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Ampere's Law and charge on the capacitor

  1. Apr 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Screen Shot 2015-04-28 at 7.25.18 PM.png

    Number 9

    2. Relevant equations

    Ampere's law; B*dl= mu(I_enclosed + I_displace)
    I_displace=e_0 * d(E*dA)/dt

    3. The attempt at a solution

    thumb_IMG_4118_1024.jpg

    There's no current enclosed, only displacement current. I don't know how to find the change in electric flux, so I don't know how I can compare B1 and B2. I know B1 is just = mu*I/(pi*r)
     
  2. jcsd
  3. Apr 29, 2015 #2

    ehild

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    You know how the charge on the capacitor and the current flowing to it are related. You also know the relations between voltage and charge, and voltage and electric field. From all of those, you get the displacement current - how is it related to the current flowing in and out of the capacitor?
     
  4. Apr 29, 2015 #3
    Will displacement current = current flowing in and out of capacitor?

    If I relate them to area ->
    B_2*pi*r = mu*e_0* (pi/4*r^2)/(pi*r^2) * I <-------- ratio of small disk (radius = r/2) over total area of capacitor radius r
    B_2= e_0/4 * mu*I/(pi*r)
    B_1,3= mu*I/(pi*r)

    Except we still have e_0 as a factor, which isn't one of the options....
     
  5. Apr 29, 2015 #4

    ehild

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    I can not follow your derivation, where you got e_0 from?
     
  6. Apr 29, 2015 #5
    Oh nevermind, I got confused... there's no e_0. Thanks! :)
     
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