∮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 96.641, 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