ch23-p016 - 16. The total electric flux through the cube is...

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[] [ ] 22 11 13 21 01 ( ) ( ) 10 2(4) 10 2(1) 6 6(1)(2) 12. yz yz x x E x x E x x dydz dy dz dy dz == Φ= = − = = + − − = ∫∫ Similarly, the net flux through the two faces parallel to the xz plane is 43 ( ) ( ) [3 (3 ) ] 0 xz xz y y E y y E y y dxdz dy dz ⎡⎤ = = −−− = ⎣⎦ , and the net flux through the two faces parallel to the xy plane is () 41 10 3 2 ( 3 ) ( 1 ) 6 . xy xy z z E z z E z z dxdy dx dy b b b b = = − = = Applying Gauss’ law, we obtain enc 0 0 0 0 ( ) (6.00 0 12.0) 24.0 xy xz yz qb ε εε = Φ+
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This note was uploaded on 09/07/2010 for the course PHYS 231 taught by Professor Kd during the Fall '10 term at The Petroleum Institute.

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