(d) For x= 26.0 mm = 2.60×10−2m, we take a Gaussian surface of the same shape and orientation, but with x> d/2, so the left and right faces are outside the slab. The total flux through the surface is again 22EaΦ=but the charge enclosed is now q = a2dρ. Gauss’ law yields 2ε0Ea2= a2d, so 1533612220(5.80 10C/m )(9.40 10 m)3.08 10 N/C.22(8.8510C/Nm)dEε−−−−××===××⋅43. We use a Gaussian surface in the form of a box with rectangular sides. The cross section is shown with dashed lines in the diagram below. It is centered at the central plane of the slab, so the left and right faces are each a distance xfrom the central plane. We take the thickness of the rectangular solid to be a, the same as its length, so the left and right faces are squares. The electric field is normal to the left and right faces and is uniform over them. Since = 5.80 fC/m3is positive, it points outward at both faces: toward the left at the left face and toward the right at the
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.