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workout 24 - HPHY 253 Workout 24 Gauss Law for Electricity...

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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) ( ² 1) = ² 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 ) ² 0 : 10 m cos 60 ° ³ = 0 : 060 m 2 : Because the electric ±eld is constant on this °at surface, we obtain Z E ± da = E ± A = EA cos 60 : 0 ° = ° 7 : 80 ° 10 4 ± (0 : 060) ² 1 2 ³ = +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 box²s 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 = ² 2340 + 2340 + [ terms of zero ] = 0 as the electric °ux through the entire box surface. FYI: It follows from Gauss²s law for electricity that the net charge within the box²s volume is zero.
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