Chapter23 - 1. The vector area A and the electric field E...

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1. The vector area G A and the electric field G E are shown on the diagram below. The angle θ between them is 180° – 35° = 145°, so the electric flux through the area is () ( ) 2 32 2 cos 1800 N C 3.2 10 m cos145 1.5 10 N m C. EA E A −− Φ= ⋅ = = × °=− × G G
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(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 N·m 2 /C. Thus the net flux through the cube is Φ = (–72 + 24 – 16 + 0 + 0 + 16) N·m 2 /C = – 48 N·m 2 /C. 2. We use Φ= z GG EdA and note that the side length of the cube is (3.0 m–1.0 m) = 2.0 m. (a) On the top face of the cube y = 2.0 m and ( ) ˆ j dA dA = G . Therefore, we have () 2 ˆˆ ˆ ˆ 4i 3 2.0 2 j 4i 18j E =− + =− G . Thus the flux is () ( ) ( ) 2 22 top top top ˆ 4i 18j j 18 18 2.0 N m C 72 N m C. dA = = − ∫∫ G G (b) On the bottom face of the cube y = 0 and dA dA G b g e j ± j . Therefore, we have E + 43 02 46 2 ±± ± ± ij i j c h . Thus, the flux is ( ) 2 bottom bottom bottom ˆ 4 i 6 j j 6 62 . 0 Nm C 2 4 Nm C . = − = = =+ G G (c) On the left face of the cube ( ) ˆ i dA dA = G . So ( ) ( ) 2 left left bottom ˆ ˆ 4i j i 4 4 2.0 N m C 16 N m C. y E dA E dA dA = + − =− G (d) On the back face of the cube ( ) ˆ k dA dA = G . But since E G has no z component 0 ⋅= G G . Thus, Φ = 0.
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3. We use Φ= ⋅ G G E A , where G AA == ± . ± jm j 2 140 b g . (a) () ( ) 2 ˆˆ 6.00 N C i 1.40 m j 0. Φ= = (b) ( ) 2 2 2.00 N C j 1.40 m j 3.92 N m C. Φ= − =− (c) ( ) 2 ˆ 3.00 N C i 400 N C k 1.40 m j 0 ⎡⎤ + = ⎣⎦ . (d) The total flux of a uniform field through a closed surface is always zero.
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4. 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 Φ = – 486 N m 2 /C. Gauss’ law then leads to 12 2 2 2 9 enc 0 (8.85 10 C /N m )( 486 N m C) 4.3 10 C. q ε −− = × = − ×
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5. We use Gauss’ law: 0 q ε Φ= , where Φ is the total flux through the cube surface and q is the net charge inside the cube. Thus, 6 52 12 2 2 0 1.8 10 C 2.0 10 N m C. 8.85 10 C N m q × = = × ×⋅
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6. The flux through the flat surface encircled by the rim is given by 2 . aE Φ=π Thus, the flux through the netting is 23 4 2 (0.11 m) (3.0 10 N/C) 1.1 10 N m /C ππ 2− Φ =−Φ=− =− × × .
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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 Φ net = q / ε 0 , and we conclude that the flux through the square is one-sixth of that. Thus, 19 92 12 2 2 0 1.6 10 C 3.01 10 N m C. 66 ( 8 . 8 5 1 0 C N m ) q × Φ= = = × ×⋅
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8. We note that only the smaller shell contributes a (non-zero) field at the designated point, since the point is inside the radius of the large sphere (and E = 0 inside of a spherical charge), and the field points towards the x direction. Thus, with
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Chapter23 - 1. The vector area A and the electric field E...

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