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hw11 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 11...

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PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS CLASS 11. 32.14, 32.19, 32.53; Q15.01, Q15.02 32.14. Since the radius of the circular loop is greater than the radius of the capacitor plates, the displacement current through the loop is the entire displacement current between the plates. Other than that, the size of the loop has nothing to do with the problem. dt dE R dt dE A dt EA d dt d i E d 2 0 0 0 0 ) ( π ε ε ε ε = = = Φ = , where R is the radius of the capacitor plates, since the electric field is confined to the space between the plates. Then, 2 2 2 12 - 2 0 m) 10 . 0 ( ) m /N C 10 (8.85 A 0 . 2 π π ε × = = R i dt dE d s V/m 10 2 . 7 12 × = dt dE 32.19. ) m 6 . 1 m)( / F 10 85 . 8 ( ) ( 2 12 0 0 0 dt dE dt dE A dt EA d dt d i E d - × = = = Φ = ε ε ε A 8 . 2 s) m / V 10 0 . 2 V)( / s m A 10 42 . 1 ( Then s. m / V 10 0 . 2 s 10 0 . 2 m) / V 10 ) 0 . 4 0 . 0 ( c) 0 Then . 0 b) A 71 . 0 s) m / V 10 0 . 5 V)( / s m A 10 42 . 1 ( Then s. m / V 10 0 . 5 s 10 0 . 4 m) / V 10 ) 0 . 6 0 . 4 ( a) V) / s m A 10 42 . 1 ( 11 11 11 6 5 10 11 10 6 5 11 = × - × = × - = × × - = = = = × - × = × - = × × - = × = - - - - - d d d d d d i i dt dE i dt dE i i dt dE dt dE i
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32.53. a)
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