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Chapter 5 Steady Electric Currents 5-1 Introduction 5-2 Current density and Ohm’s law 5-3 Electromotive force and Kirchhoff’s voltage law 5-4 Continuity of equation and Kirchhoff’s current law 5-5 Power dissipation and Joule’s law 5-6 Boundary conditions for current density 5-7 Resistance calculation §5-1 Introduction Types of electric currents: n Conduction current: Drift motion of conduction electrons and/or holes in conductors and semiconductors; very low for the average drift velocity of the electrons (1/1000 m/s); n Electrolytic current: Migration of positive and negative ions; n Convection current: Motion of electrons and/or ions (positively or negatively charged particles) in vacuum or in rarefied gas; not governed by Ohm’s law. §5-2 Current density and Ohm’s law Consider a steady motion of one kind of charges q over a differential surface Δs with velocity u and the total charges ΔQ passing this surface in a time interval Δt is Total current over the differential surface Δs: Current density: The total current I flowing through an arbitrary surface S: Noting that the charge density ρ = Nq, we have Which is the relation between the convection current density and the velocity of the charge carrier p Drift motion of charge carriers under electric field p Atoms remain neutral (ρ=0) For most of metallic conductors in which the average electron drift is proportional to electric field, we write Electron mobility µe: Cu: Al: Then we have, Where ρe=-Ne is the charge density of the drifting electrons. We rewrite it by Conductivity, σ: σ = -ρeμe; a macroscopic parameter of a medium; Unit: amper per volt-meter (A/Vm) or siemens per meter (S/m) Drude model: conductivity for metals (Microscopic Ohm’s law) Metal and Free electron gas Harmonic oscillation of electrons under the excitation of electric fields m e a el = F E _ local + F damping −iωt du me = −eE L e − meγ u dt Collision frequency ⇒ u= −iωt e EL e imeω − meγ ⇒ J = −Neu = ( Nuclei fixed; electron are free moving −iωt −Ne 2 )E L e imeω + meγ Ne 2 Ne 2 1 ⇒ σ (ω ) = = −imeω + meγ meγ (1− iω / γ ) Conductivity (static) Ne 2 σ (0) = meγ Resistivity: the reciprocal of conductivity (Ωm) Count Alessandro Giuseppe Antonio Anastasio Volta, 1745--1827, Italy André-Marie Ampère, 1775--1836, France Georg Simon Ohm, 1789—1854, German Resistance and conductance Voltage between terminals 1 and 2: The total current: Point for microscopic form of Ohm’s law Fig. 5-3 Homogeneous conductor with a constant cross section Resistance Macroscopic Ohm’s law Resistance and conductance §5-3 Electromotive force and Kirchhoff’s voltage law Ohmic material under static electric field Static electric field is conservative A steady current cannot be maintained in the same direction in a closed circuit by an electrostatic field. (D. K. Cheng, p. 206) (steady: motion and constant velocity;static: no motion) n Electric fields inside an electric battery? Impressed electric field, Ei p non-conservative electric field caused by chemical action; p Electromotive force (emf) is the integral from the negative to the positive electrode inside the battery. The conservative electrostatic field satisfies emf: or Two different electric fields: (1) Conservative fields from the charges, both inside and outside the battery; (2) Nonconservative fieldsimpressed fields, from chemical reaction, only inside the battery R Connect the battery electrodes by resistance R Emf: (integral over the closed loop) J = I/S n For one source of electromagnetic force, n For more than one source of electromagnetic force and more than one resistor, Kirchhoff’s voltage law Around a closed path in an electric circuit, the algebraic sum of the emf’s (voltage rises) is equal to the algebraic sum of the voltage drops across the resistances. §5-4 Equation of continuity and Kirchhoff’s current law From the principle of conservative of charge, the current leaving one region is the total outward flux of the current density through the enclosing surface S, Apply divergence theorem, we obtain for a stationary volume The above equation holds regardless of V, and the integrands must be equal. Equation of continuity The charge conservation law can also be given as follows For steady currents, Steady electric currents are divergenceless or solenoidal. Over any enclosed surface, we have Which can be written as Kirchhollf’s current law The algebraic sum of all currents flowing out of a junction in an electric circuit is zero. Relaxation time of charges in a metal? Combining the Ohm’s law with the continuity equation and assuming σ, we have Relaxation time, τ The solution is Eg: Copper τ = 1.52×10-19s The charge density at a given location will decrease with time exponentially. §5-5 Power dissipation and Joule’s law Power p provided by an electric field E in moving a charge q: Total power P delivered to all charge carriers in a volume dv: = J (current density) or = power density Total power converted into heat in a volume V Joule’s law In a conductor of a constant cross section, dv=dsdl, with dl measured in the current direction §5-6 Boundary conditions for current density In the absence of non-conservative energy source, we shall have We can obtain the boundary conditions for J (as in Fig. 3-23 and in Sec. 3-9): (1) Normal component of boundary current, Jn The normal components of J at two sides of an interface is continuous. (2) Tangent component of boundary current, Jt The ratio of the tangential components of J at two sides of an interface is equal to the ratio of the electric conductivities. Q5-3 Two conducting media with conductivities σ1 and σ2 are separated by an interface. The steady current density in medium 1 at point P1 has a magnitude J1 and makes an angle α1 with the normal. Determine the magnitude and direction of the current density at point P2 in medium 2. J1 cos α1 = J 2 cos α2 σ 2 J1 sin α1 = σ1 J 2 sin α2 tan α2 σ 2 = tan α1 σ1 If medium 1 is a much better conductor than medium 2 σ 1 >> σ 2 or σ2 →0 σ1 α2 →zero, and J2 emerges almost perpendicularly to the interface (normal to the surface of the good conductor). The magnitude of J2 is J 2 = J 2 2t + J 2 2 n = ( J 2 sin α 2 )2 + ( J 2 cos α 2 )2 = [( 1 1 σ2 σ J1 sin α1 )2 + ( J1 cos α1 )2 ] 2 =J1[( 2 sin α1 )2 + cos 2 α1 ] 2 σ1 σ1 Boundary between two lossy dielectrics Q5-4 (a) J normal component Kirchhoff’s voltage law J1 = J 2 V =(R1 + R2)I =( Hence, J= d1 d + 2 )I σ1S σ 2 S σ1σ 2V I V = = S ( d1 ) + ( d 2 ) σ 2 d1 + σ1d 2 σ1 σ2 ( A / m2 ) (b) To determine E1 and E2: V = E1d1 + E2 d 2 σE =σ E J1 = J 2 1 1 E1 = σ 2V σ 2 d1 + σ1d 2 E2 = σ1V σ 2 d1 + σ1d 2 2 2 (V/m) (V/m) (c) The surface charge densities on the upper and lower plates : ρs1 = ε1E1 = ε1σ 2V σ 2 d1 + σ1d 2 ρs 2 = -ε2 E 2 = - (C/m 2 ) ε2 σ1V σ 2 d1 + σ1d 2 (C/m 2 ) The surface charge densities at the interface: ρ si = D2 n − D1n = ε 2 E2 n -ε1E1n = (ε 2σ 1 -ε1σ 2 )V σ 2 d1 + σ 1d 2 (C/m2 ) From these results we see that ρ ≠ - ρ but that ρ + ρ + ρ = 0 s2 s1 s2 s1 si §5-7 Resistance calculations The basic formula for capacitance can be written as Fig. 5-7 Two conductors in a lossy dielectric medium Since the metallic conductors are equipotential media, you can choose anyone of the integral paths for calculating the electric potential difference. Note: The dimension of RC and ε/σ is [time]. Calculate resistance Conductor resistance V − ∫ L E ⋅ dl − ∫ L E ⋅ dl = R= = I ∫ J ⋅ dS ∫ σ E ⋅ dS S S The procedure for computing the resistance of a piece of conducting material between specified equipotential surfaces (or terminals): 1. Choose an appropriate coordinate system. 2. Assume a potential difference V0 between conductor terminals. 3. Find electric field intensity E within the conductor. homogeneous material: Laplace’s equation ∇ V = 0 E = -∇V 2 4. Find the total current 5. I= ∫ s J ⋅ ds = ⋅ s σ E ⋅ ds Find resistance R by taking the ratio V0/I. Q5-6 A conducting material of uniform thickness h and conductivity σ has the shape of a quarter of a flat circular washer, with inner radius a and outer radius b. Determine the resistance between the end faces. appropriate coordinate: cylindrical coordinate assume a potential difference V0 between the end faces: V=0 on the end face at y=0 V=V0 on the end face at x=0. solve Laplace’s equation subject to the following boundary conditions d 2V =0 2 dφ V=0 at Φ=0 V=V0 at Φ =π/2 V = c1φ1 + c2 , V= 2V0 π φ 1 ∂ ∂V 1 ∂ 2V ∂ 2V ∇V = (r )+ 2 2 + 2 r ∂r ∂r r ∂φ ∂z 2 The current density is ∂V 2σ V0 J = σ E = -σ∇V = -aφσ = -aφ r∂φ πr The total current I can be found by integrating J over the φ = π surface at 2 which-ds = -aφ hdr I= ∫ s 2σ V b dr 2σ hV b 0 0 J ⋅ ds = h∫ = ln π r π a a V0 π R= = I 2σ h ln(b / a) Note that, for this problem, it is not convenient to begin by assuming a total I because it is not obvious how J varies with r for a given I. Without current J , E and V cannot be determined. 0