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Fall 2016 ECE 305 Electromagnetic Theory Chapter 5 Electric Fields In Material Space Qiliang Li Dept. of Electrical and Computer Engineering, George Mason University, Fairfax, VA 1 § 5.1 Introduction Free Space / Vacuum: ε0 conductors Real material space nonconductors Nonconductors – insulators or dielectrics Susceptibility, permittivity, linearity, isotropy, homogeneity, dielectric strength, and relaxation time 2 § 5.2 Properties of Materials • Conductivity, σ unit: Siemens per meter ( S/m ) metals: σ >> 1 Insulators: σ << 1 in between Semiconductors As T 0K, some conductors have σ ∞ Called: superconductors What is “Semiconductor”? 3 Examples MOSFET conductor Semiconductor Optical fiber 4 § 5.3 Convection and Conduction Current • The current (in amperes) through a given area is the electric charge passing through the area per unit time. 1 A = 1 C/s 𝑑𝑄 𝐼= 𝑑𝑡 Current density 𝐽 = ∆𝐼 ∆𝑆 or 𝐼 = 𝑱 ∙ 𝑑𝑺 5 • Different between convection current and conduction current • Convection current does not involve conductors, therefore, does not satisfy Ohm’s law. It occurs when current flows through an insulating medium such as liquid, rarefied gas or a vacuum 6 ΔS ρV (charge density) z y Δy x ∆𝑄 𝜌𝑉 ∆𝑉 𝜌𝑉 ∆𝑆∆𝑦 ∆𝐼 = = = = 𝜌𝑉 ∆𝑆 𝑢𝑦 ∆𝑡 ∆𝑡 ∆𝑡 ∆𝐼 𝐽𝑦 = = 𝜌𝑉 𝑢𝑦 ∆𝑆 7 • In general: 𝑱 = 𝜌𝑉 𝒖 So, this equation is the general form of both convection and conduction current density For conduction current, 𝜌𝑉 can be expressed by electron density – see next page 8 • Conduction current requires a conductor 𝑭 = −𝑒𝑬 From momentum Ft = mu: 𝑚𝒖 = −𝑒𝑬 𝜏 τ: the average time interval between collisions – electron travel in a real materials space e 9 • Therefore, 𝒖 = 𝑒𝜏 − 𝑬 𝑚 If there are n electrons per unit volume 𝜌𝑉 = −𝑛𝑒 So 𝑛𝑒 2 𝜏 𝑱 = 𝜌𝑉 𝒖 = 𝑬 𝑚 or 𝑱 = 𝜎𝑬 10 § 5.4 Conductors • A perfect conductor (𝜎 = ∞) cannot contain E-field within it. Ee Ei Ee ρV=0 E=0 Ee is applied external field, Ei is internal field --isolated conductor: V= constant 11 • For a conductor whose ends are maintained at potential difference V 𝑬 ≠ 0 inside the conductor 𝑉 𝐸= 𝑙 𝐼 𝑉 𝐽 = = 𝜎𝐸 = 𝜎 𝑆 𝑙 𝐼 𝑆 = 𝑉 𝜎 𝑙 𝑅 = 𝑉 𝐼 = 𝑙 𝜎𝑆 or 𝑅 = 𝜌𝑐 𝑙 𝑆 Where ρc = 1/σ is the resistivity 12 • For a conductor with nonuniform cross section 𝑉 𝑅= = 𝐼 𝑉 𝑆 𝑬 ∙ 𝑑𝒍 𝜎𝑬 ∙ 𝑑𝑺 Power P (in watts): rate of change of energy W (in Joules) 𝑃= 𝜌𝑉 𝑑𝑉 𝑬 ∙ 𝒖 = 𝑉 𝑬 ∙ 𝜌𝑉 𝒖𝑑𝑉 = 𝑬 ∙ 𝑱𝑑𝑉 𝑉 13 Power density wP (in W/m3) 𝑑𝑃 𝑤𝑃 = = 𝑬 ∙ 𝑱 = 𝜎|𝑬|2 𝑑𝑉 For a conductor with uniform cross-section 𝑃= 𝑬 ∙ 𝑑𝒍 𝐿 𝑱 ∙ 𝑑𝑺 = 𝑉𝐼 𝑆 Or 𝑃 = 𝐼 2 𝑅 --- the common form of Joule’s law 14 Example 5.1: If 𝑱 = 1 𝑟3 2𝑐𝑜𝑠𝜃𝒂𝒓 + 𝑠𝑖𝑛𝜃𝒂𝜽 𝐴/𝑚2 Calculate the current passing through (a) A hemispherical shell of radius = 20 cm, 0 < 𝜋 𝜃 < ; 0 < 𝜙 < 2𝜋 2 𝐼= Solve: 𝑆 𝑱 ∙ 𝑑𝑺 where 𝑑𝑆 = 𝑟 2 𝑠𝑖𝑛𝜃𝑑𝜙𝑑𝜃𝒂𝒓 So 𝐼= 4𝜋 𝑠𝑖𝑛2 𝜃 (θ 0.2 2 𝜋 2 0 2𝜋 1 2𝑐𝑜𝑠𝜃 0 𝑟3 𝑟 2 𝑠𝑖𝑛𝜃𝑑𝜙𝑑𝜃|𝑟=0.2 = from 0 to π/2)=31.4 (A) 15 Continue example 5.1 (b) A spherical shell of radius 10 cm 4𝜋 𝑠𝑖𝑛2 𝜃 𝐼= (θ from 0 to π)=0 0.1 2 Example 5.4: A lead (𝜎 = 5 × 106 𝑆/𝑚) bar of square cross section has a circular hole. L=4 m Find the resistance. Solve: 𝑅 = 𝑙/𝜎𝑆 b/c: 𝑆 = 𝑑 2 − 𝜋𝑟 2 1 cm So, 𝑅 = 4 5×106 × 𝜋 9− 4 ×10−4 3 cm 16 § 5.5 Polarization in Dielectrics Dipole moment p= 𝑄𝒅 Polarization in dielectrics 𝑷 = lim 𝑁 𝑘=1 𝑄𝑘 𝒅𝒌 ∆𝑉→0 ∆𝑉 This is dipole moment per unit volume So, potential 𝑑𝑉 = 𝑷∙𝒂𝒓 𝑑𝑉 ′ 4𝜋𝜖0 𝑅 2 where P dV’ is the dipole moment (dV’ is diff vol.) b/c: 𝛻 So ′ 1 𝑅 = 𝒂𝑹 𝑅2 𝑷∙𝒂𝑹 𝑅2 = 𝜵′ ∙ 𝑷 𝑅 − 𝜵′ ∙𝑷 𝑅 17 • Therefore, 𝑉 = 𝑷∙𝒂 ′ 𝑹 𝑆 ′ 4𝜋𝜖0 𝑅 𝑑𝑆 ′ + −𝛻 ′ ∙𝑷 ′ 𝑑𝑉 𝑉 ′ 4𝜋𝜖0 𝑅 Compared to Eq. (4.68) and (4.69) The total positive bound charges on surface S: 𝑄𝑏 = 𝑷 ∙ 𝑑𝑺 The charge that remains inside S is −𝑄𝑏 = − 𝛻 ∙ 𝑷𝑑𝑉 𝑉 Total volume charge density ρt: 𝜌𝑡 = 𝜌𝑉 + 𝜌𝑝𝑉 = 𝛻 ∙ 𝜖0 𝐸 18 Therefore, 𝜌𝑉 = 𝛻 ∙ 𝜖0 𝑬 − 𝜌𝑝𝑉 = 𝛻 ∙ (𝜖0 𝑬 + 𝑷) From this equation, we have:𝐃 = (𝜖0 𝑬 + 𝑷) For some dielectrics, 𝑷 = 𝑋𝑒 𝜖0 𝑬 where Xe is electric susceptibility 19 § 5.6 Dielectric Constant and Strength 𝑫 = 𝜖0 1 + 𝑋𝑒 𝑬 = 𝜖0 𝜖𝑟 𝑬 Or 𝑫 = 𝜖0 𝜖𝑟 𝑬 = 𝜖𝑬 where 𝜖 = 𝜖0 𝜖𝑟 , the permittivity 𝜖𝑟 is the dielectric constant: 𝜖𝑟 = 1 + 𝑋𝑒 = 𝜖 𝜖0 Dielectric strength is the maximum electric field that a dielectric can tolerate or withstand without electric breakdown. 20 § 5.7 is not required § 5.8 Continuity equation and relaxation time 𝑑𝑄𝑖𝑛 𝑱 ∙ 𝑑𝑺 = − =− 𝑑𝑡 𝐼𝑜𝑢𝑡 = 𝜵∙𝑱= 𝜕𝜌𝑉 − 𝜕𝑡 b/c: 𝐽 = 𝜎𝐸 , and 𝛻 ∙ 𝐸 = So, 𝛻 ∙ 𝜎𝐸 = 𝜎𝜌𝑉 = 𝜖 −𝑡/𝑇𝑟 So, 𝜌𝑉 = 𝜌𝑉0 𝑒 𝑉 𝑑𝜌𝑉 𝑑𝑉 𝑑𝑡 𝜕𝜌𝑉 − 𝜕𝑡 𝜌𝑉 𝜖 𝜕𝜌𝑉 𝜌𝑉 = 𝜎 − 𝜕𝑡 𝜖 , where 𝑇𝑟 = 𝜖/𝜎 21 § 5.9 Boundary Conditions • Use Maxwell’s Equations 𝑬 ∙ 𝑑𝒍 = 0 𝑫 ∙ 𝑑𝑺 = 𝑄𝑒𝑛𝑐 A. Dielectric-Dielectric E1n Boundary Conditions Consider L: abcd 𝑆 𝑬 ∙ 𝑑𝒍 = 0 E1t E2 E2n ε1 E1 E2t Δh ε2 d a ΔW b c 22 𝑬 ∙ 𝑑𝒍 = 0 ∆ℎ ∆ℎ ∆ℎ = 𝐸1𝑡 ∆𝑤 − 𝐸1𝑛 − 𝐸2𝑛 − 𝐸2𝑡 ∆𝑤 + 𝐸2𝑛 2 2 2 ∆ℎ + 𝐸1𝑛 2 𝐷1𝑡 𝐷2𝑡 𝐸1𝑡 = 𝐸2𝑡 = 𝜖1 𝜖2 Use cylindrical Gaussian surface ∆𝑄 = 𝜌𝑆 ∆𝑆 = 𝐷1𝑛 ∆𝑆 − 𝐷2𝑛 ∆𝑆 𝐷1𝑛 − 𝐷2𝑛 = 𝜌𝑆 If 𝜌𝑆 = 0 (no free surface charge), 𝐷1𝑛 = 𝐷2𝑛 23 therefore, 𝜖1 𝐸1𝑛 = 𝜖2 𝐸2𝑛 From 𝐸1𝑡 = 𝐸2𝑡 and 𝜖1 𝐸1𝑛 = 𝜖2 𝐸2𝑛 𝑡𝑎𝑛𝜃1 𝑡𝑎𝑛𝜃2 = 𝜖𝑟1 𝜖𝑟2 θ1 B. Conductor-Dielectric θ2 Boundary Conditions ① No electric field may exist within a conductor 𝜌𝑉 = 0 , 𝐸=0 ②b/c E = −𝛻𝑉 = 0 there can be no potential difference between any two point in a conductor 24 ③ 𝐷𝑡 = 𝜖0 𝜖𝑟 𝐸𝑡 = 0, 𝐷𝑛 = 𝜖0 𝜖𝑟 𝐸𝑛 = 𝜌𝑆 E-field must be external to the conductor C. Conductor-Free space boundary conditions 𝐷𝑡 = 𝜖0 𝐸𝑡 = 0, 𝐷𝑛 = 𝜖0 𝐸𝑛 = 𝜌𝑆 Example 5.9 and 5.10 will discussed in the class. 25

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