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Homework Help: Capacitor+magnetic field

  1. Feb 6, 2010 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    There's a capacitor with circular plates of radius 50 mm and separated by a distance of 5 mm. We apply a sinusoidal difference of potential whose maximum value is 150 V and has a frequency of 60 Hz. Determine the amplitude of the magnetic field, between the plates and at a distance of 50 mm away from the center of the capacitor.


    2. Relevant equations
    I've no idea.



    3. The attempt at a solution

    I've no clue about what equation to use. I'm thinking about Maxwell's equation: [tex]\oint \vec B d\vec s =\mu _0 \varepsilon _0 \frac{d\Phi _E}{dt}+\mu _0 I_{\text{enclosed}}[/tex] but I'm not even sure.
    In any case, I'm almost sure I have to calculate the E field of such a capacitor. I don't know if this is right, but I reached that the E field in a point situated inside the capacitor and over the straight line passing by both center of the plates as to be worth [tex]4\pi d \sigma \int _0 ^{0.005} \frac{dr}{r^2 \sqrt{r^2+d^2}}[/tex] and I'm stuck here.
    But I'm not sure this is relevant to calculate the E field only in this line.
    I've also figured out that [tex]V(t)=150 \sin (60 t)[/tex] and that [tex]\vec E =-\nabla V[/tex].
    Any tip is greatly appreciated as I'm at a loss.
     
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  3. Feb 7, 2010 #2

    kuruman

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    I think you are making this problem more complicated than it is. Assume that the electric field is uniform between the plates (even though your point of interest is at the edge where fringing is significant). Then you can easily relate V to E to the electric flux and use Maxwell's equation as you suggested.

    If you do not wish to assume that the electric field is uniform between the plates, then to calculate it off the axis, where your point of interest lies, you need to do an elliptic integral and I don't think you want to go there.
     
  4. Feb 8, 2010 #3

    fluidistic

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    Thanks for the tip kuruman. I'm still somehow confused. Say I assume the E field is uniform between the plates, what value do I choose? As I know it's not uniform, I don't know what value (average maybe?) of the E field to take.

    And about the elliptic integral, you're right, I would not go there! But next semester starts the serious (Jackson's book) E&M course, so don't be surprised if I pop up with such an integral.
     
  5. Feb 8, 2010 #4

    ideasrule

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    I think you're still overcomplicating it. V=Ed since E is uniform, so E=V/d. Now you can find an expression for electric flux and apply Ampere's law.
     
  6. Feb 8, 2010 #5

    fluidistic

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    Thanks. I don't know if I'm on the right way, I get that the net flux through the plates is worth [tex]\frac{V_0 \sin (\omega t)}{d} \cdot 2 \pi r^2[/tex] by applying Gauss's law. Is this right? If so, I'm goint to tackle the rest.
     
  7. Feb 8, 2010 #6

    kuruman

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    This is the net electric flux through what area?
     
  8. Feb 8, 2010 #7

    ideasrule

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    Gauss' law? :confused: You just multiply electric field by area to get flux. E=V/d and A=pi*r^2, so flux is just V/d*pi*r^2.
     
  9. Feb 8, 2010 #8

    fluidistic

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    Through the two plates... oh... I should divide by 2, right? Because flux enters one plate and leave the other. So I'm wrong.
    So it would be [tex]\frac{V_0 \sin (\omega t)}{d} \cdot \pi r^2[/tex]. Am I right?

    Edit: Ok I got it! I plan on to continue!
     
  10. Feb 8, 2010 #9

    fluidistic

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    Now I guess that [tex]I_{\text{enclosed}}=0[/tex] because I don't see any circuit, but I'm not 100% sure. If I assume that, I reach [tex]\oint \vec B d \vec s=\frac{\mu _0 \varepsilon _0 V_0 \sin (\omega t)}{d}[/tex].
    I'm not able to figure out in my head the magnetic field. Where is it? What is its direction?
    For the sake of guessing, I guess I'll simply end up with [tex]Bd=\frac{\mu _0 \varepsilon _0 V_0 \sin (\omega t)}{d}[/tex], thus [tex]B=\mu _0 \varepsilon _0 V_0 \sin (\omega t)[/tex].
    That would be between the plates of the capacitor.
    And for any points outside it, does B=0 since E (hence the E flux)=0?
     
  11. Feb 9, 2010 #10

    kuruman

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    Yes, Ienclosed = 0. It is the displacement current that produces the magnetic field. To see the relation of the magnetic field to the changing electric flux, look at the equation

    [tex]
    \oint \vec B d\vec s =\mu _0 \varepsilon _0 \frac{\partial}{\partial t}\int \vec{E}\cdot\hat{n}dA[/tex]

    The line integral is over the contour loop that forms the boundary of the area through which you are calculating the electric flux. The loop is oriented so the circulation in the sense of integration matches the normal to the surface with the use of the right hand rule.
     
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