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Parallel Plate Capacitor - Finding E field between plates

  1. Feb 9, 2012 #1
    Why is it that the field magnitude between two plates in a parallel plate capacitor is given by q/(εA)? In my book it is stated that one plate is of charge +q and the other -q. But if each plate is charged, wouldn't you need to account for the electric field produced by both places making magnitude 2q/(εA)? That is, the sum of both field vectors.

    My guess here is that only one plate is charged and the other one is given an induced charge. In other words, putting a Gaussian Field over the induced one results in zero flux. But I am unsure if this is correct?
     
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  3. Feb 9, 2012 #2

    NascentOxygen

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    If you were able to place a charge on one plate, without inducing an equal but opposite charge on the other plate, then it would not meet the definition of a capacitor. It would mean that the flux lines originating with the charge were terminating somewhere else other than on the opposing plate. In a [proper] parallel plate capacitor, flux does not escape.
     
  4. Feb 9, 2012 #3
    Agreed. I think I might know where I went wrong. Could you verify me? If a plate has charge q then its surface charge density is q/(2A) where 2A is accounting for both sides of the plate (edges can be neglected for long thin plates.) If that is true then the sum of the two vector fields is indeed q/(εA). And going back to what you said about no flux escaping the capacitor, this means that each plate must be equal and opposite in magnitude in charge. I had originally thought (and still don't know if it is correct) that an induced charge is one in which when the charge inducing the other is taken away, the induced charge is also taken away. In other words, with an induced charge, there is no net change change. But if no flux escapes the capacitor, this means that each plate must have net charge in opposite magnitude. That is, if only plate A had net charge and B did not (only surface induced charge,) there would be net flux. This makes sense though as if you connect a battery source with a potential difference, electrons will flow into one side and out of the other. When disconnected, each plate has a net charge - but their sum is zero. So even if there is induced charges near the surfaces, both do indeed have equal an opposite net charges.

    Is this correct?
     
  5. Feb 10, 2012 #4

    NascentOxygen

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    I don't like that at all. Where would the flux lines go, of all those charges on the wrong side of the plate? Flux lines are all supposed to be contained between the plates, within the dielectric.
    Quite so. If the voltage on one plate of the capacitor changes, that change is conveyed in a predictable way to the other side.
    You can't have a flux line originating on a + charge and not ending on a - charge somewhere. A line of flux is defined by a pair of charges: a + where it begins, and a - where it ends. In an ideal capacitor, all lines of flux are contained between the plates.
     
  6. Feb 10, 2012 #5
    Thanks for replying back!

    Yes, I agree with this. I was trying to make sense of it by visualizing each plate separately. If this was the case, the charge would be distributed on each surface evenly, correct? When the plates are brought together though, I know this is not the case as the charge moves between the two. So I guess where I am confused is to how the electric field does not double when the surface charge density doubles on one side? Is this always the case? I guess I somewhat understand that this increased surface charge density is accounting for the other plate, but for some reason my intuition is wanting me to think the field magnitude should be doubled. Is there a more correct way of looking at this?

    Yes, that is what I believe I was trying to imply. Plate A and plate B would have to be oppositely charged for all flux lines to terminate. If plate A was neutral and isolated, and plate B was is some way charged, when bring the plates together, plate A would get an induced charge on the face near the other, but in placing a Gaussian field around the two flux would then be able to escape, correct? Or simply put, having charge pulled toward the surface of plate A would result in the other face of A having equal and opposite charge (where flux comes from) being that it was originally neutral. The flux from this should be equal to the flux that would exist in a Gaussian field around the two plates as if the neutral plate did not exist. If this is true, then it goes back to my second sentence in this paragraph - Both must be equally charged such that when one induces the other, the outer surfaces are left neutral. Am I getting closer to understanding this or further?
     
  7. Feb 10, 2012 #6

    NascentOxygen

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    If you had a charge on one plate, but not on the other, than that would not be a capacitor. It would have capacitance, since it is storing charge, but you'd call it stray capacitance, like you get with a connector, or a track on a PC board. It's not something we could pack up and sell as a capacitor. The only way you could have a charge evenly distributed over both sides of a flat plate is is there is no other charged body anywhere nearby, OR if there are charged plates on both sides of it.
    I don't see anything doubling. There are no charges on the outside of the ideal parallel plate capacitor. They are all on the inside, attracted by the proximity of opposite charges.
    Do you mean a Gaussian surface? There is no flux escaping the dielectric, otherwise it would not be an ideal capacitor. In a capacitor, both plates start off uncharged, and if you add a charge to one plate it repels an exactly equal charge off the other plate and into the circuit it connects to. If it is not connected to a conductive path, then it will not be possible to add a charge to any plate of the capacitor.
     
  8. Feb 11, 2012 #7
    Yes Gaussian surface. I know it would not be an ideal capacitor, I was trying to further enforce the idea that both must be equally charged rather than one simply being induced only. Here, I have drawn a diagram of what I was saying. However, maybe my idea of an induced charge is still not correct.
     

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  9. Feb 11, 2012 #8

    ehild

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    When the idea of capacitor was introduced in our studies, it was said that it consist of two parallel metal plates, one charged, the other grounded. The charges of one plate attract opposite charges in the grounded one, and repel like charges which go to the ground. It was also shown, that disconnected from the ground, this plate is really charged.

    Instead of connecting to the ground, the capacitor plates can be connected to + and - terminals of a battery, and then both plates become charged with equal and opposite charges.

    Now forget about capacitors, you have only a planar arrangement of charges. The electric field is E= q/(2εA)=σ/2ε, pointing away from the plane if q>0.
    (σ is the surface charge density).

    Take a metal plate, with the same charge q. You know that the charges move to the surface, and the electric filed is zero inside the metal. Allow unequal surface charge on the sides, but σ1+σ2=q/A
    You can imagine two planes of charge instead of the metal. One produces the uniform electric field E1=σ1/(2ε), pointing away from the plane on both sides, the other plate produces E2=σ2/(2ε) .
    The contributions of the planes add up and produce the resultant electric field. On the left of the metal sheet, it is Eleft=-E1-E2=-(σ1+σ2)/(2ε). In the volume in the place of the metal, Emetal=E1-E2=(σ1-σ2)/(2ε)=0, and on the right Eright=(σ1+σ2)/(2ε).
    The electric field is zero inside the metal, so there is equal surface charge on both sides: σ1=σ2=q/2.
    At the end, the metal plate behaves like a single plane of charges, producing electric field of magnitude E=q/(2ε) on both sides.

    Place an other, uncharged metal plate parallel with the first one. Allow again surface charges, but the sum of them is zero in this case: σ3+σ4=0. Replace both metals with a pair of charged planes. Collect the contributions from each planes to the electric field.
    On the left of the figure, Eleft=-(σ1+σ2+σ3+σ4)/2ε.
    Inside the grey volume, Eg=(σ1-σ2-σ3-σ4)/ε=0.
    Between the metal sheets, Emiddle=(σ1+σ2-σ3-σ4)/2ε
    Inside the blue volume, Eb=(σ1+σ2+σ3-σ4)/2ε=0.
    On the right, Eright=σ1+σ2+σ3+σ4)/2ε.


    σ3+σ4=0, so σ4=-σ3.
    From the requirement of zero electric field inside the metals,
    σ1-σ2=0, that is σ1=σ2=q/(2A).
    and
    Eb=(σ1+σ2+σ3-σ4)/2ε=0=>q+2σ3=0=>σ3=-q/(2A),σ4=-σ3=q/(2A).

    Eleft=-(σ1+σ2+σ3+σ4)/2ε=-q/(2εA).
    Eright=(σ1+σ2+σ3+σ4)/2ε=q/(2εA).

    Emiddle=(σ1+σ2-σ3-σ4)/2ε=q/(2εA).


    The new uncharged metal does not change the field of the charged one. The surface charge density is -q/2 next to the charged metal and opposite on the surface away from it.

    You can apply this method when the second metal plate has some arbitrary charge.

    ehild
     

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    Last edited: Feb 11, 2012
  10. Feb 11, 2012 #9

    tiny-tim

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    Hi Sefrez! :smile:
    The electric displacement field (in equilibrium) is normal to the surface of any conductor, and has magnitude equal to the surface charge density: D = σ.

    (This is from the "discontinuous" or "impulsive" version of Gauss' law:
    divD = ρf becomes D - 0 = ∆D = ∆ρf = σ :wink:)​


    The plate of a capacitor is like any other conductor …

    if it has charge Q and area A, then:

    D = σ = Q/A if the charge is all on one side

    D = σ = Q/2A if the charge is on both sides (twice the area! :wink:)​

    (and bringing a second plate close of course pulls all the charge onto one side)
     
    Last edited: Feb 11, 2012
  11. Feb 11, 2012 #10

    ehild

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    Hi tyni-tim,

    Just to think about the problem a bit further, let be two metal sheets of area A parallel with each order. Give Q1 charge to the left sheet, Q2 to the right one.
    What is the electric field outside and in between the plates?
    What is the surface charge density on the metal surfaces?

    The method of superposition outlined in my previous post is an easy way to solve the problem. :smile:

    ehild

    E(left)=-(Q1+Q2)/(2Aε)
    E(right=(Q1+Q2)/(2Aε)
    E(middle)=(Q1-Q2)(2Aε)
    σ1=(Q1+Q2)/(2A)
    σ2=(Q1-Q2)/(2A)
    σ3=(-Q1+Q2)/(2A)
    σ4=(Q1+Q2)/(2A)
     
    Last edited: Feb 11, 2012
  12. Feb 11, 2012 #11
    Genius. Look at the metal plates as if they were two planes as charge is only on the surface. :) So basically what you are saying is that even if another neutral plate is present, it does not affect the field from the charged one because what ever charge is pulled (or pushed) to the inside, the outside is in opposite magnitude. Hence it cancels out the other surface?

    Hmm, I do not believe I have heard anything about the discontinuous or impulsive version. However, σ = Q/2A before another plate is present and then σ/A (meaning the field "does" double) does make sense. :)
     
  13. Feb 12, 2012 #12

    ehild

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    I do not understand your last sentence but it is true that a neutral metal plate parallel with the charged one does not alter the electric field, except in the volume it occupies. From a distance, the opposite charges at both surfaces of the neutral plate cancel each other, so the field should be the same as without the neutral plate. If the plates are of infinite extent the field is homogeneous. So it is the same close to the plate and far away.
    Try to solve the problem I wrote in post #10.


    ehild
     
    Last edited: Feb 12, 2012
  14. Feb 12, 2012 #13
    That is what I was saying. :) I did your problem earlier today, looking at it this way really makes it easier to understand!
     
  15. Feb 12, 2012 #14

    ehild

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    And I learnt this method here in Physicsforums. :smile:

    ehild
     
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