QUIZ XVI GAUSS LAW Given that the ELECTRIC FIELD due to an infinite thin sheet of uniformly distributed charge is uniform perpendicular to the sheet of charge and of magnitude E 2 0 where is the surface charge density associated with the charged sheet determine the ELECTRIC FIELD magnitude and direction due to the charged sheet distribution displayed below where the surface charge density of each positively charged sheet is every negatively charged sheet uy 5 d 4 c 3 b 2 a 1 a b c d e in Region 1 in Region 2 in Region 3 in Region 4 in Region 5 2 2 2 2 2 KEY CONCEPTS RELATIONSHIPS LAWS GAUSS LAW UNIFORM FIELD SUPERPOSITION THEOREM SOLUTION The ELECTRIC FIELD at any point in any Region is the superposition of the uniform ELECTRIC FIELDS from the four thin infinite sheets recalling that the ELECTRIC FIELD due to a positively charged sheet is directed perpendicular to and away from the sheet for negative sheets perpendicular to and directed toward the negative sheet Labeling the sheets a b d Then in any Region at any point E E a E b E c E d Region 1 E 1 2 0 uy 4 0 uy 2 0 uy 4 0 a uy yielding E 1 2 0 uy b Region 2 E 2 2 0 uy 4 0 uy 2 0 uy 4 0 uy yielding E 2 2 0 uy Region 3 E 3 2 0 uy 4 0 uy 2 0 uy 4 0 c uy yielding E 3 0 d Region 4 E 4 2 0 uy 4 0 uy 2 0 uy 4 0 uy yielding E 4 0 uy Region 5 E 5 2 0 uy 4 0 uy 2 0 uy 4 0 e uy yielding E 5 2 0 uy
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