=(10−936π F⋅m−1)(2)100 V(5 mm)ln(15 mm5 mm)=322×10−9 C/m2=322 nC/m2

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 5 mm for ρ, and 15 mm for b in Equation (4) to find the density ρs when ρ=15 mm.

ρs=−(10−936π F⋅m−1)(2)100 V(15 mm)ln(15 mm5 mm)=−107.3×10−9 C/m2=−107.3 nC/m2

Conclusion:

Thus, the values of V, E, and D are 36.91 V_, 9.102aρ kV/m_, and 161aρ nC/m2_ respectively. The values of density ρs for ρ values of 5 mm and 15 mm are 322 nC/m2_ and −107.3 nC/m2_ respectively.

Answer 7

Chapter 6, Problem 14P

To determine

Calculate the values of V, E, and D and find the values of density ρs for ρ values of 5 mm and 15 mm.

Answer 8

Expert Solution & Answer

Answer to Problem 14P

The values of V, E, and D are 36.91 V_, 9.102aρ kV/m_, and 161aρ nC/m2_ respectively. The values of density ρs for ρ values of 5 mm and 15 mm are 322 nC/m2_ and −107.3 nC/m2_ respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Calculation:

Refer to FIGURE 6.31 in the textbook.

Write the expression for the voltage V.

V=Voln(ρa)ln(ba)

V=Voln(aρ)ln(ab) (1)

Write the expression for the electric field.

E=−∇V

Substitute Voln(aρ)ln(ab) for V.

E=−∇(Voln(aρ)ln(ab))=−Voρln(ba)aρ

E=Voρln(ab)aρ (2)

Write the expression for the displacement field vector.

D=εE

Substitute εoεr for ε and −Voρln(ba)aρ for E.

D=εoεr(−Voρln(ba)aρ)

D=−εoεrVoρln(ba)aρ (3)

Write the expression for the expression for density.

ρs=±εoεrVoρln(ab) (4)

Consider that Vo=100 V, b=5 mm, a=15 mm, and εr=2.

Substitute 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (1).

V=100ln(1510)ln(155)=36.91 V

Substitute 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (2).

E=100 V(10 mm)ln(15 mm5 mm)aρ=9.102aρ kV/m

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 10 mm for ρ, and 5 mm for b in Equation (3).

D=−(10−936π F⋅m−1)(2)100 V(10 mm)ln(5 mm15 mm)aρ=161aρ×10−9 C/m2 {∵1 C=1 F⋅1 V}=161aρ nC/m2

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 5 mm for ρ, and 5 mm for b in Equation (4) to find the density ρs when ρ=5 mm.

ρs=(10−936π F⋅m−1)(2)100 V(5 mm)ln(15 mm5 mm)=322×10−9 C/m2=322 nC/m2

Substitute 2 for εr, 10−936π F⋅m−1 for εo, 100 V for Vo, 15 mm for a, 5 mm for ρ, and 15 mm for b in Equation (4) to find the density ρs when ρ=15 mm.

ρs=−(10−936π F⋅m−1)(2)100 V(15 mm)ln(15 mm5 mm)=−107.3×10−9 C/m2=−107.3 nC/m2

Conclusion:

Thus, the values of V, E, and D are 36.91 V_, 9.102aρ kV/m_, and 161aρ nC/m2_ respectively. The values of density ρs for ρ values of 5 mm and 15 mm are 322 nC/m2_ and −107.3 nC/m2_ respectively.