Bài giảng Engineering electromagnetic - Chapter VIII: Poisson’s & Laplace’s Equations - Nguyễn Công Phương

Nguy ễn Công Ph ươ ng Engineering Electromagnetics Poisson’s & Laplace’s Equations 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 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Poisson’s & Laplace’s Equations 1. Poisson’s Equation 2. Laplace’s Equation 3. Uniqueness Theorem 4. Examples of the Solution of Laplace’s Equation 5. Examples of the Solution of Poisson’s Equation 6. Product Solution of Laplace’s Equation 7. The Iteration Method Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Poisson's Equation (1) ∇ = ρ Gauss’s Law: .D v = ε →∇ =∇ε =−∇ ε ∇ = ρ DE .D .() E . (V ) v 0 ρ Gradient: E = −∇V →∇∇. V =− v ε (Poisson's Equation) ∂V ∂ V ∂ V ∇=V a + a + a ∂xx ∂ y y ∂ z z ∂A ∂A ∂A ∇=.A x +y + z ∂x ∂ y ∂ z ∂ 2 2 2 ∂∂VV ∂Vy  ∂  ∂ ∂∂∂V V V →∇∇=. V x  +  +z  =++ ∂∂xx  ∂∂ yy  ∂∂ zz  ∂ xy2 ∂ 2 ∂ z 2 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Poisson's Equation (2) ρ ∇∇. V =− v ε 2 2 2 2 2 2 ∂V ∂ V ∂ V ρ ∂V ∂ V ∂ V ∇=2V + + =− v ∇∇=. V + + ∂2 ∂ 2 ∂ 2 ε ∂x2 ∂ y 2 ∂ z 2 x y z Define ∇. ∇ = ∇2 (rectangular) 1∂∂V  1 ∂∂2 V 2 V ρ ρ  + + =− v (cylindrical) ρρ∂∂ ρρϕ  2 ∂∂ 2z 2 ε 1∂∂V 1 ∂∂  V  1 ∂ 2 V ρ r 2 + sin θ  + =− v rrrr2∂∂ 2 sinθθ ∂∂  θ  r 2 sin 2 θϕ ∂ 2 ε (spherical) Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Ex. Poisson’s Equation (3) Find the Laplacian of the following scalar fields: a) A= 2 xy2 z 3 cos2 ϕ b) B = ρ 20sin θ c) C = r3 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Poisson’s & Laplace’s Equations 1. Poisson’s Equation 2. Laplace’s Equation 3. Uniqueness Theorem 4. Examples of the Solution of Laplace’s Equation 5. Examples of the Solution of Poisson’s Equation 6. Product Solution of Laplace’s Equation 7. The Iteration Method Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Laplace’s Equation ∂2V ∂ 2 V ∂ 2 V ρ Poisson's Equation: ∇=2V + + =− v ∂x2 ∂ y 2 ∂ z 2 ε ρ = v 0 ∂2V ∂ 2 V ∂ 2 V ∇=2V + + = 0 (Laplace’s equation, rectangular) ∂x2 ∂ y 2 ∂ z 2 1∂∂V  1 ∂∂2 V 2 V ρ  + + = 0 (cylindrical) ρ∂∂ ρ ρ  ρ2 ∂∂ ϕ 2z 2 1∂∂V 1 ∂∂  V  1 ∂ 2 V r2 + sinθ  + = 0 rrrr2∂∂ 2 sinθ ∂∂ θ  θ  r 2 sin 2 θ ∂ ϕ 2 (spherical) Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Poisson’s & Laplace’s Equations 1. Poisson’s Equation 2. Laplace’s Equation 3. Uniqueness Theorem 4. Examples of the Solution of Laplace’s Equation 5. Examples of the Solution of Poisson’s Equation 6. Product Solution of Laplace’s Equation 7. The Iteration Method Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Uniqueness Theorem (1) ∂2V ∂ 2 V ∂ 2 V ∇=2V + + = 0 ∂x2 ∂ y 2 ∂ z 2 Assume two solutions V1 & V2, : ∇2V = 0 1 →∇2 (V − V ) = 0 ∇2 = 1 2 V2 0 → = = Assume the boundary condition Vb V1b V 2 b V b ∇.D()V = V ( ∇ .D ) + D. () ∇ V = − V V1 V 2 = ∇ − D (V1 V 2 ) →∇ −∇− = − ∇∇− + .[(VV12 ) ( VV 12 )] ( VV 12 )[ . ( VV 12 )] +∇ − ∇ − (VV12 )(. VV 12 ) Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Uniqueness Theorem (2) ∇ −∇− =− ∇∇− +∇− ∇− .[(VVVV1212 )( )]( VV 12 )[ . ( VV 12 )]( VV 12 )( . VV 12 ) →∇−∇−.[(VV ) ( VVdv )] = ( VV −∇∇− )[ . ( VVdv )] + ∫V 1212 ∫ V 12 12 +∇(V − V )(. ∇ V − Vdv ) ∫V 12 12 Divergence theorem: DSD.d= ∇ . dv ∫S ∫ V →∇−∇− = −∇− .[(VV1212 ) ( VVdv )] [( VV 12b b ) ( VV 12 b b )] . d S ∫V ∫ S = = V1b V 2 b V b →∇.[(V −∇− V ) ( V V )] dv = 0 ∫V 1 2 1 2 →=0(VV − )[( ∇∇−. VVdv )] +∇− ( VV )( . ∇− VVdv ) ∫V 12 12 ∫ V 1212 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Uniqueness Theorem (3) (VV− )[( ∇∇−. VVdv )] +∇−∇− ( VV )( . VVdv )0 = ∫V 12 12 ∫ V 1212 ∇∇ − =∇2 − = . (VV12 ) ( VV 12 )0 →∇− ∇− = =[ ∇ − ]2 (V V )(. V V ) dv 0 (V1 V 2 ) dv ∫V 12 12 ∫V [∇ −]2 ≥ (V1 V 2 ) 0 →∇ −2 = →∇ − = [ (V1 V 2 )] 0 (V1 V 2 ) 0 → − = ∂V ∂ V ∂ V V1 V 2 const ∇=V a + a + a ∂xx ∂ y y ∂ z z V1 = V2 At boundary V1 = Vb1, V2 = Vb2 → const = V – V = 0 = = b1 b2 V1b V 2 b V b Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Poisson’s & Laplace’s Equations 1. Poisson’s Equation 2. Laplace’s Equation 3. Uniqueness Theorem 4. Examples of the Solution of Laplace’s Equation 5. Examples of the Solution of Poisson’s Equation 6. Product Solution of Laplace’s Equation 7. The Iteration Method Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Examples of the Solution of Laplace’s Equation (1) Ex. 1 Assume V = V(x) d2 V ∂2V ∂ 2 V ∂ 2 V → = 0 →V = Ax + B ∇=2V + + = 0 dx 2 ∂x2 ∂ y 2 ∂ z 2 = V= V 1 x x 1 = V= V 2 x x 2  − = V1 V 2 A  x− x Vxx(− )( − Vxx − ) → 1 2 →V = 1 22 1  − −  Vx21 Vx 12 x1 x 2 V x B = →V = 0  x− x V = 0  1 2 x=0 d V= V x= d 0 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Examples of the Solution of Laplace’s Equation (2) Ex. 1 x V = V(x) V x V = 0 →V = 0 Conductor surface x = d x=0 d V= V x= d 0 E = −∇V V → = − 0 V Conductor surface x = 0 E a x → = − ε 0 d D a x d DE= ε V V V →D = D =− ε 0 a →D = − ε 0 →ρ =D =− ε 0 Sx=0 d x N d S N d −εV V S Q ε S →Q =ρ dS = 0 dS = − ε 0 →C = = ∫SS ∫ S d d V0 d Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Examples of the Solution of Laplace’s Equation (3) Ex. 2 Assume V = V(ρ) (cylindrical) 1 ∂ ∂ V  1∂∂V  1 ∂∂2 V 2 V →ρ  = 0 ∇=2V ρ  + += 0 ρ∂ ρ ∂ ρ  ρ∂∂ ρ ρ  ρ2 ∂∂ ϕ 2z 2 1 d dV  d dV  dV →ρ  = 0 →ρ  = 0→ρ = A ρd ρ d ρ  dρ d ρ  dρ →V = Aln ρ + B  V = 0 ln(b /ρ ) = + = A →V = V Vρ= AaBVln 0  lna− ln b 0 a →  ln(b / a ) V= AbBln += 0 ( ba > ) Vln b ρ=b B = − 0  lna− ln b Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Examples of the Solution of Laplace’s Equation (4) Ex. 2 Assume V = V(ρ) (cylindrical) ln(b /ρ ) 2 2 → = 1∂∂V  1 ∂∂ V V V V 0 ∇=2V ρ  + += 0 ln(b / a ) ρ∂∂ ρ ρ  ρ2 ∂∂ ϕ 2z 2 V0 →E = −∇V = a ρ ρ ln(b / a ) ε V0 →D ρ= = = ρ N( a ) aln( b / a ) S εV2 π aL Qε2 π L →Q =ρ dS = 0 →C = = ∫S S aln( b / a ) V0 ln( b / a ) Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Examples of the Solution of Laplace’s Equation (5) Ex. 3 z Assume V = V(φ) (cylindrical) ∂∂ ∂∂2 2 Air gap 2 1V  1 V V ∇=V ρ  + += 0 ρρ∂∂ ρ  ρϕ2 ∂∂ 2z 2 1 ∂2V ∂2V → = → = →V = Aϕ + B 2 2 0 2 0 ρ∂ ϕ ∂ϕ α = ϕ B 0 →V = V V= B = 0  0 α ϕ =0 →  V =α + = A = 0 Vϕ= α A BV 0  α V →E = −∇V = − 0 a αρ ϕ Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Examples of the Solution of Laplace’s Equation (6) Ex. 4 Assume V = V(θ) (spherical) ∂∂ ∂∂ ∂ 2 21 2 V 1  V  1 V ∇=V r + sinθ  + = 0 rrrr2∂∂ 2 sinθθ ∂∂  θ  r 2 sin 2 θϕ ∂ 2 1 ∂ ∂ V  →sinθ  = 0 r 2 sin θ∂ θ ∂ θ  ∂ ∂ V  dV →sinθ  = 0 →sin θ = A ∂θ ∂ θ  dθ Assume r ≠ 0; θ ≠ 0; θ ≠ π dθ dθ θ  →dV = A →V = A + B =Aln tan  + B sin θ ∫ sin θ 2  Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Examples of the Solution of Laplace’s Equation (7) Ex. 4 θ  Assume V = V(θ) →V = Aln tan  + B 2  α V = 0 θ= π / 2 V = V0 =α < π Vθ= α V 0 ( / 2) Air gap θ   V = 0 ln tan  ∂ 2  1 V V 0 →V = V →E =−∇V =− aθ =− a θ 0 α  r ∂θ α  ln tan  rsinθ ln tan  2  2  εV →ρ =D = ε E =− 0 S N α  rsinα ln tan  2  Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Examples of the Solution of Laplace’s Equation (8) Ex. 4 εV Assume V = V(θ) →ρ = − 0 S α  r sinα ln tan  α 2  V = V εV 0 →=Qρ dS =− 0 dS ∫SS ∫ S α  Air gap rsinα ln tan  2  dS= rsin α d ϕ dr V = 0 −εV∞ 2π rsin α ddr ϕ −2πε V ∞ →Q = 0 ∫ ∫ = 0 ∫ dr α  0 0 r α  0 sinα ln tan  ln tan  2  2  Q2πε r →C = ≐ 1 V α  0 ln cot  Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn 2  Poisson’s & Laplace’s Equations 1. Poisson’s Equation 2. Laplace’s Equation 3. Uniqueness Theorem 4. Examples of the Solution of Laplace’s Equation 5. Examples of the Solution of Poisson’s Equation 6. Product Solution of Laplace’s Equation 7. The Iteration Method Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn ρ v Examples of the Solution of ρ v0 Poisson’s Equation (1) 1 p-type n-type ρ= ρ x x 0,5 v2 v 0 sech th –5 –4 –3 –2 –1 a a − ρ x− x v =2 = e e 1 2 3 4 5 ( sechx− ; th x − ) –0,5 x/ a eexx+ ee xx + ρ –1 ∇2 = − v ε Poisson’s Equation : V Ex ε ρ dV2 2ρ xx –5 –4 –3 –2 –1 2 v0a → = − v0 sech th 2 ε E 1 2 3 4 5 dx a a x –0,5 x/ a dV2ρ a x → =v0 sech + C –1 dxε a 1 2ρ a x →=−Ev0 sech − C dV x ε 1 → = E = − a C1 0 x → → dx If x ± ∞ then Ex 0 2ρ a x →E = − v0 sech x ε a Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn ρ v Examples of the Solution of ρ v0 Poisson’s Equation (2) 1 p-type n-type ρ= ρ x x 0,5 v2 v 0 sech th a a ρ –5 –4 –3 –2 –1 ρ v 1 2 3 4 5 Poisson’s equation : ∇2V = − v ε –0,5 x/ a –1 2ρ a x ε →E = − v0 sech Ex x ε ρ a –5 –4 –3 –2 –1 2 v0a ρ 2 4 v0a x/ a E 1 2 3 4 5 →V =arctg e + C x –0,5 ε 2 x/ a 2 –1 4ρ a π πε V Supp. V = 0 →0 =v0 + C 0,5 x=0 2 ρ 2 ε 4 2 v0a 0,25 ρ 2 π –5 –4 –3 –2 –1 4 v0a x/ a  →V =arctg e −  V 1 2 3 4 5 ε 4  –0,25 x/ a –0,5 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn ρ v Examples of the Solution of ρ v0 Poisson’s Equation (3) 1 ρ= ρ x x p-type n-type v2 v 0 sech th 0,5 a a ρ –5 –4 –3 –2 –1 4ρ a2 π  v 1 2 3 4 5 V=v0 arctg e x/ a − –0,5 x/ a ε   4  –1 2πρ a2 V= V − V = v0 0 x→∞ x →−∞ ε xx∞ xx Q==ρρ dv2 sech th dv = S 2 ρ sech th dx = 2 ρ aS ∫Vvv ∫ V 0aa ∫ 0 v 0 aa v 0 ρ ε → = 2 v0V 0 Q S ρ ε ε S π →C = S v0 = π π dQdV dQ 2V0 2 a I= = C0 →= C dt dt dV 0 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Poisson’s & Laplace’s Equations 1. Poisson’s Equation 2. Laplace’s Equation 3. Uniqueness Theorem 4. Examples of the Solution of Laplace’s Equation 5. Examples of the Solution of Poisson’s Equation 6. Product Solution of Laplace’s Equation 7. The Iteration Method Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Product Solution of Laplace’s Equation (1) • Previous examples assumed that V varies with one of the three coordinates • The product solution can be used to solve for V(x, y) ∂2V ∂ 2 V ∂ 2 V ∇=2V + + = 0 ∂2V ∂ 2 V ∂x2 ∂ y 2 ∂ z 2 → + = 0 ∂x2 ∂ y 2 V= Vxy( , ) • Assume V = XY , X = X(x), Y = Y(y) ∂2X ∂ 2 Y dX2 dY 2 →Y + X = 0 →Y + X = 0 ∂x2 ∂ y 2 dx2 dy 2 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Product Solution of Laplace’s Equation (2) dX2 dY 2 1dX2 1 dY 2 1dX2 1 dY 2 Y+ X = 0 → + = 0 → = − dx2 dy 2 X dx2 Y dy 2 X dx2 Y dy 2 1 d2 X involves no y X dx 2 1 d2 Y − involves no x Y dy 2  2 1 d X = α 2  2 →  X dx  2 −1 d Y = α 2  Y dy 2 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Product Solution of Laplace’s Equation (3) V= V(,)ρ ϕ = R ()() ρ Φ ϕ ρ ∂ ∂R  1 ∂2 Φ 1∂∂V  1 ∂∂2 V 2 V →ρ  + = 0 ρ  + + = 0 R ∂ρ ∂ ρ  Φ∂ ϕ 2 ρ∂∂ ρ ρ  ρ2 ∂∂ ϕ 2z 2 ρ d dR   ρ  = α 2 R dρ d ρ  →  2Φ −1 d = α 2  Φ dϕ 2 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Product Solution of Laplace’s Equation (4) V= V(,)ρθ = R ()() ρ Θ θ ∂∂ ∂∂ ∂ 2 12 V 1  V  1 V r + sinθ  + = 0 rrrr2∂∂ 2sinθ ∂∂ θ  θ  r 222 sin θ ∂ ϕ ρ∂2R2 ρ ∂ R  1 ∂ 2 Θ 1 ∂Θ  → +++   = 0 R2∂ρ 2 R ∂ ρ  Θ∂ θ 2 Θ∂tg θθ   ρ∂2 R2 ρ ∂ R  + =n( n + 1)  R2∂ρ 2 R ∂ ρ →  1∂2 Θ 1 ∂Θ  + =−+n( n 1) Θ∂θ2 Θtg θ ∂ θ Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Product Solution of Laplace’s Equation (5) Ex. Solve for the potential & EFI in the vicinity of a conducting cylinder (posses an infinite length) in a uniform EFI E0. The permittivities of the enviroment & the cylinder are ε1 & ε2 respectively. EFI is perpendicular to the cylinder’s axis. ρ d dR   ρ  = α 2 R dρ d ρ  V= V(,)ρϕ = R ()() ρ Φ ϕ →  2Φ −1 d = α 2  Φ dϕ 2 Φ=ϕ ϕ + ϕ =± α Assume ( )∑Ap cos p ∑ B p sin pp , V()ϕ=− V ();() ϕ V ϕ =− V ( π − ϕ ) →Φ=ϕ ϕ α = ()A1 cos, 1 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Product Solution of Laplace’s Equation (6) Ex. Solve for the potential & EFI in the vicinity of a conducting cylinder (posses an infinite length) in a uniform EFI E0. The permittivities of the enviroment & the cylinder are ε1 & ε2 respectively. EFI is perpendicular to the cylinder’s axis.  2Φ −1 d = α 2 →Φ=ϕ ϕ α =  ()A1 cos, 1  Φ dϕ 2  ρ d dR   ρ= α 2 ρ 2 ρ k   d dR  k B 2 R dρ d ρ →ρ =k = α → = ±    ρ ρ  ρ k k 1 k Rd d  B Assume R(ρ ) = B ρ k k α =1 →ρ =+ ρ + − ρ − 1 R( ) B1 B 1 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Product Solution of Laplace’s Equation (7) Ex. Solve for the potential & EFI in the vicinity of a conducting cylinder (posses an infinite length) in a uniform EFI E0. The permittivities of the enviroment & the cylinder are ε1 & ε2 respectively. EFI is perpendicular to the cylinder’s axis.  1 d 2Φ − =α2 →Φ( ϕ ) = A cos ϕ  Φ dϕ 2 1  ρ    dρ dR =→ α2 ρ =+ ρρ + − − 1   R( ) B1 B 1 R dρ d ρ  V= V(,)ρϕ = R ()() ρ Φ ϕ → =+ρ ϕ + − ρ − 1 ϕ =+ρ ϕ + − ρ − 1 ϕ VAB11cos AB 11 cos CCcos cos Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Product Solution of Laplace’s Equation (8) Ex. Solve for the potential & EFI in the vicinity of a conducting cylinder (posses an infinite length) in a uniform EFI E0. The permittivities of the enviroment & the cylinder are ε1 & ε2 respectively. EFI is perpendicular to the cylinder’s axis. =+ρ ϕ + − ρ − 1 ϕ y Exterior: VCC1 1cos 1 cos E0 =+ρ ϕ + − ρ − 1 ϕ Interior: VCC2 2cos 2 cos ε1 V= E x ρ −∞0 x →−∞ →+ = − ε = = − + C1 E 0 2 θ VV−∞1θ = π ρ →−∞ Cx 1 , x→−∞ x C − V= V =2 → ∞ gèc täa ®é 2 ρ→0 ρ →C − = 0 ρ→0 2 EFI is finite at the origin Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Product Solution of Laplace’s Equation (9) Ex. Solve for the potential & EFI in the vicinity of a conducting cylinder (posses an infinite length) in a uniform EFI E0. The permittivities of the enviroment & the cylinder are ε1 & ε2 respectively. EFI is perpendicular to the cylinder’s axis. =+ρ ϕ + − ρ − 1 ϕ y Exterior: VCC1 1cos 1 cos E0 Interior: VCC=+ρcos ϕ + − ρ − 1 cos ϕ 2 2 2 ε += − − = 1 ρ CEC1 0, 2 0 ε 2 θ − − VEC= −ρcos ϕ + ρ1 cos ϕ x →  1 0 1 = + ρ ϕ VC2 2 cos Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Product Solution of Laplace’s Equation (10) Ex. Solve for the potential & EFI in the vicinity of a conducting cylinder (posses an infinite length) in a uniform EFI E0. The permittivities of the enviroment & the cylinder are ε1 & ε2 respectively. EFI is perpendicular to the cylinder’s axis. = −ρ ϕ + − ρ − 1 ϕ y VEC1 0cos 1 cos E0 = + ρ ϕ VC2 2 cos ε1 V= V ρ 1ρ=a 2 ρ = a ε2 →−+(EaCa− −1 )cosϕ = Ca + cos ϕ θ 0 1 2 x ∂ ∂ a εV1= ε V 2 1∂ρ 2 ∂ ρ ρ=a ρ = a →−−ε− −2 ϕ = ε + ϕ 101(ECa )cos 22 Ca cos Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Product Solution of Laplace’s Equation (11) Ex. Solve for the potential & EFI in the vicinity of a conducting cylinder (posses an infinite length) in a uniform EFI E0. The permittivities of the enviroment & the cylinder are ε1 & ε2 respectively. EFI is perpendicular to the cylinder’s axis. = −ρϕ + − ρ − 1 ϕ V1 E 0cos C 1 cos = + ρ ϕ V2 C 2 cos − −1 + (−EaCa + )cosϕ = Ca cos ϕ −ε− ε + 2 ε 0 1 2 →=−C E1 2 aC2 , =− E 1 ε− −− −2 ϕε = + ϕ 10εε+ 20 εε + 101(ECa )cos 22 Ca cos 12 12  ε− ε 2  =−−1 2 a ρ ϕ ρ ≥ V1 E 0 1  cos , as a  ε+ ε ρ 2  →  1 2 2ε V= − E1 ρcos ϕas ρ ≤ a  2 0 ε+ ε  1 2 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Product Solution of Laplace’s Equation (12) Ex. Solve for the potential & EFI in the vicinity of a conducting cylinder (posses an infinite length) in a uniform EFI E0. The permittivities of the enviroment & the cylinder are ε1 & ε2 respectively. EFI is perpendicular to the cylinder’s axis.  ε− ε a2  V=−− E11 2 ρ cos ϕ , as ρ ≥ a 1 0 ε+ ε ρ 2   1 2    2ε V= − E1 ρcos ϕas ρ ≤ a  2 0 ε+ ε  1 2 ∂−εε2 ∂−  εε 2 V1 1 2 a V 1 1 2 a EEρ =−=−1 cos,ϕ EEϕ =−=−−  1 sin ϕ 1∂+ρ 0 εερ2 1 ∂+ ϕ 0 εερ 2 1 2 1 2 ∂V2ε ∂ V 2 ε EE=−=21cosϕ , E =−=− 21 E sin ϕ 20ρ ∂+ρεε 2ϕ ∂+ ϕεε 0 1 2 1 2 2ε →E = E = E 1 2 2z 0 ε+ ε 1 2 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Poisson’s & Laplace’s Equations 1. Poisson’s Equation 2. Laplace’s Equation 3. Uniqueness Theorem 4. Examples of the Solution of Laplace’s Equation 5. Examples of the Solution of Poisson’s Equation 6. Product Solution of Laplace’s Equation 7. The Iteration Method Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn The Iteration Method (1) • To solve for Laplace’s equation as V = V(x, y) • A numeric method Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn y The Iteration Method (2) x ∂2V ∂ 2 V ∂ 2 V ∇=2V + + = 0 ∂2 ∂ 2 ∂ 2 x y z V2 = V Vxy( , ) b V V0 V ∂2V ∂ 2 V 3 1 → + = 0 c a ∂x2 ∂ y 2 h d ∂V V− V ≈ 1 0 ∂ V4 xa h h ∂ − V V0 V 3 ≈ ∂2V V− V − V + V ∂x h → ≈ 1 0 0 3 c 2 2 ∂ ∂ ∂x h V− V ∂2V ∂x ∂ x ≈ a c ∂ 2 x h Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn y The Iteration Method (3) x ∂2V ∂ 2 V + = 0 ∂x2 ∂ y 2 V2 ∂2V V− V − V + V ≈ 1 0 0 3 V3 V0 V1 ∂x2 h 2 h ∂2 V− V − V + V V ≈ 2 0 0 4 ∂ 2 2 V y h h 4 ∂2V ∂ 2 V VV+ + VV + − 4 V →+≈123 4 0 = 0 ∂x2 ∂ y 2 h 1 → V≈() VVVV +++ 04 1234 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn The Iteration Method (4) Ex. 1 1 Air gapV = 100 Air gap V=() VVVV +++ 04 1234 1 43.8 53.2 43.8 ()0+ 100 ++ 0 0 = 25 4 1 18.8 25 18.8 ()100+ 50 ++ 0 25 = 43.8 V = 0 V = 0 4 6.2 9.4 6.2 1 ()0+ 25 ++ 0 0 = 6.2 4 1 ()43.8+ 100 + 43.8 + 25 = 53.2 V = 0 4 1 1 Step 1 ()25+ 43.8 ++ 0 6.2 = 18.8 ()6.2+ 25 + 6.2 + 0 = 9.4 4 4 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn The Iteration Method (5) Ex. 1 1 Air gapV = 100 Air gap V=() VVVV +++ 04 1234 4352.8 43 1 43.8 53.2 43.8 ()100+ 50 ++ 0 25 = 43,8 4 18.6 18.6 1 18.8 25 18.8 ()53.2+ 100 ++ 0 18.8 = 43 V = 0 V = 0 4 6.2 9.4 6.2 1 ()43.8+ 100 + 43.8 + 25 = 53.2 4 1 ()43+ 100 ++ 43 25 = 52.8 V = 0 4 1 1 Step 2 ()25+ 43.8 ++ 0 6.2 = 18.8 ()25+ 43 ++ 0 6.2 = 18.6 4 4 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn The Iteration Method (6) Ex. 1 1 Air gapV = 100 Air gap V=() VVVV +++ 04 1234 4352.8 43 1 43.8 53.2 43.8 ()0+ 100 ++ 0 0 = 25 4 18.624.9 18.6 1 18.8 25 18.8 ()18.6+ 52.8 + 18.6 + 9.4 = 24.9 V = 0 V = 0 4 7.09.8 7.0 6.2 9.4 6.2 1 ()0+ 25 ++ 0 0 = 6.2 4 1 ()9.4+ 18.6 ++ 0 0 = 7.0 V = 0 4 1 1 Step 2 ()6.2+ 25 + 6.2 + 0 = 9.4 ()7.0+ 25 + 7.0 + 0 = 9.8 4 4 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn The Iteration Method (7) Ex. 1 Air gapV = 100 Air gap =1 () +++ V0 VVVV 1234 4 42.9 52.7 42.9 4352.8 43 1 43.8 53.2 43.8 ()52.8+ 100 ++ 0 18.6 = 42.9 4 18.7 25.0 18.7 18.624.9 18.6 18.8 25 18.8 1 V = 0 V = 0 ()42.9++ 100 42.9 + 24.9 = 52.7 7.1 9.8 7.1 4 7.09.8 7.0 6.2 9.4 6.2 1 ()24.9+ 42.9 ++ 0 7.0 = 18.7 4 1 ()18.7+ 52.7 + 18.7 + 9.8 = 25.0 V = 0 4 1 1 Step 3 ()9.8+ 18.7 ++ 0 0 = 7.1 ()7.1+ 25 + 7.1 + 0 = 9.8 4 4 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn The Iteration Method (8) 10 V Ex. 2 1 2 (0)= (0) == (0) = 3 V1 V 2... V 10 0 0 V 4 5 6 (1)=1 (0) +++ (0) = V1() V 210 0 V 4 2.5000V 20 V 4 7 8 (1)1 (0) (1) (0) 0 V V=() V +++10 VV = 3.1250V 9 10 24 3 15 ... 0 V 1 V(1)=() VV (0) +++ (1)0 V (0) = 0.2344V 0 V 74 85 9 ... 1 V(1)=()20 + V (1) + V (1) += 0 6.7358V 104 8 9 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn The Iteration Method (9) 10 V Ex. 2 k 0 1 24 23 1 2 3 V (k ) (V) 0 2.5000 5.6429 5.6429 1 0 V 4 5 6 V (k ) (V) 0 3.1250 9.1735 9.1735 2 20 V (k ) 7 8 V3 (V) 0 8.2813 13.1111 13.1111 (k ) 0 V V4 (V) 0 0.6250 3.3957 3.3957 9 10 (k ) 0 V V5 (V) 0 0.9375 7.9405 7.9405 (k ) V6 (V) 0 7.3047 13.2710 13.2710 0 V (k ) V7 (V) 0 0.2344 5.9219 5.9219 (k ) V8 (V) 0 6.8848 13.0324 13.0324 (k ) 0 0.0586 3.7147 3.7147 V9 (V) (k ) 0 6.7358 8.9368 8.9368 V10 (V) Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn