ch23_p41 - 41. We use a Gaussian surface in the form of a...

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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 x from 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 ρ = 5.80 fC/m 3 is positive, it points outward at both faces: toward the left at the left face and toward the right at the right face. Furthermore, the magnitude is the same at both faces. The electric flux through each of these faces is Ea 2 . The field is parallel to the other faces of the Gaussian surface and the flux through them is zero. The total flux through the Gaussian surface is 2 2. Ea Φ= The volume enclosed by the Gaussian surface is 2 a 2 x and the charge contained within it is 2 2 qa x = . Gauss’ law yields 2 ε 0 Ea 2
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ch23_p41 - 41. We use a Gaussian surface in the form of a...

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