Bài giảng Engineering electromagnetic - Chapter VII: Dielectrics & Capacitance - Nguyễn Công Phương

Nguy ễn Công Ph ươ ng Engineering Electromagnetics Dielectrics & Capacitance 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 Dielectrics & Capacitance 1. Dielectric Materials 2. Boundary Conditions for Perfect Dielectric Materials 3. Capacitance 4. Using Field Sketches to Estimate Capacitance 5. Current Density & Flux Density Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Dielectric Materials (1) +– d Q – + E E • Dipole moment: p = Qd • Q: the positive one of the 2 bound charges • d: the vector from the negative to the positive charge Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Dielectric Materials (2) • Dipole moment: p = Qd • If there are n dipoles per unit volume, then the total dipole moment in ∆v: n∆ v = ptotal ∑ p i i=1 • The polarization: n∆ v = 1 Plim ∑ p i ∆v →0 ∆ v i=1 • Unit: C/m2 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Dielectric Materials (3) + ∆S + 3 E + Density: n molecules/m + + 1 ∆v = dcos θ ∆ S + – d cos θ ∆S – + + 2 – 1 ∆ = ∆ + + θ d cos θ Qb nQ v – – 2 →∆=θ ∆ – d Qb nQdcos S – – = ∆ – nQd. S →∆ = ∆ – Qb P. S p= Q d →P = nQ d →Q = − P. d S b ∫S → =(ε + ) Gauss’s law: Q= ε E. d S Q∫ 0E P. d S T ∫S 0 S QT = Qb + Q → Q = QT – Qb (Q: the total free charge) Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Dielectric Materials (4) Q=(ε E + P). d S ∫S 0 → =ε + D0 E P Gauss’s law: Q= D. d S ∫S Divergence theorem: DSD.d= ∇ . dv ∫S ∫ v → ∇.D = ρ = ρ v Qv dv ∫V Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Dielectric Materials (5) • D = ε0E + P • In an isotropic material, E & P are always parallel, regardless of the orientation of the field • P = χeε0E • χe : the electric susceptibility • → D = ε0E + P = ε0E + χeε0E = ( χe + 1) ε0E • εr = χe + 1: the relative permitivity • → D = ε0εrE = εE • ε = ε0εr : the permitivity Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Dielectrics & Capacitance 1. Dielectric Materials 2. Boundary Conditions for Perfect Dielectric Materials 3. Capacitance 4. Using Field Sketches to Estimate Capacitance 5. Current Density & Flux Density Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Boundary Conditions for Perfect Dielectric Materials (1) ∆w = D ∆h Etan 1 ∫ EL.d 0 N1 ∆S → ∆− ∆= Region 1, ε1 Etan 2 Etan1 w E tan 2 w 0 → = DN2 EEtan1 tan 2 Region 2, ε2 DD D ε →tan1 =EE = = tan 2 →tan 1 = 1 εtan1 tan 2 ε ε 1 2 Dtan 2 2 ∆Q =ρ ∆ S S →DD − = ρ ∆ = ∆ − ∆ N1 N 2 S →DD = QDN1 SD N 2 S N1 N 2 No free charge on the interface → ρ = 0 S E ε →εEE = ε →N1 = 2 1 N1 2 N 2 ε EN 2 1 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Boundary Conditions for Perfect Dielectric Materials (2) ∆w ∆h E DN1 ∆S tan 1 Region 1, ε1 Etan 2 DN2 = Region 2, ε2 EEtan1 tan 2 D ε tan 1= 1 ε If we know the field on one side (e.g E1 or D1) Dtan 2 2 of a boundary, we cand find quickly the field = DDN1 N 2 on the other side (E2 & D2) E ε N1 = 2 ε EN 2 1 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Boundary Conditions for Perfect Dielectric Materials (3) D1 = DN1 DDN1 N 2 θ1 = θ Dtan 1 DDN1 1cos 1 Region 1, ε1 = θ ε1 > ε2 DDN 2 2cos 2 θ DN2 →θ = θ D2 2 DD1cos 1 2 cos 2 Region 2, ε2 D ε Dtan 2 tan 1= 1 D ε θ ε tan 2 2 →tan 1 = 1 →θ DD= sin θ tan θ ε 2 → tan 1 1 1 2 2 D2 = θ DDcosθ= cos θ DDtan 2 2sin 2 1 1 2 2 →ε θε = θ 21DDsin 1 12 sin 2 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Boundary Conditions for Perfect Dielectric Materials (4) D1 DN1 ε  θ1 2 Dtan 1 θ= atan tan θ Region 1, ε1 2 1 ε  1  ε1 > ε2 D D θ2 N2 2 2 ε  Region 2, ε2 DD=cos2θ + 2 sin 2 θ Dtan 2 21 1ε  1 1  ε  2 EE=sin2θ + 1 cos 2 θ 211ε  1 2  Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Boundary Conditions for Perfect Dielectric Materials (6) Ex. 2 Given the region z < 0 with εr1 = 3.2 & D1 = –30 ax + 50 ay + 70 az nC/m . The region z > 0 possesses εr2 = 2. Find DN1, Dtan 1, Dtan 1, θ1, DN2, Dtan 2, D2, θ2 ? = = 2 DN1 D 1 z 70 nC/m = − + 2 Dtan1 30 a x 50 a y nC/m = =−2 + 2 = 2 Dtan1D tan 1 ( 30) 50 58.3nC/m = =−222 + + = 2 D1D 1 ( 30) 50 70 91.1nC/m θ =Dtan 1 =58.3 = o 1 atan atan 39.8 D1z 70 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Boundary Conditions for Perfect Dielectric Materials (7) Ex. 2 Given the region z < 0 with εr1 = 3.2 & D1 = –30 ax + 50 ay + 70 az nC/m . The region z > 0 possesses εr2 = 2. Find DN1, Dtan 1, Dtan 1, θ1, DN2, Dtan 2, D2, θ2 ? ==2 →= 2 DNN2 D 170nC/mD Nz 2 70 a nC/m D εD ε ε 2 tan11=→ tan 11 =→D = 2 D =−+(30 aa 50 ) ε εtan2 ε tan 1 x y Dtan22D tan 22 1 3.2 = − + 2 18.75ax 31.25 a y nC/m = +=− + + 2 DDD2tanN 2 2 18.75 a x 31.25 aa y 70 z nC/m ε  2  θ=atan tan θ 2 = atan tan 39.8o = 27.5 o 2 1 ε    1  3.2  Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Dielectrics & Capacitance 1. Dielectric Materials 2. Boundary Conditions for Perfect Dielectric Materials 3. Capacitance 4. Using Field Sketches to Estimate Capacitance 5. Current Density & Flux Density Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Capacitance (1) + + + + + + + +Q + Q dielectric, ε + Capacitance: C = + Conductor 2 ε – + + V0 E.d S – – – + + ∫S – – + + + →C = – – Q – = ε + Conductor 1 – QE. d S – –– ∫S − E.d L – – – + ∫− – – = − V0 ∫− E. d L • V0 : work to carry a unit positive charge from the surface 1 to the surface 2 • C depends on the physical dimensions (of the system of conductors) & on the permittivity • Unit: F (farad), C/V, practically µF, nF, pF Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Capacitance (2) ρ S E= a Conductor surface, –ρ ε z S z = d = ρ E DS a z Conductor surface, +ρS z = 0 bottom 0 ρ ρ V= − E. d L = − S dz = S d 0 ∫top ∫d ε ε = ρ ε S QS S C = d Q C = V0 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Capacitance (3) = 1 ε 2 W Edv Conductor surface, –ρS z = d E ∫V 2 ρ E = S ε E 2 S d ερ Conductor surface, +ρS z = 0 → = 1 S WE ∫ ∫ 2 dzdS 2 0 0 ε 1 ρ 2 1 ε S ρ 2d 2 = S Sd = S 2 ε 2 d ε 2 ε S 1 1 1 Q2 C = → W= CV2 = QV = d E 20 2 0 2 C ρ V= S d 0 ε Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn ρ = a ρ = b Capacitance (4) ρL ρ b V = L ln ab 2πε a 2πε L Q= ρ L → C = L L b Q ln C = a Vab Q 1 1  V = −  ab πε 4πε Q 4 a b  →C = 1 1 a = Q − C a b Vab b Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Capacitance (5) = + Area, S V0 Ed 11 Ed 22 DDEE= →ε = ε N1 N 2 11 22 ε V 2 d2 → = 0 Conducting d E1 ε plates ε d+ d 1 1 d1 1 2 ε 2 →ρ == ε = V0 S1DE 1 11 d1+ d 2 ε ε 1 1 1 2 C = = =ρ = ρ d d 1 1 QS S S 1 S 1+ 2 + Q εS ε S CC C = 1 2 1 2 V0 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Capacitance (6) Area, S Conducting 1 1 plates C = = ε2 d2 d1+ d 2 1+ 1 d ε ε 1S 2 S CC1 2 ε1 d1 Conducting plates εS+ ε S S S CCC=11 22 = + 1 2 d 1 2 d ε1 ε2 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Capacitance (7) y P(x, y, 0) ρ R R2 V = L ln 01 1 2πε R R1 1 (– a, 0, 0) −ρ R V = L ln 02 (a, 0, 0) x 2 πε 2 R2 – ρL ρ R R  z +ρL →=+=V V V L ln01 − ln 02 1 2 πε   2 R1 R 2  ρ R R = L ln 01 2 πε 2 R02 R 1 = ρ (x+ a ) 2 + y 2 R01 R 02 →V = L ln 2πε (x− a ) 2 + y 2 R=( xa − ) 2 + y 2 1 ρ +2 + 2 = L (x a ) y R=( xa + ) 2 + y 2 ln 2 4πε (x− a ) 2 + y 2 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Capacitance (8) y P(x, y, 0) R2 ρ +2 + 2 R L (x a ) y 1 V = ln (– a, 0, 0) 4πε (x− a ) 2 + y 2 (a, 0, 0) x – ρ Choosing an equipotential surface V1, we define: L z +ρL πε ρ = 4V1 / L K1 e (x+ a ) 2 + y 2 K +1 →K = →−x22 ax1 ++= ya 2 2 0 1 2 2 − (x− a ) + y K1 1 2 2 K +1  2a K  →−x a1 += y 2 1  −   −  KK11   1 1  Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Capacitance (9) y P(x, y, 0) R πε ρ 2 = 4V1 / L K1 e R1 (– a, 0, 0) 2   2 K +1  2a K1 (a, 0, 0) →−x a1  += y 2   x − −  – ρ KK11   1 1  L z +ρL • The V = V1 surface is independent of z → it is a cylinder • It intersects the xy plane in a circle of radius: 2a K b = 1 − K1 1 K +1 & this circle is centered at (x = h, y = 0) where h= a 1 − K1 1 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn y Capacitance (10) h V0 = 0 V1 The V1 surface intersects the xy plane in a circle of radius b x 2a K1 b = z K −1 1 K +1 & centered at (x = h, y = 0) where h= a 1 − K1 1  =2 − 2 a h b →  4πε V If h, b & V1 are given h+ h2 − b 2 →ρ = 1  K = L  1 b ln K1 then a, ρL & K1 can be found πε ρ = 4V1 / L K1 e ρ πε πε 2πε L → =LL = 4 L = 2 L = Cplane, cylinder −1 VK1ln 1 ln[(h+ h2 − b 2 )/ b ] cosh (h / b ) Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Ex. Capacitance (11) Given the system, find the location & the y The equivalent magnitude of the equivalent line charge, & the line charge V0 = 0 location of the 50V equipotential surface. =22 −= 22 −= a h b 13 5 12 m


2 2 x h+ h − b 13 + 12 V = 100 V K = = = 5 h = 13 m 1 1 b 5 b = 5 m →K = 25 1 π × ×−12 × πε →ρ =4 8.854 10 100 = ρ = 4 V1 L 3.46 nC /m L ln 25 ln K1 πε π × × −12 =2 = 2 8.85410 = Cplane, cylinder − − 34.6 pF/m cosh1 (h / b ) cosh 1 (13/ 5) Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Ex. Capacitance (12) Given the system, find the location & the y The equivalent magnitude of the equivalent line charge, & the line charge V0 = 0 location of the 50V equipotential surface. 4πεV / ρ = 2 L K2 e


π × ××−12 × − 9 x =e4 8.854 10 50/3.46 10 = 5.00 h = 13 m V1 = 100 V 2a K 2× 12 5 b = 5 m →=b 2 = = 13.42 m 2 − − K2 1 5 1 K +1 5+ 1 h= a 2 =12 = 18 m 2 − − K2 1 5 1 =→= = V325V b 3 29.06m, h 3 31.44m Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn y Capacitance (13) h V0 = 0 V1 πε = 2 L Cplane, cylinder 2 2 ln[(h+ h − bb )/ ] b x b≪ h z 2πε L →CC = = y plane, cylinder plane, wire 2h ln h b πε L →C = wire, wire 2h x ln b z Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Dielectrics & Capacitance 1. Dielectric Materials 2. Boundary Conditions for Perfect Dielectric Materials 3. Capacitance 4. Using Field Sketches to Estimate Capacitance 5. Current Density & Flux Density Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Using Field Sketches to Estimate Capacitance (1) • A conductor boundary is an equipotential surface • The electric field intensity E & the electric flux D are both perpendicular to the equipotential surfaces • E & D are perpendicular to the conductor boundaries & posses zero tangential values • The lines of electric flux, or streamlines, begin & terminate on charge & therefore, in a charge-free, homogeneous dielectric, begin & terminate only on the conductor boundaries Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Using Field Sketches to Estimate Capacitance (2) ∆L E & D are both tan = 1 ∆ perpendicular to the LN equipotential surfaces 1 ∆ψ E = ε ∆ ∆ψ ∆ Ltan →1 = V ∆V ε ∆L ∆ L B E = tan N ∆L ∆L A N tan B’ ∆L 1 ∆ ψ ∆LN A’ →tan =const = ∆ε ∆ LN V Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Using Field Sketches to Estimate Capacitance (3) Q C = V0 = ∆= ∆ ψ QNQNQ Q = ∆ V0 NV V N ∆ψ →C = Q ∆ NV V ∆L 1 ∆ ψ tan =const = = 1 ∆ε ∆ LN V NN∆L →C =Qεtan = ε Q ∆ NLNV N V Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Using Field Sketches to Estimate Capacitance (4) 0 15 30 46 62 100 V 80 Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Dielectrics & Capacitance 1. Dielectric Materials 2. Boundary Conditions for Perfect Dielectric Materials 3. Capacitance 4. Using Field Sketches to Estimate Capacitance 5. Current Density & Flux Density Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn Current Density & Flux Density JE= σ σ DE= ε ε Eσ= −∇ V σ Eε= −∇ V ε = = σ  − I∫J. d S ∫ Eσ . d S Vσ ∫ E.Lσ d S S R =0 = = −  I σ Eσ .d S Vσ0 ∫ E.L σ d  ∫S →  ε Q= ε Eε . d S  Q ∫ Eε .d S ∫S C = = S = −  Vε 0 − E.Lε d Vε0 ∫ E.L ε d  ∫ ε →RC = σ Dielectrics & Capacitance - sites.google.com/site/ncpdhbkhn