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Calculating displacement current and magnetic field.

  1. Oct 16, 2016 #1
    1. The problem statement,

    A capacitor is there in free space consisting of 2 circular plates of radius ##r## separated by a distance ##z## which is a function of time. ##z(t) = z_0 + z_1 cos (\omega t)##; ##z_0(<<r)## and ##z_1(<z_0)## are constants. The separation ##z(t)## is varied in such a way that the potential difference ##V_0## between the plates remains constant.
    • Calculate the displacement current density & displacement current between the plates through a concentric of radius ##\frac{r}{2}##
    • Calculate ##\vec H## between the plates at a distance of ##\frac{r}{2}## from the axis of the capacitor.


    Now, ##\oint_{\frac{r}{2}} \vec H \cdot d\vec l = \int_S \vec J \cdot d\vec a + \int_S \vec J_d \cdot d\vec a## where ##J_d## is displacement current density. I hope to use this equation and equate ##J=0## and find out ##J_d## in the region between the plates. What I don't understand is how to use the condition ##V_0 = constant## to solve this problem. I calculated the capacitance ##C(t) = \frac{A\epsilon}{z_0+z_1 cos(\omega t)}## and that's it. I'm stuck.

    EDIT 1: I think the way to go is to use ##\vec E(t) = -\nabla V(t)## where ##V(t) = \frac{Q}{C(t)}## but I'm not sure if ##Q## is also a function of time. Is it?
     
  2. jcsd
  3. Oct 16, 2016 #2

    vela

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    How do you propose to find the displacement current?
     
  4. Oct 16, 2016 #3
    I updated the question. See Edit 1. I think I have to find ##\partial_t \vec E## first but I'm not sure if ##Q=Q(t)##.
     
  5. Oct 16, 2016 #4

    vela

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    The V(t) in ##\vec{E} = -\nabla V(t)## and ##V(t) = \frac{Q}{C(t)}## aren't the same quantity. The V in the first expression is also a function of the spatial coordinates, otherwise you'll get 0 when you take the gradient. The V(t) in the second expression is the potential difference across the capacitor; you were told it's equal to a constant ##V_0##. (That should answer your question about whether Q is a constant or not.)

    You're on the right track though. You want to relate the electric field inside the capacitor to the potential difference across the plates.
     
  6. Oct 16, 2016 #5
    But the question seems contradictory to me. if ##V_0## is constant then how can a displacement current exist? Does the constant V_0 imply a fixed charge on the capacitor?
     
  7. Oct 16, 2016 #6

    vela

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    The charge on the capacitor is equal to Q = C(t)V0. Do you still think the charge is fixed?
     
  8. Oct 16, 2016 #7
    Obviously no. That was my first thought! But I don't see another way to explain ##V_0## = constant. Moreover I don't even see why that is a necessary condition in this question.
     
  9. Oct 16, 2016 #8
    SO I did this :
    ## - \int _{0}^{z_0+z_1cos(\omega t)} \vec E \cdot d\vec l = V_0 \implies E= -\frac{V_0}{z_0+z_1cos(\omega t)} ## or ##\partial_t E = \frac{V_0z_1sin(\omega t)}{z_0+z_1cos(\omega t)}##. Is this the right way?
     
  10. Oct 16, 2016 #9

    vela

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    Yes, but the denominator should be squared in your final result.
     
  11. Oct 16, 2016 #10
    Oops. Sorry for that!
     
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