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Plate-mesh-plate capacitor double charge?

  1. Jun 10, 2014 #1
    need advice please...

    what happened when I insert additional mesh between plates inside capacitor's dielectric ?

    first I charge between plate 1 (-) and the mesh (+). I know this no problem from this link http://personal.ee.surrey.ac.uk/Personal/D.Jefferies/perfcap.html [Broken]

    after fully charged, I try to charge plate 1 (-) and plate 2 (+)...

    is there any charge between plate 1 and plate 2 ?

    thank you
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 10, 2014 #2

    NascentOxygen

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    Hi stevan. This sounds like a good homework-type question, so should be posted in the homework sub-forum.

    What do you think will happen?
     
  4. Jun 10, 2014 #3
    nooo :D
    I'm 32 years old programmer and I just curious about this problem

    because I read that if I use mesh for capacitor, it don't affect much to the capacity

    so I think what if I add more plate to get more charge...

    if I change the mesh with a plate too, I know nothing happened when I do the second charge...because it's same as 2 capacitors in series...

    1--------| |------2------| |-----------3

    so when I charge first capacitor (pin 1 negative, pin 2 positive),I can get charge,
    but when I charge again both capacitor at once (pin 1 negative, pin 3 positive), no current will flow because the voltage is equal


    but this time, the center plate is mesh, so in this design, I see 2 capacitor in one....
    plate 1 - dielectric - mesh
    plate 1 - dielectric - plate 2

    I know, if I charge plate 1 (-) and mesh (+), then charge mesh(+) and plate 2(-), I can get double charge...

    but this time I charge plate 1 (-) and mesh (+), then charge plate 1 (-) and plate 2 (+).....
    I don't know if I can charge plate 1 and plate 2 when the mesh already charged


    maybe you judge me from my bad english :D, sorry, because english isn't my daily language
     
    Last edited: Jun 10, 2014
  5. Jun 10, 2014 #4

    NascentOxygen

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    Taken in isolation, this question is easy to answer ....

    The uniform field between 2 plates is almost unaffected if you insert into the dielectric a conducting plate of negligible thickness so that it lies parallel to the plates. The conductor in this case simply joins together points already of equipotential, so nothing is changed by inserting that third plate.
     
  6. Jun 10, 2014 #5
    thank you sir.
    so, with configuration plate 1 - mesh - plate 2,
    when at first I charge plate 1(-) and plate 2(+), there is current flow to the plates until the voltage is equal with the power source...

    but, if I first charge plate 1(-) and mesh (+), and then charge plate 1(-) and plate 2(+)...

    I can't simulate this case in multisim or other electronic simulator program...because there is no capacitor with mesh inside :D

    maybe someone can give me other discussion that have correlation with this case ?
     
  7. Jun 10, 2014 #6

    Born2bwire

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    If the mesh is unconnected, then it does not have an impact as explained above. But if you are talking about charging the mesh, then it requires a circuit connection and then it is just the case of two capacitors in series with the plate separations defined by the placement of the mesh.
     
  8. Jun 10, 2014 #7

    NascentOxygen

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    No current would flow here because there exists zero potential difference now, assuming you are using the same battery, so the extra capacitor would remain uncharged.

    It sounds to me that for your experiment to work as you envisage the "mesh" you have in mind is going to be of such open weave that under a microscope it would resemble wire netting as seen on a garden fence, with voids so wide as to allow through a significant proportion of flux lines between the outer plates. Otherwise, you'll find a tight-weave mesh will act no differently to plates.

    In your case with the mesh, I think you would model it simply as a capacitor that has in parallel to it a pair of capacitors, the pair themselves identical and in series. :smile:
     
    Last edited: Jun 10, 2014
  9. Jun 10, 2014 #8
    That's exactly what I mean sir :)
    yes I use the same battery for first and second charge, and I disconnect the mesh first before I do the second charge.

    I try to add more charge through the holes in the mesh, so the outer plates can act as "another" capacitor.
    maybe distance between plate 1 and mesh is 1 mm, distance between mesh and plate2 is 1 mm.The holes in the mesh about 0.5 mm x 0.5 mm...maybe the ratio area of wires and the holes in the mesh is 50:50...so I think that's not too tight.

    I use mesh because try to get more charge by create more electric field through the holes of the mesh...
     
  10. Jun 11, 2014 #9

    NascentOxygen

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    I'll just add that I doubt that you will get more capacitance this way. The capacitance between the open mesh and ground plate will be less than if you used plates, and the presence of the charged mesh capacitor will reduce the charge added to the uppermost plate when you attempt to charge it. I'd guess you'll find no difference at all, sorry.
     
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