EE468G NOTES (4)Conductors: There are free charges Conduction current: EJrrσ= (General Ohm’s law) σ: conductivity [S/m] Insider perfect conductor: 0,0, ==∞= JErrσ, Power dissipation: ∫=VdVEP2rσ Dielectrics: Charges are tightly bound to individual nuclei. Then can only move a fraction of an atomic distance in a electric field. For isotropic dielectric, the polarization vector is related to electric field by EPerrχε0=, eχ: Susceptibility (no unit) Charges associated with the polarization vector: ]C/m[,3PVPr⋅−∇=ρ ]C/m[,ˆ2PnSPr⋅=ρ + _ Un-polarized molecule +_ d Polarized moleculeMaxwell’s equation in dielectric material []VVVVPVPEPEPEρερερρρε=+⋅∇=⋅∇+⋅∇⋅∇−=+=⋅∇rrrrrr000 Let PEDrrr+=0ε, then we have VDρ=⋅∇r, Vρ---- Free-charge density. Since EPerrχε0=, ()EEEEPEDreerrrrrrrεεχεχεεε000001 =+=+=+= rεεε0= is the dielectric permittivity rε is the relative permittivity. Air is very close to free-space, hence, for air, 1airr,≈εSolution: (),2ˆ :surfacelower on ,2ˆ :surfaceupper on ˆ00ˆ22100SP00SP000dVPadVPaPnPadVEEPzzSPVPzrερερρρεεεε−=⋅−==⋅=⇒⋅==⇒=⋅∇==−=rrrrrrr zradVEDˆ3000εεε==rr ε d z 0 0 x Example: a dielectric slab has a relative permittivity of 3=rε, the electric field in a dielectric slab is: ()zadVEˆ0=r, determine SPVPPρρ,,r.Boundary condition: Why we need boundary condition? How boundary condition is derived? 1. Maxwell’s differential equations are valid in continuous media 2. Real world consists of discontinuities in material parameters. From Maxwell equation: 0=⋅∫CldErrContinuous media ε 1ε 2ε Discontinuity in permittivityLet us choose the integral contour to be the closed path ABCDA as shown in the above figure in red dotted line. The size of the contour is chosen such that lhlh ∆<<∆→∆→∆ ,0,0. The integral on the path is the sum of four integrals integrating on AB, BC, CD, DA. ∫∫∫∫∫⋅+⋅+⋅+⋅=⋅DACDBCABCldEldEldEldEldErrrrrrrrrr ()()()ltEldEhnEhnEldEltEldEhnEhnEldEDACDBCAB∆−⋅=⋅∆−⋅+∆−⋅=⋅∆⋅=⋅∆⋅+∆⋅=⋅∫∫∫∫ˆ2ˆ2ˆˆ2ˆ2ˆ221112rrrrrrrrrrrrrr Since lh ∆<<∆, the integral on AB and CD are much smaller than that on BC and DA. ()0ˆ0ˆˆ2121=−⋅⇒=∆⋅−∆⋅=⋅∫EEtltEltEldECrrrrrr This means that the normal components of the electric field is continuous across a dielectric interface. l∆ h∆ 1Er 2Er D A B C 1ε 2ε nˆ tˆSince tˆ is an arbitrary vector that is perpendicular tonˆ, the boundary condition can also be written as Using the similar method, and starting from the Maxwell equation QSdDS=⋅∫rr We can show that the boundary condition for the electric flux Dr is (Sρ is the free charge density on the interface) Boundary condition on conducting surface: 0,2=∞= Erσ ()SDDnρ=−⋅21ˆrr ()0ˆ21=−× EEnrr nˆ Sρ SntDEρ==11,0Using boundary condition to solve boundary value problems. Step 1: Find general solution (general solution is expressed with certain constants. Step2: Determine the constants using the boundary conditions. Example 1: Determine the electric field inside the dielectric slab that is sandwiched between two infinitely large conducting plates. Solution: It is expected that the potential function depends on z only: ()()zVzyxV =,,. Hence ()00222222222==∂∂+∂∂+∂∂⇒=∇dzVdzVyVxVzV ()21220 czczVdzVd+=⇒= (General solution) On the lower plate: ()000021=+⋅⇒== cczV, (1) On the upper plate: ()0210VcdcVdzV =+⋅⇒==, (2) Solving (1) and (2), we have: dVcc /,0012==. Hence d 0V 04εε= () ()zadVzVEzdVzVˆ,00−=−∇=⇒=rExample 2: Determine the electric field inside two dielectric slab that are sandwiched between two infinitely large conducting plates. Assuming that the dielectric slabs are of equal thickness. Solution: It is expected that the potential function depends on z only: ()()zVzyxV =,,. Hence ()00222222222==∂∂+∂∂+∂∂⇒=∇dzVdzVyVxVzV ()()dzdbzbzVdzczczVdzVd<<+=<<+=⇒=2/,2/0,021221122 On the lower plate: ()0000211=+⋅⇒== cczV, (1) On the upper plate: ()02102VbdbVdzV =+⋅⇒==, (2) At 21212122,2bdbcdcVVdz +=+⇒== (3) We now have 4 constants to determine and have 3 equations. Another equation is obtained by using the continuity of the electric flux normal component across the dielectric interface. d 0V 04εε= 02εε=()())4(42charge)-free no(02,ˆ2,ˆ4,ˆ4,ˆ1010211021022222210110111111bcDDbDabEzDacVEcDacEzDacVEnnnzznzzεεεεεεεε=⇒=−=⇒===−∇==⇒===−∇=rrrrrr The 4 equations are listed again as: () ()=+⋅=+⋅=+⋅=+⋅1010212102121242200bcbdbcdcVbdbccεε >>>>> ()()−====3/340320201201VbdVbcdVc Solutions are: () () ()()()() () ()zzadVzVEVzdVzVadVzVEzdVzVˆ34,334ˆ32,3202200201101−=−∇=⇒−=−=−∇=⇒=rrExample: For the following figure, (1) Calculate the surface charge density on the upper and lower plates. (2) Calculate the polarization charge density on the two plates. (3) Calculate the total (free plus polarization) charge density on the two plates. Solution: The previous examples give the solution of the electric field to be ()zadVEˆ/0−=r, Hence ()zradVEDˆ/4000εεε−==rr (1) Free charge density on the upper plate: dVDazS/4ˆ00ερ=⋅−=r (2) Polarization charge on the upper plate: ()dVEaParzzSP/31ˆˆ000εεερ−=−⋅=⋅=rr (3) Total charge on the upper plate: 00000//3/4 dVdVdVSPSεεερρ=−=+ In the same way, we can show that the total charge on the lower plate is 00/dVSPSερρ−=+. Hence the total charge on the two plates is zero. d 0V