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Unformatted text preview: 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. Visualize: Please refer to Figure EX28.6. 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 θ Φ = × = 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 ( 29 ( 29 2 out 20 N/C 10 N/C 10 N/C cos0 40 N m /C A A Φ = + + ° = Similarly, the electric flux into the closed cube surface is ( 29 ( 29 2 in 20 N/C 15 N/C cos180 35 N m /C A A Φ = + ° = - Because the cube contains negative charge, Φ out + Φ in must be negative. This means Φ out + Φ in + Φ unknown < 0 N m 2 /C. Therefore, ( 29 ( 29 2 2 2 unknown 40 N m /C 35 N m /C 0 N m /C A A + - + Φ < ⇒ ( 29 2 unknown 5 N m /C A Φ < - That is, the unknown vector points into the front face of the cube and its field strength is greater than 5 N/C. 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 r is θ = 120 ° . Solve: The electric flux is ( 29 ( 29 2 2 2 e cos 200 N/C 15 10 m cos120 2.3 N m /C E A EA θ- Φ = × = = × ° = - r r 28.26. Model: The excess charge on a conductor resides on the outer surface....
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This note was uploaded on 04/03/2012 for the course PHYSICS AN 101 taught by Professor N.a during the Spring '12 term at Clark College.
- Spring '12