Chapter 28 - 28.1. Visualize: As discussed in Section 28.1,...

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28.1. Visualize: As discussed in Section 28.1, the symmetry of the electric field must match the symmetry of the charge distribution. In particular, the electric field of a cylindrically symmetric charge distribution cannot have a component parallel to the cylinder axis. Also, the electric field of a cylindrically symmetric charge distribution cannot have a component tangent to the circular cross section. The only shape for the electric field that matches the symmetry of the charge distribution with respect to (i) translation parallel to the cylinder axis, (ii) rotation by an angle about the cylinder axis, and (iii) reflections in any plane containing or perpendicular to the cylinder axis is the one shown in the figure.
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28.2. Visualize: The object has spherical symmetry, so the electric field is radial.
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28.3. Visualize: Figure 28.6 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.
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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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28.5. 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 EX28.5. Let A be the area 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 10 N/C 10 N/C 20 N/C 5 N/C cos0 45 N m /C AA Φ= + + + ° = Similarly, the electric flux into the closed cube surface is ( ) ( ) 2 in 15 N/C 20 N/C cos 180 35 N m /C + °= Hence, Φ out + Φ in = 10 N m 2 /C. Since the net electric flux is positive (i.e., outward), the closed box contains a positive charge.
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28.6. 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.
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Chapter 28 - 28.1. Visualize: As discussed in Section 28.1,...

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