The electric flux through each of the six faces.

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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.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, 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.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, 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.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, 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.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, 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.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, 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.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, 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.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, 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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