The electric flux through each of the six faces.

Question

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Answer 1

Question

Chapter 25, Problem 70PQ

To determine

The electric flux through each of the six faces.

Expert Solution & Answer

Answer to Problem 70PQ

The electric flux through each of the six faces are −3.2×10−2 N⋅m2/C_, 0_, 0_, 4.0×10−2 N⋅m2/C_, 0_ and 2.4×10−2 N⋅m2/C_.

Explanation of Solution

The following figure marks the faces of the cube with coordinate axes.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 70PQ

Write the expression to find the electric flux.

ΦE=∫E→⋅dA→ (I)

Write the expression for the area vector for the first face of the cube.

dA→1=dydzi^

Substitute dydzi^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,1=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dydzi^=−3.2×10−2 N⋅m2/C

Write the expression for the area vector for the second face of the cube.

dA→2=−dxdzj^

Substitute −dxdzj^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,2=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dxdzj^=0

Write the expression for the area vector for the third face of the cube.

dA→3=−dydzi^

Substitute −dydzi^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,3=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dydzi^=0

Write the expression for the area vector for the fourth face of the cube.

dA→4=dxdzj^

Substitute dxdzj^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,4=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dxdzj^=4.0×10−2 N⋅m2/C

Write the expression for the area vector for the fifth face of the cube.

dA→5=−dxdyk^

Substitute −dxdyk^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,5=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dxdyk^=0

Write the expression for the area vector for the first face of the cube.

dA→6=dxdyk^

Substitute dxdyk^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,6=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dxdyk^=2.4×10−2 N⋅m2/C

Conclusion:

Therefore, the electric flux through each of the six faces are −3.2×10−2 N⋅m2/C_, 0_, 0_, 4.0×10−2 N⋅m2/C_, 0_ and 2.4×10−2 N⋅m2/C_.

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Answer 2

Question

Chapter 25, Problem 70PQ

To determine

The electric flux through each of the six faces.

Expert Solution & Answer

Answer to Problem 70PQ

The electric flux through each of the six faces are −3.2×10−2 N⋅m2/C_, 0_, 0_, 4.0×10−2 N⋅m2/C_, 0_ and 2.4×10−2 N⋅m2/C_.

Explanation of Solution

The following figure marks the faces of the cube with coordinate axes.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 70PQ

Write the expression to find the electric flux.

ΦE=∫E→⋅dA→ (I)

Write the expression for the area vector for the first face of the cube.

dA→1=dydzi^

Substitute dydzi^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,1=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dydzi^=−3.2×10−2 N⋅m2/C

Write the expression for the area vector for the second face of the cube.

dA→2=−dxdzj^

Substitute −dxdzj^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,2=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dxdzj^=0

Write the expression for the area vector for the third face of the cube.

dA→3=−dydzi^

Substitute −dydzi^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,3=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dydzi^=0

Write the expression for the area vector for the fourth face of the cube.

dA→4=dxdzj^

Substitute dxdzj^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,4=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dxdzj^=4.0×10−2 N⋅m2/C

Write the expression for the area vector for the fifth face of the cube.

dA→5=−dxdyk^

Substitute −dxdyk^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,5=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dxdyk^=0

Write the expression for the area vector for the first face of the cube.

dA→6=dxdyk^

Substitute dxdyk^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,6=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dxdyk^=2.4×10−2 N⋅m2/C

Conclusion:

Therefore, the electric flux through each of the six faces are −3.2×10−2 N⋅m2/C_, 0_, 0_, 4.0×10−2 N⋅m2/C_, 0_ and 2.4×10−2 N⋅m2/C_.

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Answer 3

Question

Chapter 25, Problem 70PQ

To determine

The electric flux through each of the six faces.

Expert Solution & Answer

Answer to Problem 70PQ

The electric flux through each of the six faces are −3.2×10−2 N⋅m2/C_, 0_, 0_, 4.0×10−2 N⋅m2/C_, 0_ and 2.4×10−2 N⋅m2/C_.

Explanation of Solution

The following figure marks the faces of the cube with coordinate axes.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 70PQ

Write the expression to find the electric flux.

ΦE=∫E→⋅dA→ (I)

Write the expression for the area vector for the first face of the cube.

dA→1=dydzi^

Substitute dydzi^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,1=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dydzi^=−3.2×10−2 N⋅m2/C

Write the expression for the area vector for the second face of the cube.

dA→2=−dxdzj^

Substitute −dxdzj^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,2=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dxdzj^=0

Write the expression for the area vector for the third face of the cube.

dA→3=−dydzi^

Substitute −dydzi^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,3=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dydzi^=0

Write the expression for the area vector for the fourth face of the cube.

dA→4=dxdzj^

Substitute dxdzj^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,4=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dxdzj^=4.0×10−2 N⋅m2/C

Write the expression for the area vector for the fifth face of the cube.

dA→5=−dxdyk^

Substitute −dxdyk^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,5=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dxdyk^=0

Write the expression for the area vector for the first face of the cube.

dA→6=dxdyk^

Substitute dxdyk^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,6=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dxdyk^=2.4×10−2 N⋅m2/C

Conclusion:

Therefore, the electric flux through each of the six faces are −3.2×10−2 N⋅m2/C_, 0_, 0_, 4.0×10−2 N⋅m2/C_, 0_ and 2.4×10−2 N⋅m2/C_.

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Answer 4

Question

Chapter 25, Problem 70PQ

To determine

The electric flux through each of the six faces.

Expert Solution & Answer

Answer to Problem 70PQ

The electric flux through each of the six faces are −3.2×10−2 N⋅m2/C_, 0_, 0_, 4.0×10−2 N⋅m2/C_, 0_ and 2.4×10−2 N⋅m2/C_.

Explanation of Solution

The following figure marks the faces of the cube with coordinate axes.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 70PQ

Write the expression to find the electric flux.

ΦE=∫E→⋅dA→ (I)

Write the expression for the area vector for the first face of the cube.

dA→1=dydzi^

Substitute dydzi^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,1=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dydzi^=−3.2×10−2 N⋅m2/C

Write the expression for the area vector for the second face of the cube.

dA→2=−dxdzj^

Substitute −dxdzj^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,2=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dxdzj^=0

Write the expression for the area vector for the third face of the cube.

dA→3=−dydzi^

Substitute −dydzi^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,3=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dydzi^=0

Write the expression for the area vector for the fourth face of the cube.

dA→4=dxdzj^

Substitute dxdzj^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,4=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dxdzj^=4.0×10−2 N⋅m2/C

Write the expression for the area vector for the fifth face of the cube.

dA→5=−dxdyk^

Substitute −dxdyk^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,5=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dxdyk^=0

Write the expression for the area vector for the first face of the cube.

dA→6=dxdyk^

Substitute dxdyk^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,6=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dxdyk^=2.4×10−2 N⋅m2/C

Conclusion:

Therefore, the electric flux through each of the six faces are −3.2×10−2 N⋅m2/C_, 0_, 0_, 4.0×10−2 N⋅m2/C_, 0_ and 2.4×10−2 N⋅m2/C_.

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Answer 5

Chapter 25, Problem 70PQ

To determine

The electric flux through each of the six faces.

Expert Solution & Answer

Answer to Problem 70PQ

The electric flux through each of the six faces are −3.2×10−2 N⋅m2/C_, 0_, 0_, 4.0×10−2 N⋅m2/C_, 0_ and 2.4×10−2 N⋅m2/C_.

Explanation of Solution

The following figure marks the faces of the cube with coordinate axes.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 70PQ

Write the expression to find the electric flux.

ΦE=∫E→⋅dA→ (I)

Write the expression for the area vector for the first face of the cube.

dA→1=dydzi^

Substitute dydzi^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,1=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dydzi^=−3.2×10−2 N⋅m2/C

Write the expression for the area vector for the second face of the cube.

dA→2=−dxdzj^

Substitute −dxdzj^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,2=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dxdzj^=0

Write the expression for the area vector for the third face of the cube.

dA→3=−dydzi^

Substitute −dydzi^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,3=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dydzi^=0

Write the expression for the area vector for the fourth face of the cube.

dA→4=dxdzj^

Substitute dxdzj^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,4=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dxdzj^=4.0×10−2 N⋅m2/C

Write the expression for the area vector for the fifth face of the cube.

dA→5=−dxdyk^

Substitute −dxdyk^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,5=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dxdyk^=0

Write the expression for the area vector for the first face of the cube.

dA→6=dxdyk^

Substitute dxdyk^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,6=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dxdyk^=2.4×10−2 N⋅m2/C

Conclusion:

Therefore, the electric flux through each of the six faces are −3.2×10−2 N⋅m2/C_, 0_, 0_, 4.0×10−2 N⋅m2/C_, 0_ and 2.4×10−2 N⋅m2/C_.

Answer 6

Chapter 25, Problem 70PQ

To determine

The electric flux through each of the six faces.

Expert Solution & Answer

Answer to Problem 70PQ

The electric flux through each of the six faces are −3.2×10−2 N⋅m2/C_, 0_, 0_, 4.0×10−2 N⋅m2/C_, 0_ and 2.4×10−2 N⋅m2/C_.

Explanation of Solution

The following figure marks the faces of the cube with coordinate axes.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 70PQ

Write the expression to find the electric flux.

ΦE=∫E→⋅dA→ (I)

Write the expression for the area vector for the first face of the cube.

dA→1=dydzi^

Substitute dydzi^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,1=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dydzi^=−3.2×10−2 N⋅m2/C

Write the expression for the area vector for the second face of the cube.

dA→2=−dxdzj^

Substitute −dxdzj^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,2=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dxdzj^=0

Write the expression for the area vector for the third face of the cube.

dA→3=−dydzi^

Substitute −dydzi^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,3=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dydzi^=0

Write the expression for the area vector for the fourth face of the cube.

dA→4=dxdzj^

Substitute dxdzj^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,4=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dxdzj^=4.0×10−2 N⋅m2/C

Write the expression for the area vector for the fifth face of the cube.

dA→5=−dxdyk^

Substitute −dxdyk^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,5=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dxdyk^=0

Write the expression for the area vector for the first face of the cube.

dA→6=dxdyk^

Substitute dxdyk^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,6=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dxdyk^=2.4×10−2 N⋅m2/C

Conclusion:

Therefore, the electric flux through each of the six faces are −3.2×10−2 N⋅m2/C_, 0_, 0_, 4.0×10−2 N⋅m2/C_, 0_ and 2.4×10−2 N⋅m2/C_.

Answer 7

Chapter 25, Problem 70PQ

To determine

The electric flux through each of the six faces.

Answer 8

Expert Solution & Answer

Answer to Problem 70PQ

The electric flux through each of the six faces are −3.2×10−2 N⋅m2/C_, 0_, 0_, 4.0×10−2 N⋅m2/C_, 0_ and 2.4×10−2 N⋅m2/C_.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The following figure marks the faces of the cube with coordinate axes.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 70PQ

Write the expression to find the electric flux.

ΦE=∫E→⋅dA→ (I)

Write the expression for the area vector for the first face of the cube.

dA→1=dydzi^

Substitute dydzi^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,1=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dydzi^=−3.2×10−2 N⋅m2/C

Write the expression for the area vector for the second face of the cube.

dA→2=−dxdzj^

Substitute −dxdzj^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,2=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dxdzj^=0

Write the expression for the area vector for the third face of the cube.

dA→3=−dydzi^

Substitute −dydzi^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,3=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dydzi^=0

Write the expression for the area vector for the fourth face of the cube.

dA→4=dxdzj^

Substitute dxdzj^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,4=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dxdzj^=4.0×10−2 N⋅m2/C

Write the expression for the area vector for the fifth face of the cube.

dA→5=−dxdyk^

Substitute −dxdyk^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,5=∫−((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)dxdyk^=0

Write the expression for the area vector for the first face of the cube.

dA→6=dxdyk^

Substitute dxdyk^ for dA→ and (−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^ for E→ in equation (I).

ΦE,6=∫((−4.0 N/C⋅m)xi^+(5.0 N/C⋅m)yj^+(3.0 N/C⋅m)zk^)⋅dxdyk^=2.4×10−2 N⋅m2/C

Conclusion:

Therefore, the electric flux through each of the six faces are −3.2×10−2 N⋅m2/C_, 0_, 0_, 4.0×10−2 N⋅m2/C_, 0_ and 2.4×10−2 N⋅m2/C_.

Answer 12

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

The electric flux through each of the six faces.

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The electric flux through each of the six faces.

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