Figure 23-31 shows a closed Gaussian surface in the shape of a cube of edge leng

ID: 2019994 • Letter: F

Question

Figure 23-31 shows a closed Gaussian surface in the shape of a cube of edge length 2.5 m, with one corner at x1 = 5.0 m,y1 = 3.7 m. The cube lies in a region where the electric field vector is given by = -3.4 -4.4 y2 +2.8 N/C, with y in meters. What is the net charge (in Coulombs) contained by the cube?

Explanation / Answer

length of the edge of the cube L = 2.5 m distances x1 = 5 m and y1 = 3.7 m electric field vector E = [-3.4 i^ - 4.4y2 j^ + 2.8 k^] N/C from Gauss's law , electric flux inside a closed surface is = E.dA ........................ (1) non zero component of electric field is Enon zero = - 4.4y2 j^ N/C the face of a cube located at y1 = 3.7 m has an area is A = (2.5 m)(2.5 m) = 6.25 m2 this area directed along +j^ direction. this area directed along +j^ direction. the flux through the face is = Enon zero (A) = (- 4.4y2 N/C)(6.25 m2) = (- 4.4(3.7 m)2 N/C)(6.25 m2) = -376.475 Wb .................................................................................. the face of cube located at y = 3.7 m - 2.5 m = 1.2 m this area directed along -j^ direction. the flux through the face is = Enon zero (A) = (- 4.4y2 N/C)(6.25 m2) cos180 = (- 4.4(1.2 m)2 N/C)(6.25 m2)(-1) = 39.6 Wb the flux through the face is = Enon zero (A) = (- 4.4y2 N/C)(6.25 m2) cos180 = (- 4.4(1.2 m)2 N/C)(6.25 m2)(-1) = 39.6 Wb hence , the net electric flux is = -376.475 Wb + 39.6 Wb = -336.875 Wb using Gauss's law , charge Q = 0 = (8.85*10-12 C2/N.m2)(-336.875 Wb) = -2.98*10-9 C = -2.98 nC

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Figure 23-31 shows a closed Gaussian surface in the shape of a cube of edge leng

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