Hw3 - PHY108 FALL 08 Solutions for HW3(Due on 9/24 3 We use(a(b(c E A where A A j b1.40mg j 2 2 6.00 N C 1.40 m 0 i j 2 2.00 N C 1.40 m j j 3.92 N

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PHY108 FALL 08 Solutions for HW3(Due on 9/24) 3. We use E A , where A A . j m j 2 140 b g . (a) 2 ˆˆ 6.00 N C i 1.40 m j 0. (b) 2 2 2.00 N C j 1.40 m j 3.92 N m C. (c) 2 ˆ ˆ ˆ 3.00 N C i 400N C k 1.40 m j 0 . (d) The total flux of a uniform field through a closed surface is always zero. 7. To exploit the symmetry of the situation, we imagine a closed Gaussian surface in the shape of a cube, of edge length d , with a proton of charge 19 1.6 10 C q situated at the inside center of the cube. The cube has six faces, and we expect an equal amount of flux through each face. The total amount of flux is net = q / 0 , and we conclude that the flux through the square is one-sixth of that. Thus, = q /6 0 = 3.01 10 –9 N m 2 /C. 9. Let A be the area of one face of the cube, E u be the magnitude of the electric field at the upper face, and E l be the magnitude of the field at the lower face. Since the field is downward, the flux through the upper face is negative and the flux through the lower face is positive. The flux through the other faces is zero, so the total flux through the cube surface is ( ). u A E E The net charge inside the cube is given by Gauss’ law: 12 2 2 2 00 6 ( ) (8.85 10 C / N m )(100 m) (100 N/C 60.0 N/C) 3.54 10 C 3.54 C.
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This note was uploaded on 03/26/2010 for the course PHY 108 taught by Professor Iashvili during the Spring '08 term at SUNY Buffalo.

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Hw3 - PHY108 FALL 08 Solutions for HW3(Due on 9/24 3 We use(a(b(c E A where A A j b1.40mg j 2 2 6.00 N C 1.40 m 0 i j 2 2.00 N C 1.40 m j j 3.92 N

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