Physics 7, Fall 00 8 ovember 00 Today in Physics 7: begin electrodynamics Constitutive relations we ve seen in E&M Charges in motion, again: Ohm s law Carrier collisions and drift velocity Carrier collisions and heating Resistance Examples: resistance and current 8 ovember 00 Physics 7, Fall 00 Constitutive relations in E&M Constitutive relations are relationships between parameters of a theory that are derived empirically or that depend for their calculation on physics that is not part of the theory We have met two constitutive relations so far and are about to meet a third: P = χee M = χmh J = E The last one is called Ohm s Law In general, the proportionality constant is a second rank tensor, as are the susceptibilities, but many common media are linear in the sense that the conductivity can be considered a scalar 8 ovember 00 Physics 7, Fall 00 Ohm s Law uppose a chunk of material with cross-sectional area A and length L has equipotential ends, with potential difference V uppose furthermore that this gives rise to a uniform E in the chunk Then the current that flows is I = JA= EA A V = V L R A E L 8 ovember 00 Physics 7, Fall 00 3 (c) University of Rochester
Physics 7, Fall 00 8 ovember 00 Ohm s Law (continued) Ohm s law should strike you as strange, at first, because you also know that J = ρv = E uppose J is constant Then v and E should be constant But if E is constant, charges should accelerate at a = qe/m, rendering it impossible to have a constant v o which relation is wrong? either What happens is that collisions between free charges in a current (like electrons) with fixed or slowermoving charges (like the ions the electrons leave behind), and other free current carriers, keep the acceleration from going on for very long 8 ovember 00 Physics 7, Fall 00 4 Ohm s Law (continued) In collisions with the ions, the kinetic energy gained by the free carriers from the field is largely transferred to the ions, and the electron starts over, like a car driving down Intercampus Drive, stopping at all the speed bumps, accelerating in between We therefore reconcile Ohm s Law with the definition of current density by supposing that the collision process results in a well-defined average velocity, and writing J = ρv In fact, we don t need to suppose; we can show that that s the way it is, in a crude classical model of what is mostly a quantum phenomenon, as follows 8 ovember 00 Physics 7, Fall 00 5 Collisions and drift velocity Consider a chunk of metal with carriers in it and an electric field E present If it has been a time t i since particle i last suffered a collision, and if it left that collision at speed v i, then the momentum of this particle is pi = mvi + qeti o a snapshot of the metal would reveal an average value of carrier momentum given by p = mv = i ( i i) p = mv + qet i= i i= The starting speeds v i are endowed by collisions with the fixed ions; their energy, in turn, comes from the thermal energy (heat) of the medium 8 ovember 00 Physics 7, Fall 00 6 (c) University of Rochester
Physics 7, Fall 00 8 ovember 00 Collisions and drift velocity (continued) Thermal motions in a solid or liquid are (essentially) random in magnitude and direction, so if is large, mvi = 0 ; thus, i= qe qe v = ti = t m m i=, Drift velocity nq t and J = ρv = nqv = E E m, where n is the number density of carriers Clearly J should be linear in E if the carrier velocities are mostly thermal, and if collisions take place 8 ovember 00 Physics 7, Fall 00 7 Collisions and heating Energy is continuously added to such a medium, as the kinetic energy gained by carriers between collisions is passed on to the fixed ions during the collisions Independent of the nature of the collisions, if a carrier passes through a potential difference V, it picks up kinetic energy qv All this is dissipated in collisions If a total current I flows, then the power dissipated in collisions is d P = qv = IV( = I R) = d d = E dτ dt J a E The collisional transfer of energy from carriers accelerated by fields, to the medium in which the current flows, is called resistance 8 ovember 00 Physics 7, Fall 00 8 Resistance For the wire with cross-sectional area A, length L, and conductivity, the resistance turned out to be Units: in cgs, In MK, L L R = ρ A = A [ ] [ ρ] [ R] ρ = resistivity ot to be confused with charge density = = sec = sec sec cm = volt Ω = =Ω m amp Ω m More commonly, [ R] [ ] [ ρ] Ω cm 8 ovember 00 Physics 7, Fall 00 9 (c) University of Rochester 3
Physics 7, Fall 00 8 ovember 00 Examples: resistance and current Griffiths problem 73: (a) Two metal objects are embedded in weakly conducting material of conductivity What is the resistance between them, in terms of their capacitance? (b) uppose you connect a battery between them and charge them up to a potential difference V 0 If you then disconnect the battery, the charge will leak away Find the time dependence of the potential difference between the metal objects 8 ovember 00 Physics 7, Fall 00 0 (a) Enclose the positively charged conductor with a surface ; then I = J da= E da= 4πQ V = 4πCV R ε0 R = = in MK 4πC C 8 ovember 00 Physics 7, Fall 00 (b) Q = CV = CIR dq Q = I = dt RC Q t dq = dt Q RC Q0 0 Qt () = Qe 0 Q0 V() t = e = V0 e C = Ve = 4πt t ε0 0 Ve 0 in MK 8 ovember 00 Physics 7, Fall 00 (c) University of Rochester 4
Physics 7, Fall 00 8 ovember 00 Griffiths problem 7: A capacitor C is charged up to potential difference V 0 ; at time t = 0 it is connected to a resistor R, and begins to discharge (a) Determine the charge on the capacitor, and the current through the resistor, as a function of time (b) What was the original energy stored in the capacitor? how that the heat delivered to the resistor is equal to the energy lost by the capacitor C R 8 ovember 00 Physics 7, Fall 00 3 Q dq Q (a) V = = IR ; = I =, so, as before, C dt RC Qt () = Qe 0 = CVe 0, and dq V0 I = = CV0 e = e dt RC R (b) W = CV0 In capacitor, at start V 0 trc E= Pdt I Rdt e = = dt R 0 0 0 Dissipated In resistor V0 RC trc e CV = = 0 R 0 8 ovember 00 Physics 7, Fall 00 4 (c) University of Rochester 5