ch23-p011 - 11. (a) Let A = (1.40 m)2. Then j j = 3.00 y A...

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() ( ) ( ) 2 2 =0 1.40 ˆˆ ˆ ˆ 3.00 j j 3.00 j A j 3.00 1.40 1.40 8.23 N m C. yy yA 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 −− = × = × . (c) The electric field can be re-written as 0 ˆ 3.00 j Ey E = + G G , where 0 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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This note was uploaded on 09/07/2010 for the course PHYS 244 taught by Professor Ac during the Fall '10 term at The Petroleum Institute.

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