Two infinitely long, parallel lines of charge with linear charge densities 3.2 μ C/m and −3.2 μ C/m are separated by a distance of 0.50 m. What is the net electric field at points A , B , and C as shown in Figure P25.35? FIGURE P25.35

Question

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Answer 1

Textbook Question

Chapter 25, Problem 35PQ

Two infinitely long, parallel lines of charge with linear charge densities 3.2 μC/m and −3.2 μC/m are separated by a distance of 0.50 m. What is the net electric field at points A, B, and C as shown in Figure P25.35?

Chapter 25, Problem 35PQ, Two infinitely long, parallel lines of charge with linear charge densities 3.2 C/m and 3.2 C/m are

FIGURE P25.35

Expert Solution & Answer

To determine

The net electric field at points A, B and C.

Answer to Problem 35PQ

The net electric field at points A, B and C are −2.1×105i^ N/C_, 4.6×105i^ N/C_ and −4.8×105i^ N/C_ respectively.

Explanation of Solution

Write the expression for finding the electric field due to first line charge.

E→1=12πε0λ1r1x^ (I)

Here, λ1 is the line charge density of the first line and r1 is the distance between the line charge and the point.

Write the expression for finding the electric field due to second line charge.

E→2=12πε0λ2r2x^ (II)

Here, λ1 is the line charge density of the second line and r2 is the distance between the line charge and the point.

Write the expression for the net electric field.

E→=E→1+E→2 (III)

Here, E→ is the net charge at any point.

Conclusion:

The following figure gives the fields acting on the point A due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 1

Substitute 3.2×10−6 C/m for λ1, 0.70 m for r1, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.70 mi^=8.2×104i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.20 m for r2, −i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.20 m(−i^)=−2.9×105i^ N/C

Substitute 8.2×104i^ N/C for E→1 and −2.9×105i^ N/C for E→2 in equation (III).

E→A=8.2×104i^ N/C+(−2.9×105i^ N/C)=−2.1×105i^ N/C

The following figure gives the fields acting on the point B due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 2

Substitute 3.2×10−6 C/m for λ1, 0.25 m for r1, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.25 mi^=2.3×105i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.25 m for r2, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.25 m(i^)=2.3×105i^ N/C

Substitute 2.3×105i^ N/C for E→1 and 2.3×105i^ N/C for E→2 in equation (III).

E→B=2.3×105i^ N/C+2.3×105i^ N/C=4.6×105i^ N/C

The following figure gives the fields acting on the point C due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 3

Substitute 3.2×10−6 C/m for λ1, 0.10 m for r1, −i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.10 m(−i^)=−5.8×105i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.60 m for r2, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.60 m(i^)=9.6×104i^ N/C

Substitute −5.8×105i^ N/C for E→1 and 9.6×104i^ N/C for E→2 in equation (III).

E→C=−5.8×105i^ N/C+9.6×104i^ N/C=−4.8×105i^ N/C

Therefore, the net electric field at points A, B and C are −2.1×105i^ N/C_, 4.6×105i^ N/C_ and −4.8×105i^ N/C_ respectively.

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Answer 2

Textbook Question

Chapter 25, Problem 35PQ

Two infinitely long, parallel lines of charge with linear charge densities 3.2 μC/m and −3.2 μC/m are separated by a distance of 0.50 m. What is the net electric field at points A, B, and C as shown in Figure P25.35?

Chapter 25, Problem 35PQ, Two infinitely long, parallel lines of charge with linear charge densities 3.2 C/m and 3.2 C/m are

FIGURE P25.35

Expert Solution & Answer

To determine

The net electric field at points A, B and C.

Answer to Problem 35PQ

The net electric field at points A, B and C are −2.1×105i^ N/C_, 4.6×105i^ N/C_ and −4.8×105i^ N/C_ respectively.

Explanation of Solution

Write the expression for finding the electric field due to first line charge.

E→1=12πε0λ1r1x^ (I)

Here, λ1 is the line charge density of the first line and r1 is the distance between the line charge and the point.

Write the expression for finding the electric field due to second line charge.

E→2=12πε0λ2r2x^ (II)

Here, λ1 is the line charge density of the second line and r2 is the distance between the line charge and the point.

Write the expression for the net electric field.

E→=E→1+E→2 (III)

Here, E→ is the net charge at any point.

Conclusion:

The following figure gives the fields acting on the point A due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 1

Substitute 3.2×10−6 C/m for λ1, 0.70 m for r1, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.70 mi^=8.2×104i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.20 m for r2, −i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.20 m(−i^)=−2.9×105i^ N/C

Substitute 8.2×104i^ N/C for E→1 and −2.9×105i^ N/C for E→2 in equation (III).

E→A=8.2×104i^ N/C+(−2.9×105i^ N/C)=−2.1×105i^ N/C

The following figure gives the fields acting on the point B due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 2

Substitute 3.2×10−6 C/m for λ1, 0.25 m for r1, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.25 mi^=2.3×105i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.25 m for r2, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.25 m(i^)=2.3×105i^ N/C

Substitute 2.3×105i^ N/C for E→1 and 2.3×105i^ N/C for E→2 in equation (III).

E→B=2.3×105i^ N/C+2.3×105i^ N/C=4.6×105i^ N/C

The following figure gives the fields acting on the point C due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 3

Substitute 3.2×10−6 C/m for λ1, 0.10 m for r1, −i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.10 m(−i^)=−5.8×105i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.60 m for r2, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.60 m(i^)=9.6×104i^ N/C

Substitute −5.8×105i^ N/C for E→1 and 9.6×104i^ N/C for E→2 in equation (III).

E→C=−5.8×105i^ N/C+9.6×104i^ N/C=−4.8×105i^ N/C

Therefore, the net electric field at points A, B and C are −2.1×105i^ N/C_, 4.6×105i^ N/C_ and −4.8×105i^ N/C_ respectively.

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Answer 3

Textbook Question

Chapter 25, Problem 35PQ

Two infinitely long, parallel lines of charge with linear charge densities 3.2 μC/m and −3.2 μC/m are separated by a distance of 0.50 m. What is the net electric field at points A, B, and C as shown in Figure P25.35?

Chapter 25, Problem 35PQ, Two infinitely long, parallel lines of charge with linear charge densities 3.2 C/m and 3.2 C/m are

FIGURE P25.35

Expert Solution & Answer

To determine

The net electric field at points A, B and C.

Answer to Problem 35PQ

The net electric field at points A, B and C are −2.1×105i^ N/C_, 4.6×105i^ N/C_ and −4.8×105i^ N/C_ respectively.

Explanation of Solution

Write the expression for finding the electric field due to first line charge.

E→1=12πε0λ1r1x^ (I)

Here, λ1 is the line charge density of the first line and r1 is the distance between the line charge and the point.

Write the expression for finding the electric field due to second line charge.

E→2=12πε0λ2r2x^ (II)

Here, λ1 is the line charge density of the second line and r2 is the distance between the line charge and the point.

Write the expression for the net electric field.

E→=E→1+E→2 (III)

Here, E→ is the net charge at any point.

Conclusion:

The following figure gives the fields acting on the point A due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 1

Substitute 3.2×10−6 C/m for λ1, 0.70 m for r1, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.70 mi^=8.2×104i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.20 m for r2, −i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.20 m(−i^)=−2.9×105i^ N/C

Substitute 8.2×104i^ N/C for E→1 and −2.9×105i^ N/C for E→2 in equation (III).

E→A=8.2×104i^ N/C+(−2.9×105i^ N/C)=−2.1×105i^ N/C

The following figure gives the fields acting on the point B due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 2

Substitute 3.2×10−6 C/m for λ1, 0.25 m for r1, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.25 mi^=2.3×105i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.25 m for r2, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.25 m(i^)=2.3×105i^ N/C

Substitute 2.3×105i^ N/C for E→1 and 2.3×105i^ N/C for E→2 in equation (III).

E→B=2.3×105i^ N/C+2.3×105i^ N/C=4.6×105i^ N/C

The following figure gives the fields acting on the point C due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 3

Substitute 3.2×10−6 C/m for λ1, 0.10 m for r1, −i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.10 m(−i^)=−5.8×105i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.60 m for r2, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.60 m(i^)=9.6×104i^ N/C

Substitute −5.8×105i^ N/C for E→1 and 9.6×104i^ N/C for E→2 in equation (III).

E→C=−5.8×105i^ N/C+9.6×104i^ N/C=−4.8×105i^ N/C

Therefore, the net electric field at points A, B and C are −2.1×105i^ N/C_, 4.6×105i^ N/C_ and −4.8×105i^ N/C_ respectively.

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Answer 4

Textbook Question

Chapter 25, Problem 35PQ

Two infinitely long, parallel lines of charge with linear charge densities 3.2 μC/m and −3.2 μC/m are separated by a distance of 0.50 m. What is the net electric field at points A, B, and C as shown in Figure P25.35?

Chapter 25, Problem 35PQ, Two infinitely long, parallel lines of charge with linear charge densities 3.2 C/m and 3.2 C/m are

FIGURE P25.35

Expert Solution & Answer

To determine

The net electric field at points A, B and C.

Answer to Problem 35PQ

The net electric field at points A, B and C are −2.1×105i^ N/C_, 4.6×105i^ N/C_ and −4.8×105i^ N/C_ respectively.

Explanation of Solution

Write the expression for finding the electric field due to first line charge.

E→1=12πε0λ1r1x^ (I)

Here, λ1 is the line charge density of the first line and r1 is the distance between the line charge and the point.

Write the expression for finding the electric field due to second line charge.

E→2=12πε0λ2r2x^ (II)

Here, λ1 is the line charge density of the second line and r2 is the distance between the line charge and the point.

Write the expression for the net electric field.

E→=E→1+E→2 (III)

Here, E→ is the net charge at any point.

Conclusion:

The following figure gives the fields acting on the point A due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 1

Substitute 3.2×10−6 C/m for λ1, 0.70 m for r1, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.70 mi^=8.2×104i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.20 m for r2, −i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.20 m(−i^)=−2.9×105i^ N/C

Substitute 8.2×104i^ N/C for E→1 and −2.9×105i^ N/C for E→2 in equation (III).

E→A=8.2×104i^ N/C+(−2.9×105i^ N/C)=−2.1×105i^ N/C

The following figure gives the fields acting on the point B due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 2

Substitute 3.2×10−6 C/m for λ1, 0.25 m for r1, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.25 mi^=2.3×105i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.25 m for r2, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.25 m(i^)=2.3×105i^ N/C

Substitute 2.3×105i^ N/C for E→1 and 2.3×105i^ N/C for E→2 in equation (III).

E→B=2.3×105i^ N/C+2.3×105i^ N/C=4.6×105i^ N/C

The following figure gives the fields acting on the point C due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 3

Substitute 3.2×10−6 C/m for λ1, 0.10 m for r1, −i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.10 m(−i^)=−5.8×105i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.60 m for r2, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.60 m(i^)=9.6×104i^ N/C

Substitute −5.8×105i^ N/C for E→1 and 9.6×104i^ N/C for E→2 in equation (III).

E→C=−5.8×105i^ N/C+9.6×104i^ N/C=−4.8×105i^ N/C

Therefore, the net electric field at points A, B and C are −2.1×105i^ N/C_, 4.6×105i^ N/C_ and −4.8×105i^ N/C_ respectively.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The net electric field at points A, B and C.

Answer to Problem 35PQ

The net electric field at points A, B and C are −2.1×105i^ N/C_, 4.6×105i^ N/C_ and −4.8×105i^ N/C_ respectively.

Explanation of Solution

Write the expression for finding the electric field due to first line charge.

E→1=12πε0λ1r1x^ (I)

Here, λ1 is the line charge density of the first line and r1 is the distance between the line charge and the point.

Write the expression for finding the electric field due to second line charge.

E→2=12πε0λ2r2x^ (II)

Here, λ1 is the line charge density of the second line and r2 is the distance between the line charge and the point.

Write the expression for the net electric field.

E→=E→1+E→2 (III)

Here, E→ is the net charge at any point.

Conclusion:

The following figure gives the fields acting on the point A due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 1

Substitute 3.2×10−6 C/m for λ1, 0.70 m for r1, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.70 mi^=8.2×104i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.20 m for r2, −i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.20 m(−i^)=−2.9×105i^ N/C

Substitute 8.2×104i^ N/C for E→1 and −2.9×105i^ N/C for E→2 in equation (III).

E→A=8.2×104i^ N/C+(−2.9×105i^ N/C)=−2.1×105i^ N/C

The following figure gives the fields acting on the point B due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 2

Substitute 3.2×10−6 C/m for λ1, 0.25 m for r1, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.25 mi^=2.3×105i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.25 m for r2, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.25 m(i^)=2.3×105i^ N/C

Substitute 2.3×105i^ N/C for E→1 and 2.3×105i^ N/C for E→2 in equation (III).

E→B=2.3×105i^ N/C+2.3×105i^ N/C=4.6×105i^ N/C

The following figure gives the fields acting on the point C due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 3

Substitute 3.2×10−6 C/m for λ1, 0.10 m for r1, −i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.10 m(−i^)=−5.8×105i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.60 m for r2, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.60 m(i^)=9.6×104i^ N/C

Substitute −5.8×105i^ N/C for E→1 and 9.6×104i^ N/C for E→2 in equation (III).

E→C=−5.8×105i^ N/C+9.6×104i^ N/C=−4.8×105i^ N/C

Therefore, the net electric field at points A, B and C are −2.1×105i^ N/C_, 4.6×105i^ N/C_ and −4.8×105i^ N/C_ respectively.

Answer 8

Expert Solution & Answer

To determine

The net electric field at points A, B and C.

Answer to Problem 35PQ

The net electric field at points A, B and C are −2.1×105i^ N/C_, 4.6×105i^ N/C_ and −4.8×105i^ N/C_ respectively.

Explanation of Solution

Write the expression for finding the electric field due to first line charge.

E→1=12πε0λ1r1x^ (I)

Here, λ1 is the line charge density of the first line and r1 is the distance between the line charge and the point.

Write the expression for finding the electric field due to second line charge.

E→2=12πε0λ2r2x^ (II)

Here, λ1 is the line charge density of the second line and r2 is the distance between the line charge and the point.

Write the expression for the net electric field.

E→=E→1+E→2 (III)

Here, E→ is the net charge at any point.

Conclusion:

The following figure gives the fields acting on the point A due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 1

Substitute 3.2×10−6 C/m for λ1, 0.70 m for r1, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.70 mi^=8.2×104i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.20 m for r2, −i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.20 m(−i^)=−2.9×105i^ N/C

Substitute 8.2×104i^ N/C for E→1 and −2.9×105i^ N/C for E→2 in equation (III).

E→A=8.2×104i^ N/C+(−2.9×105i^ N/C)=−2.1×105i^ N/C

The following figure gives the fields acting on the point B due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 2

Substitute 3.2×10−6 C/m for λ1, 0.25 m for r1, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.25 mi^=2.3×105i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.25 m for r2, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.25 m(i^)=2.3×105i^ N/C

Substitute 2.3×105i^ N/C for E→1 and 2.3×105i^ N/C for E→2 in equation (III).

E→B=2.3×105i^ N/C+2.3×105i^ N/C=4.6×105i^ N/C

The following figure gives the fields acting on the point C due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 3

Substitute 3.2×10−6 C/m for λ1, 0.10 m for r1, −i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.10 m(−i^)=−5.8×105i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.60 m for r2, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.60 m(i^)=9.6×104i^ N/C

Substitute −5.8×105i^ N/C for E→1 and 9.6×104i^ N/C for E→2 in equation (III).

E→C=−5.8×105i^ N/C+9.6×104i^ N/C=−4.8×105i^ N/C

Therefore, the net electric field at points A, B and C are −2.1×105i^ N/C_, 4.6×105i^ N/C_ and −4.8×105i^ N/C_ respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The net electric field at points A, B and C.

Answer 12

Answer to Problem 35PQ

The net electric field at points A, B and C are −2.1×105i^ N/C_, 4.6×105i^ N/C_ and −4.8×105i^ N/C_ respectively.

Answer 13

Explanation of Solution

Write the expression for finding the electric field due to first line charge.

E→1=12πε0λ1r1x^ (I)

Here, λ1 is the line charge density of the first line and r1 is the distance between the line charge and the point.

Write the expression for finding the electric field due to second line charge.

E→2=12πε0λ2r2x^ (II)

Here, λ1 is the line charge density of the second line and r2 is the distance between the line charge and the point.

Write the expression for the net electric field.

E→=E→1+E→2 (III)

Here, E→ is the net charge at any point.

Conclusion:

The following figure gives the fields acting on the point A due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 1

Substitute 3.2×10−6 C/m for λ1, 0.70 m for r1, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.70 mi^=8.2×104i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.20 m for r2, −i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.20 m(−i^)=−2.9×105i^ N/C

Substitute 8.2×104i^ N/C for E→1 and −2.9×105i^ N/C for E→2 in equation (III).

E→A=8.2×104i^ N/C+(−2.9×105i^ N/C)=−2.1×105i^ N/C

The following figure gives the fields acting on the point B due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 2

Substitute 3.2×10−6 C/m for λ1, 0.25 m for r1, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.25 mi^=2.3×105i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.25 m for r2, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.25 m(i^)=2.3×105i^ N/C

Substitute 2.3×105i^ N/C for E→1 and 2.3×105i^ N/C for E→2 in equation (III).

E→B=2.3×105i^ N/C+2.3×105i^ N/C=4.6×105i^ N/C

The following figure gives the fields acting on the point C due to the first and the second line charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 35PQ , additional homework tip 3

Substitute 3.2×10−6 C/m for λ1, 0.10 m for r1, −i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (I).

E→1=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.10 m(−i^)=−5.8×105i^ N/C

Substitute 3.2×10−6 C/m for λ2, 0.60 m for r2, i^ for x^ and 8.85×10−12 C2/N⋅m2 for ε0 in equation (II).

E→2=12π(8.85×10−12 C2/N⋅m2)3.2×10−6 C/m0.60 m(i^)=9.6×104i^ N/C

Substitute −5.8×105i^ N/C for E→1 and 9.6×104i^ N/C for E→2 in equation (III).

E→C=−5.8×105i^ N/C+9.6×104i^ N/C=−4.8×105i^ N/C

Therefore, the net electric field at points A, B and C are −2.1×105i^ N/C_, 4.6×105i^ N/C_ and −4.8×105i^ N/C_ respectively.

Answer 14

Ch. 25.1 - a. List all the uppercase letters that have the...Ch. 25.2 - The terms electric force, electric field, and...Ch. 25.2 - Prob. 25.3CE Ch. 25.3 - Which of the following expressions are correct...Ch. 25.3 - Find the electric flux through the three Gaussian...Ch. 25.4 - Prob. 25.6CE Ch. 25.7 - Is it possible for the charged solid sphere in...Ch. 25 - Which word or name has the same symmetry as the...Ch. 25 - Prob. 2PQ Ch. 25 - Prob. 3PQ

Two infinitely long, parallel lines of charge with linear charge densities 3.2 μ C/m and −3.2 μ C/m are separated by a distance of 0.50 m. What is the net electric field at points A , B , and C as shown in Figure P25.35? FIGURE P25.35

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