6.641,Electromagnetic Fields,Forces,and Motion Prof.Markus Zahn Lecture 12:Electroquasistatic Forces I.EQS Energy Method of Forces a)Circuit Point of View Figure 11.6.1 (a)Electroquasistatic system having one electrical terminal pair and one mechanical degree of freedom.(b) Schematic representation of EQS subsystem with coupling to extemal mechanical system represented by a mechanical (a) (b) terminal pair. Courtesy of Hermann A.Haus and James R.Melcher.Used with permission. q=C()v i-=品[cey]-ce0+vace dt =c附0+ve da dt R=m=v是[c(y-c张+E架 d dt =ce是va d dt -Bce +1v2 dc d d dt dw d dt W=energy mechanical power storage (force x velocity) 6.641,Electromagnetic Fields,Forces,and Motion Lecture 12 Prof.Markus Zahn Page 1 of 9
6.641, Electromagnetic Fields, Forces, and Motion Lecture 12 Prof. Markus Zahn Page 1 of 9 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 12: Electroquasistatic Forces I. EQS Energy Method of Forces a) Circuit Point of View Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. q=C v ( ) ξ () () (ξ) ⎡ ⎤ ξ ξ+ ⎣ ⎦ dq d dv dC i= = C v =C v dt dt dt dt ( ) ξ ξ + ξ dv dC d =C v dt d dt ( ) ⎛ ⎞ ξ ξ + ⎜ ⎟ ⎝ ⎠ ξ d 1 dC d 2 2 =C v v dt 2 d dt ( ) ⎡ ⎤ ξ ξ + ⎢ ⎥ ⎣ ⎦ ξ d 1 1 dC d 2 2 = Cv v dt 2 2 d dt N N W energy mechanical power storage (force velocity) dW d f dt dt ξ = × ξ = + () () ξ ⎡ ⎤ ξ ξ+ ⎣ ⎦ ξ 2 in d dv P = vi = v C v = C v v dt dt d dt dC d
w=2c旧，6- -cge=r{c 2c2()d店 dEc佰 b)Energy Point of View n=v-0+e能 dt dt vdq=dWe +f:d=dWe vdq-fd -、w/ a=constant ;v=w 6q l=cons tant C dξ=0 f=0 Figure 11.6.2 Path of line integration in A Bξ state space (g,E)used to find energy at location C. Courtesy of Hermann A.Haus and James R.Melcher.Used with permission. 0 w。=-∫d+ vdq q=0 E=cons tant v=0 w.=c间 =cons tant q dq=2C(E) 1q2 f=-oW/ 1q2dc(5) g=constant 2 C2()dE v dc() d 6.641,Electromagnetic Fields,Forces,and Motion Lecture 12 Prof.Markus Zahn Page 2 of 9
6.641, Electromagnetic Fields, Forces, and Motion Lecture 12 Prof. Markus Zahn Page 2 of 9 ( ) ξ ξ ξ 1 1 2 2 dC W= C v , f = v 2 2 d ( ) ( ) ⎛ ⎞ − ⎜ ⎟ ξ ξ ξ ⎜ ⎟ ξ ⎝ ⎠ 2 2 2 1 q dC 1 d 1 = =q 2 d 2 dC C b) Energy Point of View ξ ξ + dq dWe d vi = v = f dt dt dt ξ ξ + ξ⇒ − ξ vdq = dW f d dW = vdq f d e e ξ = ξ= ∂ ∂ − = ∂ξ ∂ e e q cons tan t cons tan t W W f = ; v q Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. 0 ξ = ξ= − ξ+ e ∫ ∫ q 0 cons tan t W = f d vdq ( ) ξ q v = C ( ) ( ) ξ = ξ ξ ∫ 2 e cons tan t q 1q W = dq = C C 2 (ξ) ξ 2 1 dC = v 2 d ( ) ( ) ( ) = ∂ ⎛ ⎞ ξ − − ⎜ ⎟ ∂ξ ξ ⎜ ⎟ ξ ξ ξ ⎝ ⎠ 2 e 2 2 q cons tan t W 1 d 1 1q dC f = = q = 2d 2 d C C
II.Forces In Capacitors Area A e二3 (a) 0s=+8Ex=-+8y (Lower electrode) X q=cA=EEA=SVA =C(x)v C(x)=A X Area A Eo Vo (b) Figure 3-36 A parallel plate capacitor(a)immersed within a dielectric fluid or with (b)a free space region in series with a solid dielectric. Courtesy of Krieger Publishing.Used with permission. 6.641,Electromagnetic Fields,Forces,and Motion Lecture 12 Prof.Markus Zahn Page 3 of 9
6.641, Electromagnetic Fields, Forces, and Motion Lecture 12 Prof. Markus Zahn Page 3 of 9 II. Forces In Capacitors s x v = E= x + σ + ε ε (Lower electrode) σ ( ) ε ε s x vA q= A= E A= =C x v x ( ) εA Cx= x Courtesy of Krieger Publishing. Used with permission. Ex = V0 x
a)Coulombic force method on upper electrode: 反=，EA=-EA=- A 2 2x2 2 because E in electrode=0,E outside electrode Ex -Take average Energy method:C(x)=SA X 器或终装 2 xZ EZAZ 2EA b) area A lE。 Figure 11.6.3 Specific example of EQS b systems having one electrical and one 0 mechanical terminal pair. Courtesy of Hermann A.Haus and James R.Melcher.Used with permission. =5b EA+EA =e5+8b EE0A 6-}部 1 6-r品c--2et的 三一 (e5+8ob2 6.641,Electromagnetic Fields,Forces,and Motion Lecture 12 Prof.Markus Zahn Page 4 of 9
6.641, Electromagnetic Fields, Forces, and Motion Lecture 12 Prof. Markus Zahn Page 4 of 9 a) Coulombic force method on upper electrode: σ− − ε ε 2 2 x sx x 2 1 1 1v f = E A= E A= A 2 22 x 1 2 because E in electrode=0, E outside electrode = Ex Take average Energy method: ( ) εA Cx= x ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ ε ε 2 2 2 x 2 1 dC 1 d 1 1 v A f= v = v A = 2 dx 2 dx x 2 x ( ) ⇒ − ε ε x q qx 1 v= = f = C x A 2 A 2 x 2 2 q x ε2 2 A − ε 2 1 q = 2 A b) Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. ( ) + ξ ξ ε0 ε a b a b 1 11 A A = ;C = ,C = C CC b ξ + ε ε 0 b = A A ε ε ξ + ε ε 0 0 b = A ( ) ξ ( ) − ⎛ ⎞ − − ⎜ ⎟ ξ ξ ⎜ ⎟ ξ ⎝ ⎠ εξ+ε ε ε ε 2 2 2 0 0 0 1 q = q = b= 1 d 1 d 1q 2 2 d Ad 2 A C f ( ) ( ) ( ) ξ ⎡ ⎤ ξ − ⎢ ⎥ ξ ξ ⎣ ⎦ εε ε ε εξ+ε εξ+ε 2 2 2 2 0 0 2 0 0 1d 1d 1 A vA =v C =v = 2d 2d b 2 b f
III.Energy Conversion Cycles Figure 11.6.4 Apparatus used to demonstrate amplification of voltage as the upper electrode is raised.(The electrodes are initially charged and then the voltage source is removed so g constant.)The electrodes,consisting of foil mounted on insulating sheets,are about 1 m x I m. with the upper one insulated from the frame,which is used to control its position. The voltage is measured by the electrostatic voltmeter,which“loads'”the system with a electrostatic capacitance that is small compared to that of the electrodes and (at least on a dry day) voltmeter a negligible resistance. B C(OV -C(O) C(L) B Figure 11.6.5 Closed paths followed A (b) in cyclic conversion of energy from DV mechanical to electrical form:(a)in(q,v) (a) plane;and (b)in (f,)plane. .A-B.With v=0,the upper electrode rests on the plastic sheet.A voltage Vo is applied. B-C.With the voltage source removed so that the upper electrode is electrically isolated,it is raised to the position =L. .C-D.The upper electrode is shorted,so that its voltage returns to zero. ·D→A.The upper electrode is retured to its original position atξ=O. Is electrical energy converted to mechanical form,or vice versa? Courtesy of Hermann A.Haus and James R.Melcher.Used with permission. 6.641,Electromagnetic Fields,Forces,and Motion Lecture 12 Prof.Markus Zahn Page 5 of 9
6.641, Electromagnetic Fields, Forces, and Motion Lecture 12 Prof. Markus Zahn Page 5 of 9 III. Energy Conversion Cycles Courtesy of Hermann A. Haus and James R. Melcher. Used with permission
∮vdq=∮y%+∮ds ∮vdq=∮fd ∮vda,∮de>o Electric energy in,mechanical energy out. ∮vdq,∮fd<0 Electric power out,mechanical energy in. D ∮vdq=jvdg+了vdq=2co)g-cv C(o)Vo =C(L)V 5o-ow-9sl-别 c0、 +b C(L)b(oA) L+b ∮vdg=c(og 1-ε】 c()Vlc enery out) ∮fd=-f6L 6-20-2co% EA ∮iu=-co8品=∮vdg fd<mechanical energy out is negative means mechanical energy is put in Mechanical energy is converted to electrical energy 6.641,Electromagnetic Fields,Forces,and Motion Lecture 12 Prof.Markus Zahn Page 6 of 9
6.641, Electromagnetic Fields, Forces, and Motion Lecture 12 Prof. Markus Zahn Page 6 of 9 ξ ξ v v ∫ ∫ vdq = f d ξ ξ > v v ∫ ∫ vdq, f d 0 Electric energy in, mechanical energy out. ξ ξ < v v ∫ ∫ vdq, f d 0 Electric power out, mechanical energy in. + − ( ) ( ) v∫ ∫∫ B D 2 2 0 A C 1 1 vdq = vdq vdq = C 0 V C L V 2 2 C 0 V =C L V () () 0 ( ) ( ) − v∫ 2 0 C L 1 vdq = C 0 V 1 2 ( ) 2 C 0 C ( ) ( ) ( ) ( ) ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ − ⎥ ⎢ ⎥ ⎢⎣ ⎥⎦ ⎣ ⎦ 2 0 1 C 0 = C0V 1 L 2 CL ( ) ( ) ε A C 0 = C L ⎛ ⎞ ⎜ ⎟ + ⎝ ⎠ ε ε ε 0 0 L b b A ( ) ( ) ( ) ⎡ ⎤ ⎛ ⎞ ⎢ ⎥ ⎜ ⎟ + ⎝ ⎠ −−< ⎣ ⎦ ε ε ε ε ε ε v∫ 0 2 2 0 0 0 0 L b 1 1L vdq = C 0 V 1 = C 0 V 0 2 b2 b (electric energy out) ξ − v∫ 0 fd = f L ( ) ++ + ( ) ε ε ε 2 2 2 0 2 0 0 0 0 1q 1 1 A C 0V f = = = C0V 2A 2 A 2 εb A0 ⎡ ⎤ ⎢ ⎥ ⎢⎣ ⎥⎦ ξ − ( ) ε ε v v ∫ ∫ 2 0 0 1 L fd = C 0 V = vdq 2 b v∫ fd 0 ξ< ⇒ mechanical energy out is negative means mechanical energy is put in Mechanical energy is converted to electrical energy 0 ξ + ξ vvv ∫∫∫ vdq = dWe f d
IV.Force on a Dielectric Material 0 Eo depth Figure 11.6.6 Slab of dielectric partially C extending between capacitor plates.The spacing,a,is much less than either b or the depth c of the system into the paper. Further,the upper surface at is many spacings a away from the upper and lower edges of the capacitor plates,as is the lower surface as well. Courtesy of Hermann A.Haus and James R.Melcher.Used with permission. C()=5o(b-5)c.EEc a 黄-支但 In equilibrium: Mass density 氏=2e-o)=ga fluid weight 5=-1v2e-o) 2 pga2 6.641,Electromagnetic Fields,Forces,and Motion Lecture 12 Prof.Markus Zahn Page 7 of 9
6.641, Electromagnetic Fields, Forces, and Motion Lecture 12 Prof. Markus Zahn Page 7 of 9 IV. Force on a Dielectric Material Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. ( ) ( ) − ξ ξ ξ + ε0 b c ε c C = a a ( ) ξ ξ ξ 2 1 dC f= v 2 d ( ) ε ε − 2 0 1 c = v 2 a In equilibrium: Mass density ξ ( ) ε ε − ρξ 2 0 fluid weight 1 c f = v = g ac 2 a ( ) − ξ ρ ε ε 2 0 2 1 v = 2 ga
v(t) Figure 11.6.7 In a demonstration of the polarization force,a pair of conducting transparent electrodes are dipped into a liquid (corn oil dyed with food coloring). They are closer together at the upper right than at the lower left,so when a voltage is applied,the electric field intensity decreases with increasing distance,r,from the apex. As a result,the liquid is seen to rise to a height that varies as 1/r2.The electrodes are about 10 cm x 10 cm,with an electric field exceeding the nominal breakdown strength of air at atmospheric pressure, 3 x 10 V/m.The experiment is therefore carried out under pressurized nitrogen. Courtesy of Hermann A.Haus and James R.Melcher.Used with permission. a→ar 5=1v2e-o) 「2pga2r2 V.Physical Model of Forces on Dielectrics E(r+d) d E(r) Figure 11.8.1 An electric dipole experiences a net electric force if the positive charge g is subject to an electric field E(r+d)that differs from E(r)acting on the negative charge g. Courtesy of Hermann A.Haus and James R.Melcher.Used with permission. fpoe=q[E(G+d-E间] =q[E间+ā.E问-E(何)] =q(d.v)E =(何.) Kelvin force 6.641,Electromagnetic Fields,Forces,and Motion Lecture 12 Prof.Markus Zahn Page 8 of 9
6.641, Electromagnetic Fields, Forces, and Motion Lecture 12 Prof. Markus Zahn Page 8 of 9 Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. a r → α ( ) − ξ ρ α ε ε 2 0 2 2 1 v = 2 g r V. Physical Model of Forces on Dielectrics Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. ( ) ( ) ⎡ ⎤ + − dipole ⎢ ⎥ ⎣ ⎦ f =q E r d E r ( ) ( ) ( ) ⎡ ⎤ +∇ − ⎢ ⎥ ⎣ ⎦ =q E r d E r E r i =q d E ( ) i ∇ =p E ( ) i ∇ Kelvin force
Figure 11.9.4 In terms of the Kelvin force density,the dielectric liquid is pushed into the field region between capacitor plates because of the forces on individual dipoles in the fringing field. Courtesy of Hermann A.Haus and James R.Melcher.Used with permission. 20000 0e0 0 0000 0000 Eo Vo P-(e-eo) 000 0 + Depth d A linear dielectric is always attracted into a free space capacitor because of the net force on dipoles in the nonuniform field.The dipoles are now aligned with the electric field,no matter the voltage polarity. 6.641,Electromagnetic Fields,Forces,and Motion Lecture 12 Prof.Markus Zahn Page 9 of 9
6.641, Electromagnetic Fields, Forces, and Motion Lecture 12 Prof. Markus Zahn Page 9 of 9 Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. + - v + -