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Elecric Displacement in a plate capacitor

  1. Mar 4, 2009 #1
    There is a // plate capacitor with one plate charged (sigma) and the other charged (-sigma). Between the plates are 2 dielectrics each of thickness a. (This does not matter to my question). Why is the electric displacement equal to sigma? Every solution I find draws a gaussian surface around one plate and says DA=(sigma)A, so D=sigma. I can accept that, but it seems to me that that neglects the contribution of the second plate which should yeild a total D=2*sigma (by my incorrect thinking). Why do we only need the surface to be around one plate?
  2. jcsd
  3. Mar 6, 2009 #2


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    Hi Dhqwpoe

    so you have
    [tex]\int D.dA = Q[/tex]

    if you integrate around the whole capacitor what is the sum of charge? Remember signs... This shows the integral of D field outside the capacitor is identically zero.

    To make use of guass's law within the plate, enclose one plate, then one side of your guassian surface is within the plates as well

    A little thinking about the results and symmetry give the field as zero outside & as you described inside based on th charge sigma
  4. Mar 22, 2009 #3
    Okay I thought I understood it, but now I really don't think that I do.

    Is it true that the electric displacement caused by one plate is (sigma)?

    Then is it not true that the displacement caused by the second plate is also (sigma)?

    So then don't we add them to get that the total displacement inside? (which would be 2 sigma).

    (I get that by enclosing just one plate each time). I agree that the displacement outside is zero (enclose both plates).

    The only thing that I can think of is that the surface has 2 sides so really D=Q/(2A') where A' is the area of the part of the plate we are enclosing (and A'=A/2, where A is the area of our guassian surface). So we have the total D= Q/(2A')+Q/(2A') =Q/A'=sigma. Is that right? So if there were only one plate D would be (sigma)/2?

    Any response would be awesome. Thanks!
  5. Mar 24, 2009 #4
    The electric displacement, D, is related to the free charge, and in this case the charge on the capacitor plates. The definition is


    Taking the volume integral of both sides gives


    Using the Divergence Theorem on the left side of the equation to relate a volume integral to a surface integral gives

    [tex]\int\vec{D}\bullet d\vec{S}=Q_f[/tex]

    where Qf is the net free charge within the volume and dS is the differential surface perpendicular to the D vector. Therefore, the vector dS, which is normal to the differential surface, is parallel to the D vector. So,


    Therefore, the D vector starts and terminates on the free charges and has a magnitude equal to the free surface charge density.

    Look at lanedance's explanation again. It gives a nice description of the Gaussian surface.
  6. Mar 25, 2009 #5
    I get all that... But don't both plates contribute to the electric displacement? As in, aren't you just finding the divergence caused by one of the plates? I would do what you just did for the top plate. Then I would do the same thing for the bottom plate. And then I would add them. That yields D=2sigma_f inside and D=0 outside. Why is that wrong? Why can't you just add the electric displacement from each plate?
  7. Mar 26, 2009 #6


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    hi dhqpwoe
    why can you?

    the gusssian surface integrates over the displacement vector, so adding 2 together is inconsistent with your original integral. The 2nd plate cancels the field outside the other plate.
  8. Mar 26, 2009 #7
    Another way to think about this is the D vector is the product of a constant and the E vector,


    For a set of parallel plates without a dielectric in between the plates (epsilon = 1), the charges on one plate (assume positive) surrounded by a Gaussian surface have only E vectors leaving the surface towards the other plate thus a negative divergence exists. For the other plate, negatively charged, the E vectors are entering the surrounding Gaussian surface thus a positive divergence exists. If the Gaussian surface encloses both plates then no E vectors are leaving the surface or entering the surface (assuming no edge effects) and there is no E field outside the plates. The same vector exists at each plate so you can't add a vector to itself to get twice the magnitude.
  9. Mar 26, 2009 #8
    If we were finding E inside, we could simply find the E contribution from one plate and add to to the E contribution from the other plate, no? Why does that not work here?
    I think that I'm just thinking about the problem really wrongly or something. (Sorry for being such a trouble, too, I just really want to understand this!)
  10. Mar 27, 2009 #9


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    As I said why can you? it is inconsistent with the rules you are using.

    To find E with a guassian surface, you are essentially integrating over the field E. This is the field that exists. The integrand is dot product of the field and area vector normal to the surface. So when you have determined the field using this method, you cannot add anything to it or double it after the fact. That is incosistent with your integrand assumption in the first place.

    The only trick in all these problems is to use symmetry to simplify the integral.

    To help your understanding try doing this, think of a single plate (infinite if need be) with surface charge sigma, take a gaussian surface around it. ie reactangular box with faces parallel to the plate and infintesimal thickness in the normal direction to the plate).

    The symmetry here, is the same magnitude field will be out of the faces on both sides, normal to the face, if a face has area A, then

    [tex]\int E_{1plate}.dA =2E_{1plate}.A = Q_{enc} = A.\sigma[/tex]

    [tex]E_{1 plate} = \frac{\sigma}{2}[/tex]

    When you add a 2nd plate with charge -sigma, the field outside is given by:
    [tex]\int dA =2E_{2 plate outside}.A = Q_{enc} = 0 [/tex]

    [tex]E_{2 plate outside} = 0 [/tex]

    Now to calculate the field inside, go back to the previous guassian surface enlcosing one plate only:
    [tex]\int E.dA =A.E_{2plate} + A.E_{2 plate outside} =E_{2plate}.A + 0 = Q_{enc} = \sigma.A[/tex]

    So the field is double that of a single plate inside & zero outside, if that is what is throwing you off, but still only sigma
    [tex]E_{2 plate} = \sigma = 2.E_{1 plate}[/tex]
  11. Mar 27, 2009 #10
    Okay maybe I understood the whole time but was explaining it poorly. If there were only one plate, would D be sigma/2? The way I see it, inside each plate contributes sigma/2 and outside they cancel. That is what I meant by adding. Calculate the D contribution of each plate and then add them. Is that wrong?
  12. Mar 28, 2009 #11


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    did you read the post above - i think it is totally covered there...

    but you can do what you propose and add the solutions, as long as you consider each plate independently, it is the method of superposition
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