Gauss 2017b mk

Nguyễn Công Phương Engineering Electromagnetics Electric Flux Density, Gauss’ Law & Divergence Contents I II III IV V VI VII VIII IX X XI XII XIII XIV XV Introduction Vector Analysis Coulomb’s Law & Electric Field Intensity Electric Flux Density, Gauss’ Law & Divergence Energy & Potential Current & Conductors Dielectrics & Capacitance Poisson’s & Laplace’s Equations The Steady Magnetic Field Magnetic Forces & Inductance Time – Varying Fields & Maxwell’s Equations Transmission Lines The Uniform Plane Wave Plane Wave Reflection & Dispersion Guided Waves & Radiation Engineering Electromagnetics - sites.google.com/site/ncpdhbkhn Electric Flux Density, Gauss’ Law & Divergence Electric Flux Density Gauss’ Law Divergence Maxwell’s First Equation The Vector Operator The Divergence Theorem Electric Flux Density, Gauss’ Law & Divergence Electric Flux Density (1) • M Faraday (1837) • Phenomenon: the total charge on the outer sphere was equal in magnitude to the original charge placed on the inner sphere, regardless of the dielectric material between the spheres • Conclusion: there was a “displacement” from the inner sphere to the outer, independent of the medium: Ψ=Q • Ψ: electric flux Electric Flux Density, Gauss’ Law & Divergence Electric Flux Density (2) Sa = 4πa2 (m2) –Q a Density of the flux at the inner sphere: Ψ Q = 4π a 4π a +Q b Electric flux density: D r= a D r =b Q = a r 4π a Q = a r 4π b D a ≤ r≤ b Q = a r 4π r Electric Flux Density, Gauss’ Law & Divergence Electric Flux Density (3) –Q Q D= a (a < r < b) r 4π r b +Q r Q D= a r 4π r Q E= a r 4πε r (in free space) → D = ε 0E (in free space) E=∫ V ρv dv a r 4πε R → D=∫ Electric Flux Density, Gauss’ Law & Divergence V ρv dv a r 4π R Ex Electric Flux Density (4) Infinite uniform line charge of 10 nC/m lie along the x & y axes in free space Find D at (0, 0, 3) Electric Flux Density, Gauss’ Law & Divergence Ex Electric Flux Density (5) The x & y axes are charged with uniform line charge of 10 nC/m A point charge of 20nC is located at (3, 3, 0) The whole system is in free space Find D at (0, 0, 3) Electric Flux Density, Gauss’ Law & Divergence Ex Electric Flux Density (6) Given infinite uniform sheets (all parallel to x0y) at z = – 3, z = & z = Their surface charge density are nC/m2, nC/m2 & –9 nC/m2 respectively Find D at P(5, 5, 5) Electric Flux Density, Gauss’ Law & Divergence Electric Flux Density, Gauss’ Law & Divergence Electric Flux Density Gauss’ Law Divergence Maxwell’s First Equation The Vector Operator The Divergence Theorem Electric Flux Density, Gauss’ Law & Divergence 10 Maxwell’s First Equation (2) div D = ρ v • Apply to electrostatic & steady magnetic fields • The electric flux per unit volume leaving a vanishingly small volume unit is exactly equal to the volume charge density there Electric Flux Density, Gauss’ Law & Divergence 43 Maxwell’s First Equation (3) Ex Given D = 4xyax + z2ay C/m2, find ρv of the region about P(1,1,1) ρ v = div D ∂ ∂ ∂ ∂Dx ∂Dy ∂Dz = xy + z + = 4y div D = + + ∂x ∂y ∂z ∂x ∂y ∂z → ρv = y → ρ v , P = ρv = × = C/ m x = y = z =1 Electric Flux Density, Gauss’ Law & Divergence 44 Electric Flux Density, Gauss’ Law & Divergence Electric Flux Density Gauss’ Law Divergence Maxwell’s First Equation The Vector Operator The Divergence Theorem Electric Flux Density, Gauss’ Law & Divergence 45 ∇(1) ∂ ∂ ∂ ∇ = a x + a y + az ∂x ∂y ∂z ∂ ∂ ∂  ∇.D =  a x + a y + a z  ( Dx a x + Dy a y + Dz a z ) ∂y ∂z   ∂x ∂ ∂ ∂ = ( Dx ) + ( Dy ) + ( Dz ) ∂x ∂y ∂z ∂Dx ∂Dy ∂Dz = + + = div D ∂x ∂y ∂z → ∂Dx ∂Dy ∂Dz div D = ∇.D = + + ∂x ∂y ∂z Electric Flux Density, Gauss’ Law & Divergence 46 ∇(2) Ex Given D = e – x sinyax – e – x cosyay + 2zaz C/m2, find ∇.D ? Electric Flux Density, Gauss’ Law & Divergence 47 Electric Flux Density, Gauss’ Law & Divergence Electric Flux Density Gauss’ Law Divergence Maxwell’s First Equation The Vector Operator The Divergence Theorem Electric Flux Density, Gauss’ Law & Divergence 48 The Divergence Theorem (1) • Applies to any vector field for which the appropriate partial derivatives exist ∫ S D.dS = Q Q = ∫ ρv dv V ∇.D = ρ v → → ∫ ∫ S S D.dS = Q = ∫ ρv dv = ∫ ∇.Ddv V V D.dS = ∫ ∇.Ddv V • Theorem: the integral of the normal component of any vector field over a closed surface is equal to the integral of the divergence of this vector field throughout the volume enclosed by the closed surface Electric Flux Density, Gauss’ Law & Divergence 49 The Divergence Theorem (2) z Ex Given D = 4xyax + z2ay C/m2 & a rectangular parallelepiped Verify the divergence theorem ∫ S D.dS = ∫ ∇.DdV ( = Q) V Left side: ∫ S D.dS = ∫ ∫ front front +∫ z =3 y =2 =∫ z =0 back ∫ y =0 +∫ left +∫ +∫ top right D x=1 (dydza x ) = ∫ ∫ Dx →∫ front =∫ z =3 z =0 ∫ y =2 ydydz = ∫ y= z= z =0 z =3 z =0 x =1 x y +∫ bottom y =2 y =0 Dx x =1 dydz = (4 xy ) x =1 = y 8dz = 24 C Electric Flux Density, Gauss’ Law & Divergence 50 The Divergence Theorem (3) z Ex Given D = 4xyax + z2ay C/m2 & a rectangular parallelepiped Verify the divergence theorem ∫ S D.dS = ∫ ∇.DdV ( = Q) V Left side: ∫ S ∫ D.dS = ∫ back front =∫ z =3 z =0 ∫ +∫ back y =2 y= +∫ left +∫ right +∫ top D x =0 ( − dydza x ) = − z =3 Dx →∫ back x =0 x +∫ bottom ∫ ∫ z =0 y y =2 y =0 Dx x =0 dydz = (4 xy ) x =0 = =0 Electric Flux Density, Gauss’ Law & Divergence 51 The Divergence Theorem (4) z Ex Given D = 4xyax + z2ay C/m2 & a rectangular parallelepiped Verify the divergence theorem ∫ S D.dS = ∫ ∇.DdV ( = Q) V Left side: ∫ S D.dS = ∫ +∫ back front ∫ right =∫ z =3 z =0 ∫ +∫ left +∫ right x =1 D y=2 ( dxdza y ) = x =0 +∫ top +∫ z =3 x =1 z =0 x= →∫ right =∫ z= z= ∫ y =2 x =1 x= x bottom ∫ ∫ Dy y Dy y= = ( z2 ) dxdz y= = z2 z dxdz Electric Flux Density, Gauss’ Law & Divergence 52 The Divergence Theorem (5) z Ex Given D = 4xyax + z2ay C/m2 & a rectangular parallelepiped Verify the divergence theorem ∫ S D.dS = ∫ ∇.DdV ( = Q) V Left side: ∫ S D.dS = ∫ front ∫ left =∫ z =3 z =0 ∫ +∫ back x =1 x =0 +∫ left +∫ right +∫ Dy →∫ left = −∫ z =3 z =0 ∫ bottom z =3 z =0 x =0 ∫ x =1 x= y =0 x =1 x +∫ top D y =0 (− dxdza y ) = − ∫ y Dy y= = ( z2 ) dxdz y =0 = z2 z 2dxdz Electric Flux Density, Gauss’ Law & Divergence 53 The Divergence Theorem (6) z Ex Given D = 4xyax + z2ay C/m2 & a rectangular parallelepiped Verify the divergence theorem ∫ S D.dS = ∫ ∇.DdV ( = Q) V Left side: ∫ S D.dS = ∫ front +∫ +∫ back left ∫ top +∫ right =∫ bottom +∫ top x y +∫ bottom =0 Electric Flux Density, Gauss’ Law & Divergence 54 The Divergence Theorem (7) z Ex Given D = 4xyax + z2ay C/m2 & a rectangular parallelepiped Verify the divergence theorem ∫ S D.dS = ∫ ∇.DdV ( = Q) V Left side: ∫ S D.dS = ∫ front +∫ back = 24 + + ∫ z =3 z =0 +∫ +∫ left ∫ x =1 x =0 right +∫ top z dxdz − ∫ z =3 z =0 ∫ x y +∫ bottom x =1 x =0 z 2dxdz + + = 24 C Electric Flux Density, Gauss’ Law & Divergence 55 The Divergence Theorem (8) z Ex Given D = 4xyax + z2ay C/m2 & a rectangular parallelepiped Verify the divergence theorem ∫ S D.dS = ∫ ∇.DdV ( = Q) V Right side: ∫ V x y ∇.DdV ∂ ∂ ∂ ∂Dx ∂Dy ∂Dz = xy + z + = 4y ∇.D = + + ∂x ∂y ∂z ∂x ∂y ∂z z =3 y= → ∫ ∇.DdV = ∫ ydV = ∫z =0 ∫ y =0 V V =∫ z =3 z =0 ∫ y =2 y =0 ∫ x =1 x= ydxdydz z =3 ydydz = ∫z =0 8dz = 24 C Electric Flux Density, Gauss’ Law & Divergence 56 The Divergence Theorem (9) z Ex Given D = 4xyax + z2ay C/m2 & a rectangular parallelepiped Verify the divergence theorem ∫ S D.dS = ∫ ∇.DdV ( = Q) V ∫ Right side: ∇.DdV = 24 ∫V Left side: S D.dS = 24 C x y C Electric Flux Density, Gauss’ Law & Divergence 57 ... 8.854 × 10−12 ρ ρ Electric Flux Density, Gauss Law & Divergence = 6.37 ρ nC/m 22 z Gauss' Law (13) • The application of Gauss law (to find D) needs a gaussian surface • Problem: hard to find... respectively Find D at P(5, 5, 5) Electric Flux Density, Gauss Law & Divergence Electric Flux Density, Gauss Law & Divergence Electric Flux Density Gauss Law Divergence Maxwell’s First Equation The... Operator The Divergence Theorem Electric Flux Density, Gauss Law & Divergence 10 Gauss' Law (1) • Generalization of Faraday’s experiment • Gauss law: the electric flux passing through any closed