Answer to Question #206122 in Electrical Engineering for Simpu

Question

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).

Answer 1

(A)

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

(B)

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

using spherical coordinate system

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

(c)

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

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