MIT OpenCourseWare http://ocw.mit.edu 6.013/ESD.013J Electromagnetics and Applications, Fall 2005 Please use the following citation format: Markus Zahn, Erich Ippen, and David Staelin, 6.013/ESD.013J Electromagnetics and Applications, Fall 2005. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms6.013, Electromagnetics and Applications Prof. Markus Zahn, November 8 & 10, 2005 Lecture 16 & 17: Electroquasistatic and Magnetoquasistatic 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 ξ v() i=dq= d ⎣⎡ ()v⎦⎤ () dv + v dC (ξ)C ξ =C ξ dt dt dt dt ()dv +vdC d ξ =C ξ dt dξdt P=vi =v d ⎡C ξ v⎤= C ξ vdυ+ v2 dC d ξ in dt ⎣()⎦ () dt dξdt d1⎛ 2 ⎞ 2 dC d ξ=C ()ξ ⎜ v ⎟+v dt ⎝2 ⎠ dξdt =d1⎡ C ξ v2 ⎤+1v2 dC d ξ ⎢ ()⎥dt ⎣2 ⎦ 2 dξdt dW dξ = + f dt ξ dt W=energy mechanical power storage (force ×velocity) 6.013, Electromagnetics and Applications Lecture 16 & 17 Prof. Markus Zahn Page 1 of 23W=1C ()ξ v2, f ξ =1v2 dC 2 2dξ 1q2dC 12d ⎛ 1 ⎞ = 2 C2 ()dξ = − 2q dξ⎝⎜⎜C ()⎠⎟⎟ξ ξ b) Energy Point of View dqdWedξvi= v = +fξdt dt dt vdq = dW + f d ξ⇒ dW = vdq − f d ξ e ξ e ξ ∂W ∂Wf= − e ;v = e ξ ∂ξ ∂qq=constan t ξ=constan t ξf d Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. 0 W= − ξ +∫ vdq e ∫ = ξ cons tan t q0 = qv= ξC ()q 1 q2 W= dq = ξξe ∫ C () 2C ()ξ=cons tan t 2∂w = −1q2 d ⎜⎜⎛ 1 ()⎟⎟⎞ = 1 C2q () dC (ξ)f= − e ∂ξ qconstant = 2 dξ⎝C ξ⎠ 2 ξ dξ 1 2dC (ξ)= v 2 dξ 6.013, Electromagnetics and Applications Lecture 16 & 17 Prof. Markus Zahn Page 2 of 23II. Forces In Capacitors σs= +εEx = +εv (Lower electrode) x s x εvA () x q= σA= εE A= =C x v ()= εxA Cx 6.013, Electromagnetics and Applications Lecture 16 & 17 Prof. Markus Zahn Page 3 of 23 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.a) Coulombic force method on upper electrode: 1 121 εv2 f= σE A= −εE A= − Ax s x x2 2 2x2 1 because E in electrode=0, E outside electrode = Ex2 Take average Energy method: ()= εA Cx x 1 1 f= v2dC =1v2εAd ⎛⎞ = −1 v 2εA x ⎜⎟2 dx 2 dx x 2⎝⎠ x2 ε A 2qqx 1 2x qx22 = −1q2 v= () = εA ⇒fx = − 2 A2 ε2 A Cx εb) Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. 1 1 1 ε0A εA () =a + b;C a= ,C b= C ξ C C ξ b ξ b = +ε0A εA ε ξ+ε0b = εε 0A 1 2d ⎛ 1 ⎞−12q2d 1q 2 f = − 2q dξ⎜⎝⎜C ()⎟⎠⎟= εε0A d (εξ+ε b)= − 2 ε0Aξ 0ξξ 22 f =1 2d ξ 1 2d ⎡ εε0A ⎤ 1v ε ε 0A 2ξ 2v dξ(C ())=2v dξ⎢⎣εξ+ε 0b⎥⎦ = − 2 (εξ+ε 0b) 6.013, Electromagnetics and Applications Lecture 16 & 17 Prof. Markus Zahn Page 4 of 23III. Energy Conversion Cycles Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. 6.013, Electromagnetics and Applications Lecture 16 & 17 Prof. Markus Zahn Page 5 of 23ξ ξ>∫ ∫ vdq, f d 0 Electric energy in, mechanical energy out. ξ ξ<∫ ∫ vdq, f d 0 dw 0 + f d ξ∫vdq = ∫ e ∫ξ vdq= f d ∫ ∫ξξ Electric power out, mechanical energy in. B D ∫vdq = ∫vdq +∫vdq = 21 () 2 − 21 () 2C 0 V 0 C L V A C =CL V C0V()0 () ∫vdq= 21 () 2 ⎢⎢⎡ 1 − ()CL2C0()⎤ 1⎡ C0()⎤ C 0 V 0 ⎥=C0V ()02 ⎢1 −⎥()L ⎥ 2 ⎢⎣ CL ()⎥⎦C⎣ ⎦ ⎛Lb ε0 ⎞ ⎜+ ⎟C0 CL ()= ε A ⎝ ε⎠ () b (ε0 A ) ⎡⎛ ε0 ⎞ε⎤ ⎢⎜Lb+ ⎟⎥ ∫12C 0 V 02 ⎢⎢−⎝ ε0b ε⎠ ⎥⎥ 12C 0 V 02 εε 0 bL <vdq = () 1 = − () 0 (electric energy out) ⎢ ⎥ ⎢⎣ ⎥⎦ ∫fdξ = −f L 0 A ⎤ f= + = + C0V 0= + C 0 V 02 0 1q2 12 () 21 () ⎡⎢ε A ⎥2 ε0A 2 ε0A 2 ⎣⎢bε0 ⎥⎦ ∫fdξ= −12C 0 V ()02 εε 0Lb = ∫vdq ∫fdξ<0 ⇒mechanical energy out is negative means mechanical energy is put in Mechanical energy is converted to electrical energy 6.013, Electromagnetics and Applications Lecture 16 & 17 Prof. Markus Zahn Page 6 of 23IV. Force on a Dielectric Material Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. ()= ε0 (ba −ξ)c +ε a ξ cC ξ 1 ()2 dC ξ f= vξ 2 dξ = 21 v2 ac (−0 ) εε In equilibrium: Mass density 1v2 c (εε− 0 )= ρξ ac f= gξ 2 a fluid weight ξ =1 v2 (− 0 )εε 2 ρga2 6.013, Electromagnetics and Applications Lecture 16 & 17 Prof. Markus Zahn Page 7 of 23Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. a →αr 1 v2 (− 0 )εε ξ= 2 ρα22gr V. Physical Model of Forces on Dielectrics Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. ⎡ ⎤f =qEr +d −E r ⎡ ⎤ dipole ⎢⎣( ) ()⎥⎦ =q E r +di ∇E r −E r⎢ () () ()⎥⎣ ⎦ =q i ∇ E( ) = p i ∇ E Kelvin force ( ) 6.013, Electromagnetics and Applications Lecture 16 & 17 Prof. Markus Zahn Page 8 of 23Courtesy of Hermann A. Haus and James R. Melcher. Used with permission. 6.013, Electromagnetics and Applications Lecture 16 & 17 Prof. Markus Zahn Page 9 of 23VI. MQS Energy Method of Forces A. Circuit Approach dλ d di dL (ξ) dt =dt ⎣ ()i⎦=L ()dt + i dt v= ⎡L ξ⎤ ξ dL ξp= vi =L ()di +i2 ()ξi dt dt ⎛ 2=L () d1i2 ⎞+i dL (ξ)ξ dt ⎜⎝2 ⎟⎠ dt d1 2 ⎤ 1 2dL (ξ)= L ξi + i dt ⎣⎢⎡ 2 () ⎦⎥ 2 dt d1 ⎤ dξ = ⎢⎡ ()i2 ⎥+1 2dL (ξ)L ξ i dt ⎣2 ⎦ 2 dξ dt vi= dWm + fξ dξ⇒ Wm = 1L ()i2, fξ = 1i2 dL (ξ)ξ dt dt 2 2 dξ λ=L ξ i ⇒f = i2() ξ 21 dLdξ(ξ) 21 λ dL (ξ)= 2 L2 ()ξ dξ = −1 λ2d ⎢⎡1 ()⎦⎤ 2 dξ⎣ L ξ⎥ 6.013, Electromagnetics and Applications Lecture 16 …
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