253-wrk-24 - HPHY 253 Workout 24 Gauss&s Law...

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Unformatted text preview: HPHY 253 Workout 24 Gauss&s Law for Electricity Dr. A. E. Bak Name 1. Serway & Jewett Problem 24 : 4 . Consider a closed triangular box resting within a horizontal electric &eld of magnitude E = 7 : 80 & 10 4 N= C as shown in Figure P 24 : 4 . We take all area vectors here to point exterior to the box. (a) Calculate the electric ux through the vertical rectangular surface. This box side is at, having area A = (0 : 30 m ) (0 : 10 m ) = 0 : 030 m 2 : Because the electric eld is constant on this at surface, we obtain Z E da = E A = EA cos 180 & = & 7 : 80 & 10 4 (0 : 030) ( 1) = 2340 N m 2 = C as the electric ux through the vertical rectangular surface. (b) Calculate the electric ux through the slanted surface. This box side is at, having area A = (0 : 30 m ) : 10 m cos 60 & = 0 : 060 m 2 : Because the electric eld is constant on this at surface, we obtain Z E da = E A = EA cos 60 : & = & 7 : 80 & 10 4 (0 : 060) 1 2 = +2340 N m 2 = C as the electric ux through the slanted rectangular surface. (c) Calculate the electric ux through the entire surface of the box. The boxs remaining sides are all at. Because the electric eld is constant on each of these sides, we obtain Z E da = E A = EA cos 90 & = 0 for each remaining side : Therefore, we obtain I box E da = 2340 + 2340 + [ terms of zero ] = 0 as the electric ux through the entire box surface.as the electric ux through the entire box surface....
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This note was uploaded on 05/01/2011 for the course HPHY 253 taught by Professor Bak during the Spring '09 term at Morehouse.

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253-wrk-24 - HPHY 253 Workout 24 Gauss&s Law...

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