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Capacitor, maxwell's equations

  1. Mar 29, 2012 #1
    1. The problem statement, all variables and given/known data

    A parallel-plate capacitor with circular plates of radius 1.7 m is being charged. Consider a circular loop centered on the central axis between the plates. The loop has a radius of 2.6 m and the displacement current through the loop is 2 A.

    (a) At what rate is the electric field between the plates changing?

    2. Relevant equations

    maxwell's equations
    1) electric flux through a closed surface = charge enclosed by the surface divided be e0
    2) magnetic flux through an open surface = u0 times current through the surface
    3) EMF in a closed path = derivative of magnetic flux through the path WRT time
    4) currents cause magnetic fields


    3. The attempt at a solution

    there's an E-field in the capacitor
    C = e0*A/d
    q = C*V
    E = q/(A*e0) SO dE/dt = dq/dt * 1/(A*e0)

    I feel like I'm missing something.
    Does a changing E-field induce a current in a closed path? is that current proportional to the electric flux through the path?
     
  2. jcsd
  3. Mar 29, 2012 #2

    tiny-tim

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    hi jehan60188! :smile:
    "displacement current" isn't a current, it's ∂D/∂t, a current density (per area) see http://en.wikipedia.org/wiki/Displacement_current :wink:

    (but it's measured in units of A/m2, not A … perhaps they mean that 2A is the total current through the loop, ie the displacement current times the area? :confused:)
     
  4. Apr 3, 2012 #3
    bump, googled all week, but still no idea what to do
     
  5. Apr 5, 2012 #4

    tiny-tim

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    hi jehan60188! :smile:
    i was hoping you might find some more information (btw, what are parts (b) (c) etc of the question?), but if that's all there is, my guess :redface: is that the question means that ∂D/∂t times the area of the loop is 2 A

    (D is in coulombs per m2, so ∂D/∂t is amps per m2, so the units are correct)

    in that case, (a) is asking for ∂E/∂t, so i suppose all you need is an equation relating D and E :confused:
     
  6. Apr 5, 2012 #5
    solution
    phi = E*A
    so
    d(phi)/dt = d(E*A)/dt = dE/dt*(A)
    so
    e*d(phi)/dt = e* dE/dt*(A)
    where e = 8.85e-12
    so, we know displacement current = 2 (since capacitors only have displacement current)
    divide that by the area of the plates to get dE/dt
     
  7. Apr 6, 2012 #6

    tiny-tim

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    that's a rather confusing way of writing it :confused:

    you're saying ∂E/∂t = (1/εo)∂D/∂t = 2/εo N/C.s ?

    why do you want to divide by the area? (and why did you mention phi?)
     
  8. Apr 6, 2012 #7
    phi is the electric flux through a surface
    I_d = ε*d(phi)/dt is the definition of displacement current
    I'm going to change over to using single-quote to represent time-derivatives (so dx/dt = x')

    amperes law:

    integral(B~ds) = u*(I + I_d)
    where ~ is the dot product
    or in words: a closed path integral in a magnetic field is equal to the permeability of free space times the sum of displacement current (I_d) and enclosed current

    for uniform E-fields electric flux is E*A
    since our A is constant, phi' = A*E'

    so: ε*phi' = εAE'
    between the plates of a capacitor, there is only displacement current
    so: ε*phi' = εAE' = 2
    so: E' = 2/(Aε)
     
  9. Apr 6, 2012 #8

    tiny-tim

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    oh yes, i forgot :redface:, we decided that 2 amps was D' times area :rolleyes:

    so you're right, E' = D'/ε = D'A/Aε = 2/(Aε) N/C.s :smile:

    but you still don't need to mention phi, do you?​
     
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