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Unformatted text preview: Chapter 23 1. The vector area A and the electric field E are shown on the diagram below. The angle between them is 180 35 = 145, so the electric flux through the area is ( 29 ( 29 2 3 2 2 cos 1800 N C 3.2 10 m cos145 1.5 10 N m C. E A EA  = = = =  2. We use = E A , where A A = = . j m j 2 140 b g . (a) ( 29 ( 29 2 6.00 N C i 1.40 m j 0. = = (b) ( 29 ( 29 2 2 2.00 N C j 1.40 m j 3.92 N m C. =  =  (c) ( 29 ( 29 ( 29 2 3.00 N C i 400 N C k 1.40 m j = + = . (d) The total flux of a uniform field through a closed surface is always zero. 3. We use = z E dA and note that the side length of the cube is (3.0 m1.0 m) = 2.0 m. (a) On the top face of the cube y = 2.0 m and ( 29 j dA dA = . Therefore, we have ( 29 ( 29 2 4i 3 2.0 2 j 4i 18j E = + = . Thus the flux is ( 29 ( 29 ( 29 ( 29 2 2 2 top top top 4i 18j j 18 18 2.0 N m C 72 N m C. E dA dA dA = = =  =  =  (b) On the bottom face of the cube y = 0 and dA dA = bg ej j . Therefore, we have E = + = 4 3 0 2 4 6 2 i j i j c h . Thus, the flux is 963 CHAPTER 23 ( 29 ( 29 ( 29 ( 29 2 2 2 bottom bottom bottom 4i 6j j 6 6 2.0 N m C 24 N m C. E dA dA dA = = = = = + (c) On the left face of the cube ( 29 ( 29 i dA dA = . So ( 29 ( 29 ( 29 ( 29 2 2 2 left left bottom 4i j i 4 4 2.0 N m C 16 N m C. y E dA E dA dA = = + =  =  =  (d) On the back face of the cube ( 29 ( 29 k dA dA = . But since E has no z component E dA = . Thus, = 0. (e) We now have to add the flux through all six faces. One can easily verify that the flux through the front face is zero, while that through the right face is the opposite of that through the left one, or +16 Nm 2 /C. Thus the net flux through the cube is = (72 + 24 16 + 0 + 0 + 16) Nm 2 /C = 48 Nm 2 /C. 4. The flux through the flat surface encircled by the rim is given by 2 . a E = Thus, the flux through the netting is 2 3 4 2 (0.11 m) (3.0 10 N/C) 1.1 10 N m /C a E 2 =  =  =  =  . 5. We use Gauss law: q = , where is the total flux through the cube surface and q is the net charge inside the cube. Thus, 6 5 2 12 2 2 1.8 10 C 2.0 10 N m C. 8.85 10 C N m q  = = = 6. There is no flux through the sides, so we have two inward contributions to the flux, one from the top (of magnitude (34)(3.0) 2 ) and one from the bottom (of magnitude (20) (3.0) 2 ). With inward flux being negative, the result is = 486 N m 2 /C. Gauss law then leads to q enc = = 4.3 10 9 C....
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This note was uploaded on 04/19/2008 for the course PHYS 1100, 1200 taught by Professor Many during the Fall '08 term at Rensselaer Polytechnic Institute.
 Fall '08
 Many
 Physics

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