Engineering Electromagnetics Lecture 9 Dr.-Ing. Erwin Sitompul President University http://zitompul.wordpress.com President University Erwin Sitompul EEM 9/1 Chapter 6 Dielectrics and Capacitance Boundary Conditions for Perfect Dielectric Materials Consider the interface between two dielectrics having permittivities ε1 and ε2, as shown below. We first examine the tangential components around the small closed path on the left, with Δw<< : E dL 0 Etan1w Etan 2w 0 Etan1 Etan 2 President University Erwin Sitompul EEM 9/2 Chapter 6 Dielectrics and Capacitance Boundary Conditions for Perfect Dielectric Materials The tangential electric flux density is discontinuous, Dtan1 Dtan 2 Etan1 Etan 2 1 Dtan1 1 Dtan 2 2 2 The boundary conditions on the normal components are found by applying Gauss’s law to the small cylinder shown at the right of the previous figure (net tangential flux is zero). DN1S DN 2S Q S S DN1 DN 2 S President University • ρS cannot be a bound surface charge density because the polarization already counted in by using dielectric constant different from unity • ρS cannot be a free surface charge density, for no free charge available in the perfect dielectrics we are considering • ρS exists only in special cases where it is deliberately placed there Erwin Sitompul EEM 9/3 Chapter 6 Dielectrics and Capacitance Boundary Conditions for Perfect Dielectric Materials Except for this special case, we may assume ρS is zero on the interface: DN 1 DN 2 The normal component of electric flux density is continuous. It follows that: 1EN1 2 EN 2 President University Erwin Sitompul EEM 9/4 Chapter 6 Dielectrics and Capacitance Boundary Conditions for Perfect Dielectric Materials Combining the normal and the tangential components of D, DN1 D1 cos1 D2 cos2 DN 2 Dtan1 D1 sin 1 1 Dtan 2 D2 sin 2 2 2 D1 sin 1 1D2 sin 2 After one division, tan 1 1 tan 2 2 President University 1 2 1 2 Erwin Sitompul EEM 9/5 Chapter 6 Dielectrics and Capacitance Boundary Conditions for Perfect Dielectric Materials The direction of E on each side of the boundary is identical with the direction of D, because D = εE. E1 1EN1 2 EN 2 Etan1 Etan 2 1 2 1 2 E2 President University Erwin Sitompul EEM 9/6 Chapter 6 Dielectrics and Capacitance Boundary Conditions for Perfect Dielectric Materials The relationship between D1 and D2 may be found from: 2 D2 D1 cos 1 1 2 2 sin 2 1 The relationship between E1 and E2 may be found from: 1 2 E2 E1 sin 1 2 President University 2 cos 2 1 Erwin Sitompul EEM 9/7 Chapter 6 Dielectrics and Capacitance Boundary Conditions for Perfect Dielectric Materials Example Complete the previous example by finding the fields within the Teflon. Eout E0ax Dout 0 E0ax Pout 0 • E only has normal component Din Dout 0 E0a x 0 E0a x Din Ein 0.476E0a x r 0 r 0 0 E0a x Pin 1.1 0Ein 1.1 0 0.524 0 E0a x r 0 President University Erwin Sitompul EEM 9/8 Chapter 6 Dielectrics and Capacitance Boundary Conditions Between a Conductor and a Dielectric The boundary conditions existing at the interface between a conductor and a dielectric are much simpler than those previously discussed. First, we know that D and E are both zero inside the conductor. Second, the tangential E and D components must both be zero to satisfy: E dL 0 D E Finally, the application of Gauss’s law shows once more that both D and E are normal to the conductor surface and that DN = ρS and EN = ρS/ε. The boundary conditions for conductor–free space are valid also for conductor–dielectric boundary, with ε0 replaced by ε. Dt Et 0 DN EN S President University Erwin Sitompul EEM 9/9 Chapter 6 Dielectrics and Capacitance Boundary Conditions Between a Conductor and a Dielectric We will now spend a moment to examine one phenomena: “Any charge that is introduced internally within a conducting material will arrive at the surface as a surface charge.” Given Ohm’s law and the continuity equation (free charges only): J E v J t We have: v E t v D t President University Erwin Sitompul EEM 9/10 Chapter 6 Dielectrics and Capacitance Boundary Conditions Between a Conductor and a Dielectric If we assume that the medium is homogenous, so that σ and ε are not functions of position, we will have: v D t Using Maxwell’s first equation, we obtain; v v t Making the rough assumption that σ is not a function of ρv, it leads to an easy solution that at least permits us to compare different conductors. The solution of the above equation is: v 0e( )t President University • ρ0 is the charge density at t = 0 • Exponential decay with time constant of ε/σ Erwin Sitompul EEM 9/11 Chapter 6 Dielectrics and Capacitance Boundary Conditions Between a Conductor and a Dielectric Good conductors have low time constant. This means that the charge density within a good conductors will decay rapidly. We may then safely consider the charge density to be zero within a good conductor. In reality, no dielectric material is without some few free electrons (the charge density is thus not completely zero), but the charge introduced internally in any of them will eventually reach the surface. ρv ρ0 v 0e( )t ρ0/e ε/σ President University Erwin Sitompul t EEM 9/12 Chapter 6 Dielectrics and Capacitance Homework No homework this week. President University Erwin Sitompul EEM 9/13