Answer to Question #206122 in Electrical Engineering for Simpu

Question #206122

3.18 The electric flux density is given as D=6ya, +2xa,+ 14xya mC/m². Determine the electric flux passing through (a) a rectangular window defined by (2, 0, 0), (0, 2, 0), (0, 2, 2), and (2, 0, 2), (b) a circle of 10-cm radius in the xy plane at z = 0, and (c) a triangular region bounded by (0.0, 0), (2, 0, 0), and (0, 2, 0).


1
Expert's answer
2021-06-21T07:33:03-0400

(A)

"\\bold{\\vec{D}}=6ya_x + 2xa_y + 14xya_z mC\/m^2 \\\\\n\n\n\n\n\n\n\n\\psi = \\oint \\bold{\\vec{D}}ds\\\\\n\n\n\n\n\n\n\nds=dxdza_y+dydza_x\\\\\n\n\n\n\\psi= \\int 6ydxdy + \\int 2xdxdz \\\\\n\n\n\n\\psi= 6\\int _0^2dx\\int _0^2 ydy + 2\\int _0^2xdx\\int _0^2dz\\\\\n\n\n\n\\psi= 3(2)(2^2)+(2^2)(2)=32mC\\\\"

(B)

Since the direction of the flux is along the z - axis

using spherical coordinate system

"ds=\\rho d\\rho d\\phi a_z\\\\\n\n\n\nAlso\\\\\n\n\n\n\\bold{\\vec{D}}=14xy\\\\\n\n\n\nx=\\rho \\cos \\phi \\\\\n\n\n\ny=\\rho \\sin \\phi \\\\\n\n\n\n\n\n\n\n\\psi=\\int \\bold{\\vec{D}}ds\\\\\n\n\n\n\\psi=14\\int _0^{0.1}\\rho^3 d\\rho \\int _0^{4\\pi}\\sin \\phi \\cos \\phi d\\phi \\\\\n\n\n\n\\psi=\\frac {14}{16}({\\rho^4}_0^{0.1})(-\\cos 2\\phi_0^{4\\pi})\\\\\n\n\n\n\\psi=\\frac{7}{8}(0.1^4-0)(-\\cos8 \\pi+ \\cos 0)\\\\\n\n\n\n\\psi=0C\\\\"

(c)

Since the direction of the flux is in the z-axis

"ds=dxdya_z\\\\\n\n\n\n\\psi=14\\int _0^2xdx \\int _2^{2-x}ydy\\\\\n\n\\psi=7\\int _0^2( (y^2) _2^{2-x}))xdx\\\\\n\n\\psi=7\\int_0^2((2-x)^2-4)xdx\\\\\n\n\\psi=-46.67mC\\\\\n\n\\psi=-46.67mC"



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