L5-1 ELECTRIC FORCES ON CHARGES Lorentz Force Law: a = f/m = qE/m ≈ eV/mL [m s-2] Kinematics*: Electron kinetic energy wk: 0 o t o 2 o v a(t)dt ˆv at ˆz = z + z•v t + at /2 = = + ∫ z anode, phosphors cathode ray tube (CRT) -+ -V cathode heated filament Electron charge = -e = -1.6x10-19 [C] z deflection plates E⊥ L * For E in -z direction wk = fs = (eV/L)L = eV [Joules] Electron volt = energy of 1 electron moving 1 volt = e Joules ( )ofq E v H Newtons fqE ma = + × μ ⇒ = =L5-2 ELECTRIC FORCES ON CHARGED CONDUCTORS Force on free charges: Repulsive forces: 0 z E(z) Eo Deepest electrons experience zero force -e ⎯Eo Attractive pressure Like charges repel, unlike charges attract +ρs +ρs E = 0 ⎯E ⎯E -2 -2 -1 e s s o o o Force: f qE eE [N] Electric pressure: P = ρ <E> [N m ] Surface charge: = - E [C m ] EAverage E: < E> = [V m ] 2 Electric pressure: = = − ρ ε -22 e o o 1P = - ε E [N m ]2L5-3 ENERGY METHOD FOR FINDING FORCES Force, work, and energy: dw = f ds ⇒ f = [N] C = εοA/s w = CV2 = [J] Q ≠ f(s) if C is open circuit f = = = = -PeA [N] ⇒ Pe = [N m-2] = [J m-3] +Q -Q ⎯E +V 0 C ds f Electric fields always pull on conductors → attractive force The static pressure Pe of the electric field on the upper plate is the same with a battery attached: oe s s 1 2 P <E> E=ρ = ρ is the force externally applied to the upper plate fff s dw ds 1 2 2 2 o 1Q 1Q s 2C 2 A = ε dw ds 2 o 1Q 2Aε 2 2o o o (EA) 1( E )A2A 2 ε = εε 2 o 1 E2 εL5-4 ENERGY METHOD WITH A BATTERY Incremental work dw: *C = εA/s +Q -Q ⎯E +V 0 C ds f dQs Externally applied force f w = dw = f ds = -V dQ + (V2/2) dC To battery To capacitor 2CV2 dQ = VdC ⇒ dw = -(V2/2)dC f = Pe = -εo E2/2 (as before) Q.E.D. Force and pressure: 22 2 o o e2 A Edw V dC* V A PAds 2 ds 2 2s ε ε=− = = =−LATERAL FORCES – ENERGY METHOD Energy derivative: Energy derivative: +Q -Q +V 0 C fε >> εοE s D W L A’ = Ws +Q -Q C s fE D W A’ = Ws A’ Fringingfield 2 2 o 22 2o o e2 2 o o dwf (externally applied) dD Q Q s w 2C 2 WD (EWD) s Qs 1f ( E )Ws P A ' [N] 22WD 2 WD =− = = ε ε = = = ε = ε ε ⊥ =− =− εε= + = + = ε ε + ε ε− ε = = Δ = & 2 o D o o o 2 o e Qd( )dw 2Cf dD dD (L-D)W DWCC C s s W[D( - ) L] s ( )Ef A' [N] P A '; A' Ws (E pushes, E pulls) 2 L5-5L5-6 ROTARY ELECTROSTATIC MOTORS Energy derivative: Torque: Motor power: Peak power: P = Tω [W] Average power: Pavg = P/2 (duty cycle = ½) stator rotor T + -θ R v + -Segmentation advantage: T [Nm] = -dw/dθ ∝ A’ ∝ nRs (n = # segments) n = 4 stator +θ v -rotor =− =−θ θ ε θ= 2 2 o Qd( )dw 2CT[Nm] d d A RC , A = 2 s 2 ε = ε θ ≅ × × 22 o 22 o EQs RTherefore : T = A' [Nm] 2 22R R pressure gap-area A' 2 A’ = 2RsL5-7 ELECTROSTATIC SENSORS s → s + δ⇒ CV2/2 → C’V’2/2 initially, then: Voltage decays to Vo = VsR/(R + Rs) ⇒ Voltage pulse to amplifier, Δw ≈ wδ/s > Eb > ~10-20 [J] Minimum detectable δ≈sEb/w [m] (hears brownian motion) Cantilevered microphone: + -Rs V+ f plate area A s C, V Cantilevered capacitor C at V volts R Vs outMIT OpenCourseWare http://ocw.mit.edu 6.013 Electromagnetics and Applications Spring 2009 For information about citing these materials or our Terms of Use, visit:
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