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Capacitor Plates

  1. Oct 28, 2006 #1

    Quick question. If you have a parallel plate capacitor charged with X volts. Is there a detectable E field outside the plates at all or only between the plates?

    Jason O
  2. jcsd
  3. Oct 28, 2006 #2
  4. Oct 28, 2006 #3
    Well, if the surface area of the cap is very very large and the plate separation is very very small then the ideality of an "infinite sheet capacitor" is approached, which has no fringe field. In a typical capacitor that approaches the properties of an infinite sheet capacitor, the fringe field effect is negligible.
  5. Oct 28, 2006 #4
    What about on the back sides of the plates? Is there a detectible E-field there?
  6. Oct 28, 2006 #5


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    completely agree.

    true, but any finite plate capacitor of the same spacing with the same voltage has an edge somewhere and will have fringe fields bowing out. if there is an edge somewhere, it doesn't matter how big the plates are (as long as they're much bigger in one dimension than the spacing).

    away from the edges, there shouldn't be a significant field on the backside. you can show this by use of Gauss's Law.
  7. Oct 28, 2006 #6
    Okay thank you. I am asking about this because I am trying to model the electric field of a simple capacitor using this FEA program called FEMME. And When I setup the model, if I draw the plates of the capacitors as thin rectangles, I can specify the voltage potential on the lines that represent the rectangles but I am a little concerned about the results it is giving me. If I specify the voltage potential all around the border of the plate, then it shows an equal e-field potential around the entire plate, but I am not sure if this is the case in real life. I attached two screenshots of the simulation I did. The first one shows the charged plates with the complete borders having the potential. The second screenshot is the same as the first except that I only defined the inner two border lines with the potential. Which diagram is more accurate?

    Jason O

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  8. Oct 28, 2006 #7


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    Science Advisor
    Homework Helper

    In real life, you sometimes have to take into account the wires going to the capacitor, as well as its relation to its suroundings.
    Particuarly at high frequencies.
  9. Oct 28, 2006 #8


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    that's a cute name for a finite element analysis program.
    i am not sure what you mean by "inner two border lines" of equal potential. (?) what those two lines looklike is a matter of a threshold and scaling. you can make that drawing on the right look different (big equi-potential line gets even bigger - more outa the picture and little circle even tighter). i am assuming that there is a scale change in the "x" axis (perp to plates) that is not reflected in the "y" direction.
    if you were to keep the "x" and "y" scale equal, as the plates get bigger and/or closer (assuming they're perfectly flat and parallel) the less you will see of the edge effect. of the two pics, i dunno that there is a less accurate. they both look legit to me.
  10. Oct 29, 2006 #9


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    Staff: Mentor

    In a static situation (no time dependence of the charge or the fields), this should be correct. As you may recall, in electrostatics the potential is the same everywhere on the surface of a conductor. If the potential were not the same, (positive) charges would start to move from high potential to low potential, and the situation would not be static any more! :biggrin:

    In real life, a capacitor is often used in situations where the charges and fields are not static (e.g. when you put one in an AC circuit), and then it is may not be safe to assume that the potential is the same everywhere on the surface.
  11. Oct 29, 2006 #10
    Thanks everyone for your help and input :-)
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