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

This preview shows pages 1–2. Sign up to view the full content.

PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS CLASS 11. 32.14, 32.19, 32.53, 32.74; 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 = = = 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 × = = 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 = 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online