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Homework Help: Parallel plate capacitor and magnetic field

  1. Aug 25, 2007 #1
    1. The problem statement, all variables and given/known data
    A parallel plate air-filled capacitor is being charged as in figure. The circular plates have a radius of 4.0 cm, and at a particular instant the conduction current in the wires is .280 A.

    A. What is the displacement current density (jD) in the air space between the plates?
    B. What is the rate at which the electric field between the plates is changing?
    C. What is the induced magnetic field between the plates at a distance of 2.0 cm from the axis?
    D. What is the induced magnetic field between the plates at a distance of 1.0 cm from the axis?

    2. Relevant equations
    an image is attached

    3. The attempt at a solution
    This is my solution as is...

    A. i = .280 A
    r = 4cm (.04m)
    n = 3.14
    jD=i/nr^2 (thats n*r squared)
    = 55.73 A

    B. dE/dt=jD/ [tex]\epsilon[/tex] (permitivitty constant)
    = 55.73/8.85*10^-12
    = 6.3*10^-12 N/Cs

    C. B= ur/2nR^2 *i (u = mu) = (1.256*10^-6)(.04m)/2(3.14)(.02*.02)
    = 2*10^-5 (.28)
    = 5.6*10^-6

    D. B=ur/2nR^2 (i) =(1.256*10^-6)(.04)/2(3.14)(.01*.01) (.28)
    = 2.24*10^-13

    answer a is correct but b, c, and d are not. I have tried this multiple times and cant figure out where im going wrong. Any help would be appreciated.

    Attached Files:

    Last edited: Aug 25, 2007
  2. jcsd
  3. Aug 25, 2007 #2


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    Your exponent for part b is wrong... it should be +12.
  4. Aug 26, 2007 #3


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    I don't understand how you're doing c and d. Did you try to apply Ampere's law? Can you show the details?

    You should use the conduction current + displacement current.
  5. Aug 26, 2007 #4


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    Also for part b), the equation should be:




    You know the area [tex]\pi{R}^2[/tex] where R = 0.04m, so you can calculate dE/dt.
  6. Aug 26, 2007 #5
    correction: using B= (uo id/2piR^2) r
    to find induced magnetic field inside a circular capacitor.

    id=displacement current
    R^2= radius of plate squared
    r= radius from center (point inside capacitor)

    I have values
    uo=1.256810^-6 (permeability constant)
    id= .280
    R= 4 cm
    r(c)= 2cm
    r(d)= 1 cm

    plugging in the given values i get
    part c= 7*10^-9
    part d= 3.5*10^-9

    does this look correct?
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