EE204B_Lesson_09_E-field_in_Material-1_

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Unformatted text preview: EE204B EE204B Electromagnetics Electromagnetics Lesson Lesson 09 Electric Fields in Material Space Currents, Conductors, Polarization Instructor: Prof. Yong H. Won Lesson 09 Topics Electric Field in Material Space (Chap. 5) • • • • Material Properties Convection and Conduction Currents Conductors Polarization in Dielectrics EE204B Electromagnetics Prof. Yong H. Won Spring 2010 2 Lesson 09 Electrical Electrical Properties of Materials Materials are broadly classified as • Conductors (σ >>1) • Dielectrics (Insulators) (σ << 1) • Semiconductors Major difference is the number of electrons available for conduction of current Conductors have an abundance of free electrons Dielectrics have few free electrons Perfect cunductor (σ = ∞) Perfect dielectrics (σ = 0) di 0) EE204B Electromagnetics Prof. Yong H. Won Spring 2010 3 Lesson 09 Conductors and Dielectrics EE204B Electromagnetics Prof. Yong H. Won Spring 2010 4 Lesson 09 Currents Electron’s moving along a certain forms a current. moving along certain forms current Current in dielectrics is generally called as convection current. Current in conductors is generally called as conduction current. Current is a rate of change of electric charge. dQ I= [A] dt Current density J [A/m2] I =∫ s rr J • dS where I is the flux of J across S. the flux of EE204B Electromagnetics Prof. Yong H. Won Spring 2010 5 Lesson 09 Currents Example 9-1 [Text Practice 5.1]: If J = 10zsin2φ aρ A/m2, find the current through the cylindrical surface ρ = 2, 1 ≤ z ≤ 5m. Solution z z=5 2 aρ z=1 y x EE204B Electromagnetics I =∫ s r rr r r J • dS aρ = ∫ 10 z sin 2 φaρ • dsaρ s 2 5 1 = ∫ 10 z sin φds = ∫ s 5 2π 1 0 ∫ 2π 0 10 z sin 2 φρd φdz = 2∫ 10 zdz ∫ sin 2 φd φ = 2 × 5z 1 − cos 2φ dφ 1 0 2 = 240 × (π) = 754 [A] 25 ×∫ 2π Prof. Yong H. Won Spring 2010 6 Lesson 09 Currents Convection Currents A flow of charge, of density ρv, passes through ΔS at velocity u = uay ΔQ ρv ΔS Δy ΔI = = ρv ΔSu y = Δt Δt r r ΔI ∴Jy = = ρv u y , In general J = ρv u ΔS EE204B Electromagnetics Prof. Yong H. Won Spring 2010 7 Lesson 09 Currents r r F = −eE [force received by one electron] Under the influence of F , electron will be accelerated the influence of electron will accelerated r r r u Average drift velocity F = ma = m τ Average time interval between collisions r r r u eτ r −eE = m →u=− E τ m If there are n electrons per unit volume ρv = −ne r r r ne2 τ r E = σE ∴ J = ρv u = m J = σE EE204B Electromagnetics Prof. Yong H. Won Spring 2010 8 Conduction Currents Lesson 09 Currents Example 9-2 A wire of conductivity 5×107 S/m has 1029 free electrons per cubic meter when an electric field of 10 mV/m is applied. Determine the drift velocity of the electrons. Solution J = ρ u = ne u = 1029 ×1.6 ×10−19 × u v Meanwhile J = σE = 5 ×107 ×10 ×10−3 = 5 ×105 A/m 2 5 ×105 u = 29 = 3.125 ×10−5 m/s −19 10 ×1.6 ×10 EE204B Electromagnetics Prof. Yong H. Won Spring 2010 9 Lesson 09 Conductors Conductors Isolated conductor E = 0 why? Consider two things: ① External E-field pushes free charges : Positive charges (same direction), Negative charges (opposite) Making surface charges induced E-field Cancel Net E-field ② Perfect conductor (σ = ∞) Since J = σE, if E ≠ 0, J ∞ (Not physical) Conductor Properties 1. E=0 2. ∇V = 0 Vab = 0 (Equipotential body) 3. ρv = 0 ∇⋅E=ρv/ε=0 EE204B Electromagnetics Prof. Yong H. Won Spring 2010 10 Lesson 09 Ohm’s Law Wired conductor (Not Isolated) co EE204B Electromagnetics Prof. Yong H. Won Spring 2010 11 Lesson 09 Ohm’s Law EE204B Electromagnetics Prof. Yong H. Won Spring 2010 12 Lesson 09 Power r dW F ⋅ dl r r P= = = F ⋅u dt dt rr rr P = ∫ ρv dvE ⋅ u = ∫ E ⋅ρv udv v v rr = ∫ E ⋅ Jdv (Joule's Law) v Power density r2 dP r r wP = = E ⋅ J = σ E J/m3 dv For a conductor with uniform cross section rr rr P = ∫ E ⋅ Jdsdl = ∫ E ⋅ σEdsdl v v In conclusion r2 P = ∫ σ E dv v = ∫ Edl ∫ σEds = VI = I 2 R L s EE204B Electromagnetics Prof. Yong H. Won Spring 2010 13 Lesson 09 Polarization in Dielectrics Dielectric in an E-field EE204B Electromagnetics Prof. Yong H. Won Spring 2010 14 Lesson 09 Polarization in Dielectrics What is polarization? Consider an atom of dielectric consisting of a negative charge –Q and a positive charge +Q. When an electric field E applied, +Q is displaced from its fi di it equilibrium position in the direction of E by F = +QE, -Q is displaced in the opposite direction by F = -QE. The distorted charge distribution is equivalent to the original distribution plus a dipole. This process is called as “polarization”. The dielectric is said to be “polarized”. EE204B Electromagnetics Prof. Yong H. Won Spring 2010 15 Lesson 09 Polarization in Dielectrics Polarization The dipole moment p is p = Qd where d is the distance vector from –Q to +Q. If there are N dipoles, the dipole moment is N r r r r Q1d1 + Q2 d 2 + ⋅⋅⋅ + QN d N = ∑ Qk d k k =1 We define “polarization” P as the dipole moment per unit volume, i.e. N r r P = lim ∑Q d k =1 k k Δv →0 Δv [C/m 2 ] EE204B Electromagnetics Prof. Yong H. Won Spring 2010 16 Lesson 09 Polarization in Dielectrics Polarization The polarization induces bound charges on the surface of the dielectric and inside the volume of the dielectric dielectric and inside the volume of the dielectric. The dipoles align themselves with E, resulting in bound surface charges, In general, inside the volume the opposite charges do not cancel each other exactly leaving bound volume charges not cancel each other exactly, leaving bound volume charges. Different from free charges in conductor, these bound charges are caused by the displacement in the polarization process. They cannot move freely They cannot move freely. Two kinds of dielectrics • Nonpolar dielectric – hydrogen, oxygen.. • Polar dielectric – water, sulfur dioxide.. di EE204B Electromagnetics Prof. Yong H. Won Spring 2010 17 Lesson 09 Polarization in Dielectrics Polarization There exists a relationship between bound surface charge density (also known as polarization surface charge density) and P. rr ρ ps = P ⋅ an r an Relationship between bound volume charge density (also known as polarization volume charge density) and P. r ρ pv = −∇ ⋅ P EE204B Electromagnetics Prof. Yong H. Won Proof is in the textbook. Spring 2010 18 Lesson 09 Polarization in Dielectrics Dielectrics r r In the free space, D = ε E o r rr In dielectrics, D = ε o E + P The dipoles induced by E in dielectrics also contribute to electric flux. Generally, P is proportional to E r r where χe is electric susceptibility, dimensionless. electric susceptibility dimensionless P = χe εo E It measures how sensitive a dielectric is to E. r r r r r r D = εo E + χe εo E = εo (1 + χe ) E = εo ε r E = εE ε r = 1 + χe ε = εo ε r EE204B Electromagnetics Dielectric constant, relative permittivity Permittivity of a dielectric Prof. Yong H. Won Spring 2010 19 Lesson 09 Polarization in Dielectrics Example 9-3 [Text Example 5.7] A dielectric sphere (εr = 5.7) of radius 10 cm has a point charge of 2 pC placed at its center. Calculate 1. The surface density of polarization charge on the surface of the sphere. 2. The volume density of polarization charge. 3. The force exerted by the charge on a -4 pC point charge placed on the sphere. th EE204B Electromagnetics Prof. Yong H. Won Spring 2010 20 Lesson 09 Polarization in Dielectrics Solution E due to a point charge at the origin r E= Qr ar 4πεo r 2 r r χ Q r (ε − 1)Q r P = χe εo E = e 2 ar = r ar 2 4πε r r 4πε r r z 2 pC x y rr rr 4.7(2 ×10−12 ) r r ρ ps = P ⋅ an = P ⋅ ar = a ⋅ ar = 13.12 [pC/m 2 ] 2r s (1) 4π × 5.7 × 0.1 r (ε − 1)Q 1∂ 1∂ ρ pv = −∇ ⋅ P = − 2 (r 2 Pr ) = − 2 [r 2 r ]=0 2 (2) r ∂r r ∂r 4πε r r r r Q1Q2 r (2 ×10−12 )(−4 ×10−12 ) r F= ar = ar = −1.263ar [pN] (3) 4πεo ε r r 2 4πεo × 5.7 × (0.1) 2 EE204B Electromagnetics Prof. Yong H. Won Spring 2010 21 Lesson 09 Polarization in Dielectrics Dielectric Strength When E is strong enough, electrons can be completely pulled out of the molecules, the dielectrics the become conducting. The dielectric breakdown occurs. The dielectric strength is the maximum E that a dielectric can tolerate without the breakdown. EE204B Electromagnetics Prof. Yong H. Won Spring 2010 22 Lesson 09 Polarization in Dielectrics Linear, Isotropic, and Homogeneous Dielectrics ⎡ Dx ⎤ ⎡ ε xx ⎢ D ⎥ = ⎢ε ⎢ y ⎥ ⎢ yx ⎢ Dz ⎥ ⎢ ε zx ⎣⎦⎣ ε xy ε yy ε zy ε xz ⎤ ⎡ Ex ⎤ ⎥ ε yz ⎥ ⎢ E y ⎥ Tensor ⎢⎥ ⎥ ⎢ Ez ⎥ ε zz ⎦ ⎣ ⎦ EE204B Electromagnetics Prof. Yong H. Won Spring 2010 23 ...
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