ch23-p002 - 2. We use = E dA and note that the side length...

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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 ˆˆ ˆ 4i 18j j 18 18 2.0 N m C 72 N m C. EdA dA dA Φ= = =− = − =− ∫∫ G G (b) On the bottom face of the cube y = 0 and dA dA G =− b g e
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This note was uploaded on 09/07/2010 for the course PHY 244 taught by Professor Hd during the Spring '10 term at The Petroleum Institute.

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