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ch23_p13 - 13(a Let A =(1.40 m)2 Then j j = 3.00 y A y =0 j...

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2 ( ) ( ) ( ) ( ) ( )( )( ) 2 2 =0 1.40 ˆ ˆ ˆ ˆ 3.00 j j 3.00 j A j 3.00 1.40 1.40 8.23 N m C. y y y A y = Φ = ⋅ − + = = (b) The charge is given by ( ) ( ) 12 2 2 2 11 enc 0 8.85 10 C / N m 8.23 N m C 7.29 10 C q ε = Φ = × = × . (c) The electric field can be re-written as 0 ˆ 3.00 j E y E = + G G , where 0 ˆ ˆ 4.00i 6.00j E = − + G is a constant field which does not contribute to the net flux through the cube. Thus
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