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Unformatted text preview: d = o d E / dt = (8.85x1012 C 2 / Nm 2 ) (5.65x10 11 V m / s) = 5.00 C/s = 5.00A . (Hooray!) 2. In a region of space the electric field varies according to = (0.0500N/C) sin(2.00x10 3 t / sec) . Find the maximum displacement current through a 1.00m 2 area in the yz plane . For an area in the yz plane we define the area vector = 1.00m 2 , so E = = (0.0500N/C) sin(2.00x10 3 t / sec) (1.00 m 2 ) = (0.0500N m 2 /C) sin(2.00x10 3 t / sec). The displacement current is then I d = o d E / dt = (8.85x1012 C 2 / Nm 2 ) (0.0500N m 2 /C) (2.00x10 3 /sec) cos(2.00x10 3 t / sec) = 8.85x1010 A cos(2.00x10 3 t / sec) . This has a maximum value when cos(2.00x10 3 t / sec) = 1 , so I d, max = 8.85x1010 A ....
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 Fall '11
 JimGuinn

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