HW06solutions-1

# HW06solutions-1 - 28.4. Model: The electric flux flows out...

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28.4. Model: The electric flux “flows” out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge. Visualize: Please refer to Figure EX28.4. Let A be the area in m 2 of each of the six faces of the cube. Solve: The electric flux is defined as e cos , EA E A θ Φ= ⋅ = G G where is the angle between the electric field and a line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is () ( ) 2 out 20 N/C 20 N/C 10 N/C cos0 50 N m /C AA Φ= + + ° = Similarly, the electric flux into the closed cube surface is ( ) 2 in 15 N/C 15 N/C 15 N/C cos180 45 N m /C + + ° = The net electric flux is ( ) ( ) 22 2 50 N m/C 45 N 5 N . A −= Since the net electric flux is positive (i.e., outward), the closed box contains a positive charge.

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28.10. Model: The electric field is uniform over the entire surface. Visualize: Please refer to Figure EX28.10. The electric field vectors make an angle of 30 ° below the surface. Because the normal ˆ n to the planar surface is at an angle of 90 ° relative to the surface, the angle between ˆ n and E G is θ = 120 ° . Solve: The electric flux is () 2 22 e cos 200 N/C 15 10 m cos120 2.3 N m /C EA E A Φ= ⋅ = = × °=− G G
28.20. Visualize: Please refer to Figure EX28.20. For any closed surface that encloses a total charge Q in , the net electric flux through the closed surface is ei n 0 . Q ε Φ= For the closed surface of the torus, Q in includes only the 1 nC charge. So, the net flux through the torus is due to this charge: 9 2 e 12 2 2 1.0 10 C 113 N m /C 8.85 10 C /Nm −× =− × This is inward flux.

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28.22.
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## This note was uploaded on 11/02/2009 for the course MASTERING PHYS taught by Professor All during the Spring '09 term at Kettering.

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HW06solutions-1 - 28.4. Model: The electric flux flows out...

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