Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
15th Edition
ISBN: 9781305289963
Author: Debora M. Katz
Publisher: Cengage Custom Learning
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Chapter 25, Problem 70PQ
To determine

The electric flux through each of the six faces.

Expert Solution & Answer
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Answer to Problem 70PQ

The electric flux through each of the six faces are 3.2×102Nm2/C_, 0_, 0_, 4.0×102Nm2/C_, 0_ and 2.4×102Nm2/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=EdA                                                                                                   (I)

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

    dA1=dydzi^

Substitute dydzi^ for dA and (4.0N/Cm)xi^+(5.0N/Cm)yj^+(3.0N/Cm)zk^ for E in equation (I).

    ΦE,1=((4.0N/Cm)xi^+(5.0N/Cm)yj^+(3.0N/Cm)zk^)dydzi^=3.2×102Nm2/C

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

    dA2=dxdzj^

Substitute dxdzj^ for dA and (4.0N/Cm)xi^+(5.0N/Cm)yj^+(3.0N/Cm)zk^ for E in equation (I).

    ΦE,2=((4.0N/Cm)xi^+(5.0N/Cm)yj^+(3.0N/Cm)zk^)dxdzj^=0

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

    dA3=dydzi^

Substitute dydzi^ for dA and (4.0N/Cm)xi^+(5.0N/Cm)yj^+(3.0N/Cm)zk^ for E in equation (I).

    ΦE,3=((4.0N/Cm)xi^+(5.0N/Cm)yj^+(3.0N/Cm)zk^)dydzi^=0

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

    dA4=dxdzj^

Substitute dxdzj^ for dA and (4.0N/Cm)xi^+(5.0N/Cm)yj^+(3.0N/Cm)zk^ for E in equation (I).

    ΦE,4=((4.0N/Cm)xi^+(5.0N/Cm)yj^+(3.0N/Cm)zk^)dxdzj^=4.0×102Nm2/C

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

    dA5=dxdyk^

Substitute dxdyk^ for dA and (4.0N/Cm)xi^+(5.0N/Cm)yj^+(3.0N/Cm)zk^ for E in equation (I).

    ΦE,5=((4.0N/Cm)xi^+(5.0N/Cm)yj^+(3.0N/Cm)zk^)dxdyk^=0

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

    dA6=dxdyk^

Substitute dxdyk^ for dA and (4.0N/Cm)xi^+(5.0N/Cm)yj^+(3.0N/Cm)zk^ for E in equation (I).

    ΦE,6=((4.0N/Cm)xi^+(5.0N/Cm)yj^+(3.0N/Cm)zk^)dxdyk^=2.4×102Nm2/C

Conclusion:

Therefore, the electric flux through each of the six faces are 3.2×102Nm2/C_, 0_, 0_, 4.0×102Nm2/C_, 0_ and 2.4×102Nm2/C_.

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Chapter 25 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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