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Gauss' Law for a Nonuniform Field
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[QUOTE="Redfire66, post: 5316839, member: 456103"] [h2]Homework Statement [/h2] A closed rectangular surface with dimensions a = b and c where the faces perpendicular to the field are a*b. The left edge of the closed surface is located at position x = a, for c > a.The electric field throughout the region is nonuniform and given by E = 3*x [B]xhat[/B] N/C, [h2]Homework Equations[/h2] Flux = Integral of E(dot)dA = qenclosed/Epsilon0 [h2]The Attempt at a Solution[/h2] I'm just wondering about how this works, if there is no enclosed charge, then there shouldn't be a net flux. I'm pretty sure the flux at one end is greater as it reaches the x = c face than at x = a, is it possible to have a nonzero net flux? [/QUOTE]
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Gauss' Law for a Nonuniform Field
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