Current and conductors 2017 mk

Nguyễn Công Phương Engineering Electromagnetics Current & Conductors 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 Current & Conductors - sites.google.com/site/ncpdhbkhn Current & Conductors Current & Current Density Metallic Conductors Conductor Properties & Boundary Conditions The Method of Images Semiconductors Applications Current & Conductors - sites.google.com/site/ncpdhbkhn Current & Current Density (1) • Current: dQ I= dt • Unit A (ampère) • Current is defined as the motion of positive charges Current & Conductors - sites.google.com/site/ncpdhbkhn Current & Current Density (2) • Current: rate of movement of charge crossing a given reference plane (of one coulomb per second) • Current density: J (A/m2) • The increment of current ΔI crossing an incremental surface ΔS normal to the current density: ΔI = JNΔS • If the current density is not perpendicular to the surface: ΔI = J.ΔS • Total current: I = J.dS ∫ S Current & Conductors - sites.google.com/site/ncpdhbkhn Current & Current Density (3) ∆Q = ρ v ∆ v = ρ v ∆S ∆L ∆Q = ρ v ∆S ∆x ∆x → ∆I = ρv ∆S ∆Q ∆I = ∆t ∆t = ρv ∆Sv x ∆I = J x ∆S ∆Q = ρ v ∆v z ∆S y x ∆∆xL → J x = ρv vx J = ρv v Current & Conductors - sites.google.com/site/ncpdhbkhn Current & Current Density (4) Ex z z=2 Given J = 2ρzaρ + 7zsin2φaφ mA/m2 Find the total current leaving the circular band z=1 I = ∫ J.dS = ∫ J ρ =3 dS S ρ=3 S x J ρ =3 = × za ρ + z sin ϕ aϕ = za ρ + z sin ϕ aϕ dS = ρ dϕ dzaρ = 3dϕ dzaρ z z+dz y dρ dz z φ x φ+dφ Current & Conductors - ρ sites.google.com/site/ncpdhbkhn ρ+dρ ρdφ Current & Current Density (5) Ex z z=2 Given J = 2ρzaρ + 7zsin2φaφ mA/m2 Find the total current leaving the circular band I = ∫ J.dS = ∫ J ρ =3 dS S z=1 ρ=3 S x J ρ =3 = × za ρ + z sin ϕ aϕ y = za ρ + z sin ϕ aϕ → J ρ = dS = 18 zdϕ dz dS = ρ dϕ dzaρ = 3dϕ dzaρ →I =∫ z= z =1 ϕ = 2π ∫ϕ =0 18 zdϕ dz = ∫ z =2 z =1 2π × 18 zdz = 169 mA Current & Conductors - sites.google.com/site/ncpdhbkhn Current & Current Density (6) The current leaving a closed surface: I = ∫ S J.dS The total charge in the surface: Qi The law of conservation of charge dQi → I = ∫ J.dS = − S dt • in circuit analysis, I = dQ/dt because this is an entering current • in electromagnetism, I = – dQ/dt because this is a leaving one Current & Conductors - sites.google.com/site/ncpdhbkhn Current & Current Density (7) dQi I = ∫ J.dS = − S dt ∫ S J.dS = ∫ (∇.J) dv (div theo.) dQi → ∫ (∇.J ) dv = − V dt V Qi = ∫ ρv dv V d → ∫ (∇.J ) dv = − ∫ ρv dv = V dt V ∂ρv ∫V − ∂t dv ∂ρ v ∂ρ v → (∇.J)∆v = − ∆v → ∇.J = − ∂t ∂t Current & Conductors - sites.google.com/site/ncpdhbkhn 10 Conductor Properties & Boundary Conditions (1) • Given some electrons in the interior of a conductor • They will begin to accelerate away from each other, until they reach the surface of the conductor • Characteristic 1: the charge density inside a conductor is zero, the exterior surface has a surface charge density • Within a conductor: no charge → no current → no electric field intensity (Ohm) • Characteristic 2: the electric field intensity within the conductor is zero Current & Conductors - sites.google.com/site/ncpdhbkhn 21 Conductor Properties & Boundary Conditions (2) D ΔS Δh DN ∫ E.dL = Dt →∫ b a +∫ c b +∫ d c +∫ a d =0 Ewithin conductor = EN E a Δw b Δh Δh d Δw c Et Conductor ∆h  ∆h    → ( Et ∆w ) −  E N , at b + + E   N ,at a =0     ∆h → → Et ∆ w = → Et = → Dt = ε E t = → Dt = Et = →∫ +∫ = DN ∆S ; ∫bottom ∫ S D.dS = Q ∫top top bottom = 0; +∫ sides ∫sides =Q =0 → DN ∆S = Q = ρ S ∆S → DN = ρ S = ε E N Current & Conductors - sites.google.com/site/ncpdhbkhn 22 Conductor Properties & Boundary Conditions (3) D Dt = Et = ΔS Δh DN DN = ε E N = ρ S Dt EN E a Δw b Δh Δh d Δw c Et Conductor x Vxy = − ∫ E.dL = y Characteristics of conductors in static field: The static EFI inside a conductor is zero The static EFI at the surface of a conductor is everywhere directed normal to that surface The conductor surface is an equipotential surface Current & Conductors - sites.google.com/site/ncpdhbkhn 23 Conductor Properties & Boundary Conditions (4) Ex Given V = x2 – 10yz V & P(2, 1, 2) lies on a conductor – free space boundary Find V, E, D, ρS at P, & the equation of the conductor surface VP = 2 − 10 × × = −16 V → −16 = x − 10 yz E = −∇V = −∇( x − 10 yz ) = −2 xa x + 10 za y + 10 ya z V/m ( → E P = −2 xa x + 10 za y + 10 ya z ) x =2, y =1, z =2 = −40a x + 20a y + 10a z V/m D P = ε E P = 8.854 × 10−12 ( −40a x + 20a y + 10a z ) nC/ m2 ρ S , P = DN DN , P = DP = 8.854 × 10 −12 40 + 202 + 102 = 406 pC/m → ρ S , P = 406 pC/m Current & Conductors - sites.google.com/site/ncpdhbkhn 24 Current & Conductors Current & Current Density Metallic Conductors Conductor Properties & Boundary Conditions The Method of Images Semiconductors Applications Current & Conductors - sites.google.com/site/ncpdhbkhn 25 The Method of Images (1) +Q +Q Equipotential surface, V = Equipotential surface, V = –Q • Dipole: the plane between the charges is zero potential • That plane can be represented by a vanishingly thin conducting plane, infinite in extent • → the dipole can be substituted for a system of a charge and a conducting plane, & then the fields above the conducting plane obtain equivalence Current & Conductors - sites.google.com/site/ncpdhbkhn 26 The Method of Images (2) +Q +Q Equipotential surface, V = Equipotential surface, V = –Q +Q +Q Equipotential surface, V = Equipotential surface, V = –Q Current & Conductors - sites.google.com/site/ncpdhbkhn 27 The Method of Images (3) Ex ρL ρL +1 –5 Equipotential surface, V = Coulomb’ s law Gauss’ law Laplace’s equation &E=– V +1 –5 ? Equipotential surface, V = –1 Current & Conductors - sites.google.com/site/ncpdhbkhn +5 – ρL 28 The Method of Images (4) Ex Given Q at (0, 0, d) Find the potential & EFI at P ? V+ Q = V− Q = P(x, y, z) Q 4πε0 R1 = Q d Q 4πε x + y + ( z − d )2 −Q −Q = 4πε R2 4πε x + y + ( z + d )2 Equipotential surface, V = R1 Q d  Q  1   V= − 2 2 4πε0  x + y + ( z − d )2 x + y + ( z + d )   P(x, y, z) R2 d –Q  y  z+d z−d   Q  x x  y  E = −∇V = −  −  a x +  −  a y +  −  a z  4πε  R2 R1  R1    R2 R1   R2 Current & Conductors - sites.google.com/site/ncpdhbkhn 29 y The Method of Images (5) Ex nC Find the potential at P ? R1 = 12 + 12 = 1.41 R2 = 32 + 12 = 3.16 y –5 nC R3 = + = 4.24 2 R4 = + = 3.16 2 × 10−9  1 1  VP = − + −   = 14.03 V 4πε0  R1 R2 R3 R4  Current & Conductors - sites.google.com/site/ncpdhbkhn P R2 R3 x nC R1 P x R4 –5 nC nC 30 z The Method of Images (6) Ex Q A point charge Q at a distance d from a center of a grounded conducting sphere of radius a Find the image charge? y Problem: find q & b x R1 = (d − R cosθ )2 + ( R sin θ )2 = R + d − 2Rd cosθ z R2 = ( R cos θ − b)2 + ( R sin θ )2 = R + b2 − Rb cosθ VP = Q q Q mQ Q − = − = 4πε R1 4πε R2 4πε R1 4πε R2 4πε R = a → VP = → m − =0 R1 R2  m  −   R1 R2  d a a  m = → q = − Q  d d → a b =  d Current & Conductors - sites.google.com/site/ncpdhbkhn Q P R2 q b R1 θ R a x 31 c Current & Conductors Current & Current Density Metallic Conductors Conductor Properties & Boundary Conditions The Method of Images Semiconductors Applications Current & Conductors - sites.google.com/site/ncpdhbkhn 32 Semiconductors • Germani, silicon J • Conductivity of conductors: E σ = – ρe μe • Conductivity of semiconductors: σ = – ρe μe + ρh μh • h: hole J E • At 300K: – – – – – – – – – – – – – – μe, Germani: 0.36 m2/Vs; μh, Germani: 0.17 m2/Vs – μe, Silicon: 0.12 m2/Vs; μh, Silicon: 0.025 m2/Vs Current & Conductors - sites.google.com/site/ncpdhbkhn 33 Current & Conductors Current & Current Density Metallic Conductors Conductor Properties & Boundary Conditions The Method of Images Semiconductors Applications Current & Conductors - sites.google.com/site/ncpdhbkhn 34 Applications – Faraday’s Cage https://lifeonthebluehighways.com/2013/04/20/faradays-cage/ Current & Conductors - sites.google.com/site/ncpdhbkhn 35 ... Current & Conductors - sites.google.com/site/ncpdhbkhn Current & Conductors Current & Current Density Metallic Conductors Conductor Properties & Boundary Conditions The Method of Images Semiconductors... Applications Current & Conductors - sites.google.com/site/ncpdhbkhn Current & Current Density (1) • Current: dQ I= dt • Unit A (ampère) • Current is defined as the motion of positive charges Current & Conductors. .. m     = r m/ s r ρv  r   r  r Current & Conductors - sites.google.com/site/ncpdhbkhn 11 Current & Conductors Current & Current Density Metallic Conductors Conductor Properties & Boundary