Bài giảng Engineering electromagnetic - Chapter IV: Electric Flux Density, Gauss’ Law & Divergence - Nguyễn Công Phương

Nguy ễn Công Ph ươ ng Engineering Electromagnetics Electric Flux Density, Gauss’ Law & Divergence Contents I. Introduction II. Vector Analysis III. Coulomb’s Law & Electric Field Intensity IV. Electric Flux Density, Gauss’ Law & Divergence V. Energy & Potential VI. Current & Conductors VII. Dielectrics & Capacitance VIII.Poisson’s & Laplace’s Equations IX. The Steady Magnetic Field X. Magnetic Forces & Inductance XI. Time – Varying Fields & Maxwell’s Equations XII. The Uniform Plane Wave XIII.Plane Wave Reflection & Dispersion XIV.Guided Waves & Radiation Electric Flux Density, Gauss’ Law & Divergence 2 Electric Flux Density, Gauss’ Law & Divergence 1. Electric Flux Density 2. Gauss’ Law 3. Divergence 4. Maxwell’s First Equation 5. The Vector Operator s 6. The Divergence Theorem Electric Flux Density, Gauss’ Law & Divergence 3 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 2 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 4 Electric Flux Density (2) –Q 2 2 Sa = 4 πa (m ) a Density of the flux at the inner sphere: Ψ +Q b = Q 4πa2 4 π a 2 Electric flux density: Q D= a r= a 4π a2 r Q D= a r= b 4πb2 r Q D= a a≤ r ≤ b 4π r 2 r Electric Flux Density, Gauss’ Law & Divergence 5 Electric Flux Density (3) –Q = Q D a r (a < r < b) 4π r 2 +Q b r Q D= a π 2 r 4 r → = ε DE0 = Q E a r ρ πε 2 (in free space) vdv 4 0r → D= a ∫V π 2 r (in free space) ρ dv 4 R E= v a ∫V πε 2 r 4 0R Electric Flux Density, Gauss’ Law & Divergence 6 Ex. 1 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 7 Ex. 2 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 8 Ex. 3 Electric Flux Density (6) Given 3 infinite uniform sheets (all parallel to x0y) at z = – 3, z = 2 & z = 3. Their surface charge density are 4 nC/m2, 6 nC/m2 & –9 nC/m2 respectively. Find D at P(5, 5, 5). Electric Flux Density, Gauss’ Law & Divergence 9 Electric Flux Density, Gauss’ Law & Divergence 1. Electric Flux Density 2. Gauss’ Law 3. Divergence 4. Maxwell’s First Equation 5. The Vector Operator s 6. 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 surface is equal to the total charge enclosed by that surface Q Electric Flux Density, Gauss’ Law & Divergence 11 Gauss' Law (2) ∆Ψ = flux crossing ∆S ∆S = DScos θ∆S DS, normal DS θ = DS .∆S Q P ∆S →Ψ=dψ = D . d S ∫ ∫ closed S surface Ψ=D.d S = charge enclosed = Q ∫ S S Electric Flux Density, Gauss’ Law & Divergence 12 Gauss' Law (3) Ψ= = = DS .d S charge enclosed Q ∫ S = QQ∑ n = ρ Q∫ L dL Q= ρ dS ∫S S Q= ρ dV ∫V v D.d S = ρ dv ∫SS ∫ V v Electric Flux Density, Gauss’ Law & Divergence 13 Gauss' Law (4) = Q E a r Q z 4πε r 2 →D = a 0 4π r 2 r = ε D DE0 r = a S dS → = Q DS a r (at the surface) θ 4π a2 Q x φ y → = Q ∫DSS.d ∫ 2 aS r d S S 4π a Q 2 2 → = θ θ ϕ dS = r sin θdθdφ = a sin θdθdφ ∫DS .d S ∫ sin d d S S 4π 2 → dS = a sin θdθdφar Electric Flux Density, Gauss’ Law & Divergence 14 Gauss' Law (5) = Q θ θ ϕ ∫DS .d S ∫ sin d d z S S 4π ϕ=2 π θπ = Q r = a DS = ∫ ∫ sin θd θ d ϕ dS ϕ=0 θ = 0 4π π 2π Q Q θ =( − cosθ ) d ϕ ∫0 π x φ y 4 0 2π Q = ∫ dϕ 0 2π = Q Electric Flux Density, Gauss’ Law & Divergence 15 Ex. 1 Gauss' Law (6) Given a point charge 1 nC at (2, 0, 3) & another point charge 2 nC at (4, – 5, 6). Find the total electric flux leaving the enclosed surface formed by the six planes x, y, z = ± 8. Electric Flux Density, Gauss’ Law & Divergence 16 Gauss' Law (7) • Coulomb’ law is to find E [= f(Q)] • Sometimes it is difficult to find E using Coulomb’s law • Gauss may find D (→ E) for a given Q Q= D. d S ∫S S • The solution is easy if we are able to find a closed surface satisfying 2 conditions: – DS is everywhere either normal or tangential to the closed surface, so that DS.dS becomes DSdS or zero, respectively – On that portion of that surface for which DS.dS ≠ 0, DS = const • (Gaussian surface) Electric Flux Density, Gauss’ Law & Divergence 17 z Gauss' Law (8) E = ? ρL L D= Dρ a ρ ; Dρ = f (ρ ) y Q= D. d S x ρ ∫cylinder S = + + DS dS0 dS 0 dS ∫sides ∫ top ∫ bottom =ϕ = π = z L 2 ρ ϕ = πρ → = = Q DS d dz DS 2 L DDS ρ ∫z=0 ∫ ϕ = 0 2πρ L = ρ QL L ρ ρ →D = L →E = L ρ πρ ρ πε ρ 2 2 0 Electric Flux Density, Gauss’ Law & Divergence 18 Gauss' Law (9) ρ = a ρ = b 2 coaxial cylindrical conductors. The outer surface of the inner cylinder has a ρS . = πρ Q DS 2 L (total charge of a right circular cylinder of L & ρ (a < ρ < b)) z= L ϕ = 2 π Q=ρϕ ad dz = 2 πρ aL ∫z=0 ∫ ϕ = 0 S S (total charge of the inner cylinder of length L) aρ aρ →D = S D=S a (a <ρ < b ) S ρ ρ ρ ρ==Q ρ S = ρπ.2 aa .1 = 2 πρ Ll=1 SS l = 1 s ρ L → D= a ρ 2πρ Electric Flux Density, Gauss’ Law & Divergence 19 Gauss' Law (10) ρ = a ρ = b = − QQouter cylinder innercylinder = π ρ Qouter cylinder2 bL S ,outer cylinder = π ρ Qinnercylinder2 aL S ,innercylinder →ρ = − a ρ S,outer cylinderb S ,inner cylinder Electric Flux Density, Gauss’ Law & Divergence 20 Gauss' Law (11) ρ = a ρ = b R Ψ = + = = π R, R> b QQ outer cylinder inner cylinder 0 DS, R 2 RL → = DS, R 0 The coaxial cable/capacitor has no external field & there is no field within the inner cylinder Electric Flux Density, Gauss’ Law & Divergence 21 Ex. 2 Gauss' Law (12) Consider a coaxial cable of 1-m length, its inner radius is 1mm, the outer one is 4mm. Conductors are separated by air. The total charge on the inner cylinder is 40nC. Find charge density on each conductor, E & D. Q 40× 10 −9 ρ = inner cylinder = = 6.37µ C/m 2 S,inner cylinder 2πaL 2π × 10−3 × 1 Q −40 × 10 −9 ρ = outer cylinder = = − 1.59µ C/m 2 S ,outer cylinder 2πbL 2π × 4 × 10−3 × 1 aρ ×−3 × × − 6 = S, inner cylinder =1 10 6.37 10 = 6.37 2 Dρ − − nC/m 103<ρ < 4.10 3 ρ ρ ρ D × −9 = ρ =6.37 10 = 719 2 Eρ − − − V/m 103<ρ < 4.10 3 ε × 12 ρ ρ 0 8.854 10 Electric Flux Density, Gauss’ Law & Divergence 22 D Gauss' Law (13) z 0 • The application of Gauss’ ∆z law (to find D) needs a ∆x ∆y gaussian surface y • Problem: hard to find such x surface • Solution : choose a very D = D0 = Dx0ax + Dy0ay + Dz0az small closed surface (approaching zero) Electric Flux Density, Gauss’ Law & Divergence 23 D Gauss' Law (14) z 0 Q= DS. d ∆z ∫ S S ∆x ∆y x y DS.d = + + + + + ∫S ∫∫∫∫∫∫frontback left right top bottom Because the closed surface is very small, D is almost constant over the surface ∆ ∆ ∆ ∆ ∆ ≐ DSfront. front ≐ D. y z a ≐ Dx,front y z ∫front front x Electric Flux Density, Gauss’ Law & Divergence 24 D Gauss' Law (15) z 0 ∆ ∆ ≐ Dx,front y z ∆z ∫front ∆x ∆y ∆x + × (rate of change of D with x) DDx,front ≐ x 0 x y 2 x ∆ ∂ x Dx ∂ + Dx ≐ Dx0 2 ∂x ∂x ∆x ∆x ∂ D  → +x ∆∆ D0 2 ∫ ≐ Dx0  y z front 2 ∂x  Electric Flux Density, Gauss’ Law & Divergence 25 D Gauss' Law (16) z 0 ∆ ∂  +x Dx ∆ ∆ ∫ ≐ Dx0  y z ∆z front 2 ∂x  ∆x ∆y ∆ ≐ −∆ ∆ y ≐ Dback. S back Dback .(y z a x ) x ∫back − ∆ ∆ ≐ Dx, back y z ∆x ∂ D DD≐ − x x,back x 0 2 ∂x ∆ ∂  → − +x Dx ∆ ∆ ∫ ≐ Dx0  y z back 2 ∂x  Electric Flux Density, Gauss’ Law & Divergence 26 Gauss' Law (17) ∆ ∂  +x Dx ∆ ∆ ∫ ≐ Dx0  y z front 2 ∂x  ∆ ∂  −+x Dx ∆∆ ∫ ≐ Dx0  y z back 2 ∂x  ∂D →∫ + ∫ ≐ x ∆x ∆ y ∆ z front back ∂x ∂D ∫+ ∫ ≐ y ∆x ∆ y ∆ z right left ∂y Similarly: ∂D ∫+ ∫ ≐ z ∆x ∆ y ∆ z top bottom ∂z Electric Flux Density, Gauss’ Law & Divergence 27 Gauss' Law (18) ∂D ∫+ ∫ ≐ x ∆x ∆ y ∆ z front back ∂x ∂D ∫+ ∫ ≐ y ∆∆∆x y z right left ∂y ∂D ∫+ ∫ ≐ z ∆x ∆ y ∆ z top bottom ∂z DS.d = + + + + + ∫S ∫∫∫∫∫∫frontback left right top bottom ∂ ∂DDDy ∂  →∫ DS.d≐ x + + z  ∆ xyz ∆ ∆ S ∂x ∂ y ∂ z  Electric Flux Density, Gauss’ Law & Divergence 28 D Gauss' Law (19) z 0 ∂ ∂DDDy ∂  ∆z D.d S ≐ x + +z  ∆ xyz ∆ ∆ ∫S ∂ ∂ ∂ ∆x x y z  ∆y ∆ = ∆ ∆ ∆ v xyz x y ∂ ∂DDDy ∂  →∫ D.d S ≐ x + +z  ∆ v S ∂x ∂ y ∂ z  ∂∂D ∂  DDx+y + z × ∆ Qenclosed in ∆v ≐   v ∂x ∂ y ∂ z  Electric Flux Density, Gauss’ Law & Divergence 29 Ex. 3 Gauss' Law (20) Find the approximate value for the total charge inclosed in an incremental volome – 10 3 – x – x 2 of 10 m located at the origin. Given D = e sin yax – e cos yay + 2 zaz C/m . ∂∂D ∂  DDx+y + z ×∆ Qenclosed in ∆v ≐   v ∂x ∂ y ∂ z  ∂D − ∂D x = − ex sin y →x = 0 ∂ ∂ x x x=0 ∂ ∂D Dy y = e−x sin y → = 0 ∂ ∂ y y y=0 ∂D z = 2 ∂z →( ++) −10 = Qenclosed in ∆v ≐ 0 0 2 10 0.2 nC Electric Flux Density, Gauss’ Law & Divergence 30 Ex. 4 Gauss' Law (21) dE A sphere of radius R has a uniform surface charge 1 density ρ . Find E at P? S P = ρ = ρ2 θθϕ α dQ1S dS 1 S Rsin d d r θ dS = rsin θdrd φaθ ρ dQ 1 z S R 2 dS = r sin θdθdφar dr dS = rdrd θaφ y rd θ x rsin θdφ Electric Flux Density, Gauss’ Law & Divergence 31 Ex. 4 Gauss' Law (21) dE1 A sphere of radius R has a uniform surface charge (Method 1) density ρ . Find E at P? S P = ρ = ρ2 θθϕ α dQ1S dS 1 S Rsin d d r dQ dE = 1 cos α 1 πε 2 θ 4 0RQ P ρ dQ 1 1 S R ρR2 sin θθϕ d d = S cos α 4πε R2 0 Q1 P ρ2 θθϕ− θ =S Rsin dd × rR cos 4πε R2 2+ 2 − θ 0 Q1 P r R2 rR cos π π 2 ρR2 sin θ ( rR− cos θθϕ ) dd →E = S = ??? P ∫ ∫ πε2+ 2 − θ 3/2 ϕ=0 θ = 0 40 (r R 2cos) rR Electric Flux Density, Gauss’ Law & Divergence 32 Ex. 4 Gauss' Law (22) dE dE1 A sphere of radius R has a uniform surface charge (Method 2) density ρS. Find E at P? dE dQ P 2 2 α ε = r 0E.d S Q ∫ S θ ε= ε π 2 dQ 0E.d S 0 EPr (4) r ρ 1 ∫ S S R = ρ π 2 QS (4 R ) →ε π2 = ρπ 2 0EPr(4 r ) S (4 R ) ρ R2 →E =S , r > R Pr ε 2 0r Electric Flux Density, Gauss’ Law & Divergence 33 Gauss' Law (23) Ex. 5 dE An infinitely long cylinder of radius a has a ρ S uniform surface charge density ρS. Find E? dE1 dE2 ε = 0E.d S Q ∫ S ρ ε= ε π L2 0E.d S 0 Er (2) rL ∫ S ρ L1 Q= ρ(2 π aL ) S a →ε π = ρπ 0Er(2 rL ) S (2 aL ) ρ a →E =S , r > a r ε 0r Electric Flux Density, Gauss’ Law & Divergence 34 Electric Flux Density, Gauss’ Law & Divergence 1. Electric Flux Density 2. Gauss’ Law 3. Divergence 4. Maxwell’s First Equation 5. The Vector Operator s 6. The Divergence Theorem Electric Flux Density, Gauss’ Law & Divergence 35 Divergence (1) ∂ ∂D Dy ∂D  ∫ D.d S = Q≐ x ++ z  ∆ v S ∂x ∂ y ∂ z  ∂ ∂ ∂ D.d S D Dy D  ∫ Q →++xz  ≐ S = ∂∂∂xyz  ∆ v ∆ v ∂∂ ∂ D.d S DDDy  ∫ Q →++=xz  lim S = lim ∂∂∂xyz  ∆→v0 ∆ v ∆→ v 0 ∆ v ∂∂ ∂ A.d S AAAy  ∫ →x + +z  = lim S ∂∂∂xyz  ∆v → 0 ∆ v ∫ A.d S Divergence of A=divA = lim S ∆v → 0 ∆v Electric Flux Density, Gauss’ Law & Divergence 36 Divergence (2) ∫ A.d S Divergence of A =divA = lim S ∆v → 0 ∆v • Definition : the divergence of the vector flux density A is the outflow of flux from a small closed surface per unit volume as the volume shrinks to zero • Divergence is an operation which is performed on a vector, but the result is a scalar • Divergence only tells us how much flux is leaving a small volume (on a per-unit-volume basis), not direction Electric Flux Density, Gauss’ Law & Divergence 37 Divergence (3) ∫ A.d S Divergence of A =divA = lim S ∆v → 0 ∆v ∂D ∂D ∂D div D =x +y + z (Descartes) ∂x ∂ y ∂ z ∂ ∂D ∂ 1 1 ϕ Dz divD = (ρDρ ) + + (Cylindrical) ρρ∂ ρϕ ∂ ∂ z ∂ ∂ ∂ 12 1 1 Dϕ divD = (r D ) + (sinθ Dθ ) + (Spherical) rrr2 ∂r sinθθ ∂ r sin θϕ ∂ Electric Flux Density, Gauss’ Law & Divergence 38 Ex. 1 Divergence (4) –x –x Find divergence at the origin, given D = e sin yax – e cos yay + 2 zaz C/m2. Electric Flux Density, Gauss’ Law & Divergence 39 Ex. 2 Divergence (5) Find the divergence of the following vectors: =2 3 + + a)Axy z ( aaax y z ) z b)A=ρ cos ϕ a + sin ϕ a ρ ρ z =2 θ ϕ + + c)Ar sincos( aaar θ ϕ ) Electric Flux Density, Gauss’ Law & Divergence 40 Electric Flux Density, Gauss’ Law & Divergence 1. Electric Flux Density 2. Gauss’ Law 3. Divergence 4. Maxwell’s First Equation 5. The Vector Operator s 6. The Divergence Theorem Electric Flux Density, Gauss’ Law & Divergence 41 Maxwell’s First Equation (1) ∫ D.d S Q D.d S = Q →S = ∫S ∆v ∆ v ∫ D.d S Q →limS = lim ∆→v0∆v ∆→ v 0 ∆ v D.d S Q = ∫S = ρ divD lim lim v ∆v → 0 ∆v ∆v → 0 ∆v → = ρ Maxwell’s first equation div D v Electric Flux Density, Gauss’ Law & Divergence 42 Maxwell’s First Equation (2) = ρ div D 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 Ex. Maxwell’s First Equation (3) Find ρv of the region about a point charge Q located at the origin. Q D= a 4π r 2 r ∂ ∂ ∂ 12 1 1 Dϕ divD = (r D ) + (sinθ Dθ ) + rrr2 ∂r sinθθ ∂ r sin θϕ ∂ DDθ=0, ϕ = 0 1 ∂ Q  →divD =r 2  = 0 r2∂ r4π r 2  →ρ = v 0 = ρ div D v Electric Flux Density, Gauss’ Law & Divergence 44 Electric Flux Density, Gauss’ Law & Divergence 1. Electric Flux Density 2. Gauss’ Law 3. Divergence 4. Maxwell’s First Equation 5. The Vector Operator s 6. The Divergence Theorem Electric Flux Density, Gauss’ Law & Divergence 45 ∇(1) ∂ ∂ ∂ ∇=a + a + a ∂xx ∂ y y ∂ z z ∂ ∂ ∂  ∇= ++ ++ .D aaax y z  . ()DDD xx aaa yy zz ∂x ∂ y ∂ z  ∂ ∂ ∂ =()DDD + () + () ∂xx ∂ y y ∂ z z ∂D ∂D ∂D =x +y + z = div D ∂x ∂ y ∂ z ∂D ∂D ∂D → divDD=∇= . x +y + z ∂x ∂ y ∂ z Electric Flux Density, Gauss’ Law & Divergence 46 ∇(2) Ex. – x – x 2 Given D = e sin yax – e cos yay + 2 zaz C/m , find ∇.D? Electric Flux Density, Gauss’ Law & Divergence 47 Electric Flux Density, Gauss’ Law & Divergence 1. Electric Flux Density 2. Gauss’ Law 3. Divergence 4. Maxwell’s First Equation 5. The Vector Operator s 6. 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 DS.d= Q ∫S = ρ → = =ρ = ∇ Q dv DSD.d Qv dv . dv ∫V v ∫S ∫ V ∫ V ∇.D = ρ v → DSD.d= ∇ . dv ∫S ∫ 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 z 3 Ex. The Divergence Theorem (2) 2 2 Given D = 4xy ax + z ay C/m & a rectangular parallelepiped. Verify the divergence theorem. 2 ∫DSD.d= ∫ ∇ . dV () = Q S V 1 y Left side: x D.d S = + ++ ++ ∫S ∫∫∫∫∫∫frontback left right top bottom z=3 y = 2 z=3 y = 2 = D.(dydz a ) = D dydz ∫front ∫z=0 ∫ y = 0 x=1 x ∫z=0 ∫ y = 0 x x=1 D=(4 xy ) = 4 y x x=1 x = 1 z=3 y = 2 z=3 → = 4 ydydz = 8dz = 24 C ∫front ∫z=0 ∫ y = 0 ∫z=0 Electric Flux Density, Gauss’ Law & Divergence 50 z 3 Ex. The Divergence Theorem (3) 2 2 Given D = 4xy ax + z ay C/m & a rectangular parallelepiped. Verify the divergence theorem. 2 ∫DSD.d= ∫ ∇ . dV () = Q S V 1 y Left side: x D.d S = + ++ ++ ∫S ∫∫∫∫∫∫frontback left right top bottom = = = = =z3 y 2 − = − z3 y 2 D= .(dydz a x ) D dydz ∫back ∫z= 0 ∫ y = 0 x 0 ∫z=0 ∫ y = 0 x x=0 D=(4 xy ) = 0 x x=0 x = 0 → = 0 ∫back Electric Flux Density, Gauss’ Law & Divergence 51 z 3 Ex. The Divergence Theorem (4) 2 2 Given D = 4xy ax + z ay C/m & a rectangular parallelepiped. Verify the divergence theorem. 2 ∫DSD.d= ∫ ∇ . dV () = Q S V 1 y Left side: x D.d S = + ++ ++ ∫S ∫∫∫∫∫∫frontback left right top bottom = = = = = z3 x 1 z3 x 1 D= .(dxdz a y ) = D dxdz ∫right ∫z=0 ∫ x = 0 y 2 ∫z=0 ∫ x = 0 y y=2 D=( z2 ) = z 2 y y=2 y=2 z=3 x = 1 → = z2 dxdz ∫right ∫z=0 ∫ x = 0 Electric Flux Density, Gauss’ Law & Divergence 52 z 3 Ex. The Divergence Theorem (5) 2 2 Given D = 4xy ax + z ay C/m & a rectangular parallelepiped. Verify the divergence theorem. 2 ∫DSD.d= ∫ ∇ . dV () = Q S V 1 y Left side: x D.d S = + ++ ++ ∫S ∫∫∫∫∫∫frontback left right top bottom = = = = =z3 x 1 − = − z3 x 1 D.(dxdz a y ) D dxdz ∫left ∫z= 0 ∫ x = 0 y=0 ∫z=0 ∫ x = 0 y y=0 D=( z2 ) = z 2 y y=0 y=0 z=3 x = 1 → = − z2 dxdz ∫left ∫z= 0 ∫ x = 0 Electric Flux Density, Gauss’ Law & Divergence 53 z 3 Ex. The Divergence Theorem (6) 2 2 Given D = 4xy ax + z ay C/m & a rectangular parallelepiped. Verify the divergence theorem. 2 ∫DSD.d= ∫ ∇ . dV () = Q S V 1 y Left side: x D.d S = + ++ ++ ∫S ∫∫∫∫∫∫frontback left right top bottom = = 0 ∫top ∫ bottom Electric Flux Density, Gauss’ Law & Divergence 54 z 3 Ex. The Divergence Theorem (7) 2 2 Given D = 4xy ax + z ay C/m & a rectangular parallelepiped. Verify the divergence theorem. 2 ∫DSD.d= ∫ ∇ . dV () = Q S V 1 y Left side: x D.d S = + ++ ++ ∫S ∫∫∫∫∫∫frontback left right top bottom zx==31 zx == 31 =++24 0z2 dxdz − z 2 dxdz ++ 0 0 ∫∫zx==00 ∫∫ zx == 00 = 24 C Electric Flux Density, Gauss’ Law & Divergence 55 z 3 Ex. The Divergence Theorem (8) 2 2 Given D = 4xy ax + z ay C/m & a rectangular parallelepiped. Verify the divergence theorem. 2 ∫DSD.d= ∫ ∇ . dV () = Q S V 1 y Right side: x ∇.DdV ∫V ∂D ∂D ∂D ∂ ∂ ∂ ∇=.D x +y + z =4xy + z 2 + 0 = 4y ∂x ∂ y ∂ z ∂x ∂ y ∂ z z=3 y = 2 x = 1 → ∇.DdV = 4 ydV = 4ydxdydz ∫V ∫ V ∫z=0 ∫ y = 0 ∫ x = 0 z=3 y = 2 z=3 = 4 ydydz = 8dz = 24 C ∫z=0 ∫ y = 0 ∫z=0 Electric Flux Density, Gauss’ Law & Divergence 56 z 3 Ex. The Divergence Theorem (9) 2 2 Given D = 4xy ax + z ay C/m & a rectangular parallelepiped. Verify the divergence theorem. 2 ∫DSD.d= ∫ ∇ . dV () = Q S V 1 y Left side: D.d S = 24 C x ∫S Right side: ∇.DdV = 24 C ∫V Electric Flux Density, Gauss’ Law & Divergence 57