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hw03 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 3...

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PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS CLASS 3. 23.08, 23.11, 23.53, 23.67, 23.87; Q25.02, Q25.06, Q25.07 23.08 a) Call the length, width, and height of the room l, w, and h . Then, from Gauss’ law, the charge enclosed is given by: ) ( 2 ) ( 0 0 lh wh lw E A d E q + + - = = ε ε , where E is the magnitude of the electric field at each surface, and the negative sign is because the field points inward. Then, the charge per volume in the room is: + + - = + + - = = w l h E lwh lh wh lw E V q 1 1 1 2 ) ( 2 0 0 ε ε ρ + + × - = - m 0 . 2 1 m 0 . 3 1 m 5 . 2 1 ) N/C 600 )( F/m 10 85 . 8 ( 2 12 ρ ρ = – 1.3 ×10 -8 C/m³ b) C/electron 10 6 . 1 C/m 10 3 . 1 19 3 8 - - × - × - = = e n ρ n = 8.1 ×10 10 electrons/m³ 23.11. The cube is sketched below. a) The flux through the three sides that touch the charge is zero since the electric field is parallel to each of those faces. b ) By symmetry it can be seen that the flux through each of the other three faces is the same, since all are equivalent relative to the position of the charge. The total flux that passes through the three of them combined, is one-eighth of the flux from the charge, since eight cubes could be made to touch at the position of the charge. Therefore, the
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