ch23_p3 - 3. We use = E dA and note that the side length of...

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3. We use Φ= z G G 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 = = − ³³ ³ 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 . = − = = =+ ³ 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 = + − =− ³ G (d) On the back face of the cube ˆ k dA dA G . But since
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This note was uploaded on 09/06/2009 for the course PHYSICS 14358 taught by Professor Bolton during the Spring '09 term at Kansas State University.

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