Chapter 27 Solutions

Chapter 27 Solutions - Chapter 27 Solutions (solutions...

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Chapter 27 Solutions (solutions marked with * are graded problems) 27.3. Visualize: Figure 27.6 shows the electric field for an infinite plane of charge. For two parallel planes, this is the only shape of the electric field vectors that matches the symmetry of the charge distribution. 27.5. 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 Ex27.5. Let A be the area of each of the six faces of the cube. Solve: The electric flux is defined as e cos E A EA ! " = # = r r , where θ 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 10 N/C 10 N/C 10 N/C 5 N/C cos0 35 N m /C A A ! = + + + ° = Similarly, the electric flux into the closed cube surface is ( ) ( ) ( ) 2 in 15 N/C 20 N/C cos 180 35 N m /C A A ! = + ° = " Hence, Φ out + Φ in = 0 N m 2 /C. Since the net electric flux is zero, the closed box contains no charge. 27.9. Model: The electric field is uniform over the entire surface. Visualize: Please refer to Figure Ex27.9. 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 r is = 120 ° . Solve: The electric flux is ( ) ( ) 2 2 2 e cos 200 N/C 1.0 10 m cos120 1.0 N m /C E A EA " # = $ = = % ° = " r r 27.11. Model: The electric field is uniform over the rectangle in the xy plane. Solve: (a) The area vector is perpendicular to the xy plane. Thus ( ) ( ) 4 2 ˆ ˆ 2.0 cm 3.0 cm 6.0 10 m A k k ! = " = " r The electric flux through the rectangle is ( )( ) 4 2 2 2 e ˆ ˆ ˆ 50 100 6.0 10 N m /C 6.0 10 N m /C E A i k k ! ! " = # = + # $ = $ r r (b) The electric flux is ( ) ( ) 4 2 2 e ˆ ˆ ˆ 50 100 6.0 10 N m /C 0 N m /C E A i j k ! " = # = + # $ = r r Assess: In (b), E r is in the plane of the rectangle. That is why the flux is zero.
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27.17. Solve: For any closed surface that encloses a total charge Q in , the net electric flux through the surface is ( )( ) 6 19 2 in e 12 2 2 0 55.3 10 1.60 10 C 1.0 N m /C 8.85 10 C /N m Q ! " " # " # $ = = = " # 27.19.
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This note was uploaded on 04/03/2008 for the course PHYS 100 taught by Professor All during the Spring '08 term at Cal Poly.

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Chapter 27 Solutions - Chapter 27 Solutions (solutions...

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