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Homework Help: Charge Distribution in an infinite plate

  1. Apr 17, 2013 #1
    I was trying to calculate the the charge distribution (surface charge density = σ in function of r) in a very long circular metallic plate.

    I know σ not constant if we get closer to the rim of the plate

    Let's say we want to calculate the E field in point Q that is x distant from the center, and we choose a point P, that is in a distance r from the center and in an angle θ to the line OQ (O is the center). This point has a area dA = r dr dθ.

    PQ² = z² = r² + x² - 2 r x Cos(θ)

    The field (in OQ axe) is
    dE = k dA σ/z²
    We know that E = 0 in any point and also ∫ σ r dr = Q/2π

    Integrating only θ (and simplifying a lot of stuff) we get
    ∫ σ r dr = Q/2π
    ∫ σ dr/(r²-x²) = 0
    Both integral from 0 ro R

    How can I find σ from this?
  2. jcsd
  3. Apr 17, 2013 #2

    Simon Bridge

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    In what way would a circular plate be considered "long"?

    So you know the field at some distance z perpendicular to the plate, and, from that, you want to derive the surface charge density? Since the plate is infinite - you cannot. There will always be a range of charge distributions that would give rise to that field.

    Perhaps you want to assume a uniform charge density?
    Then is that not an infinite sheet of charge?
  4. Apr 17, 2013 #3
    Is there a difference between a plate and a sheet? (sorry, I'm not american)

    The "z" I've mentioned is not a distance z perpendicular to the plate.
    I'm calculating the field at the plate (or sheet) plane

    http://img560.imageshack.us/img560/4998/42508231.png [Broken]
    Last edited by a moderator: May 6, 2017
  5. Apr 17, 2013 #4
    But I think what you said is still valid. There should be a lot of configurations in which the field is zero anywhere. In a metallic plate, though, the energy should be minimal. How can I use that?

    Energy = ∫VdQ = min
  6. Apr 17, 2013 #5

    Simon Bridge

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    Well .. "plate" has connotations of a lump of metal (though I suppose "sheet" has connotations of beds?)
    Shall we say "plane of charge" to be strictly geometric? (I'm not American either :))

    Your follow up suggests you mean an infinite planar conductor ... this should have a uniform planar charge distribution on both surfaces (front and back - metals have "thickness"). But since real metals are not infinite in extent either then it is safe to go with the geometric abstraction.

    For a uniform planar charge distribution, that is infinite in extent, is there any reason to think that any particular point in the plane will be any different from any other point in the plane? If not then you can put x=0 in your diagram.

    Apart from that, the link I gave you still holds - just put z=0 (in the link).
    But for a uniform distribution, you already know the answer.

    You could minimize the action by considering what would happen if the charge distribution were not quite uniform. That seems like a lot of trouble to go through when there is a perfectly good derivation in the link.

    [edit]Hang on - you mean you have a disk of charge? Or a charged disk?
    A charged conducting disk will have a ring of uniform linear charge density about the circumference. The field in the plane of the disk is still zero.
    Last edited: Apr 17, 2013
  7. Apr 17, 2013 #6
    Yes I do! Actually it was there where I was trying to get.

    Let's us assume that we have a metallic circular and very long plate/sheet/plane of charge and we charge it with a charge Q. Let's us assume what you said so:

    I) After charge distribution, we get a constant surface density of charge at any point

    If we choose any point inside the plane we know that the E field at this point should be zero (as is is a metallic plane). We have electric field in the plane (let's say xy) direction and in the line perpendicular to the plane (let's say z direction). Both fields have to be zero. For the z field to be zero we should have (as you mentioned) the same surface charge density at both sides of the plane, as it have thickness. So the z field is fine. We have to care about the xy field only. Let's say the thickness is very small so we can assume σ (surface charge density) = 2x the surface charge density at a single side, and that will not change the xy field calculation. So we have to have xy field zero. If (I) were right, then we should have xy field zero at a insulated circular and very long plate/sheet/plane of charge charged with the same amount of charge Q with a charge distribution such that at any point the σ is the same (note that the charge distribution does not change from the metallic one, so the e field should be the same). But we know this is wrong!

    At the rim of the insulated plate we have an electric potential of σa/πε0, where a is the radius of the plate, and at the center we have σa/2ε0

    If there is a electrical potential difference, then the electric field is not zero
    And so (I) have to be wrong!
    I know σ is approximately constant in the middle of the plate, and for a very very very long plate it will change only closer to the rim, but it is not all constant
    So I want to calculate it in function the distance r from the center of the plate/sheet/disc
    Last edited: Apr 17, 2013
  8. Apr 17, 2013 #7
    From dE = k dA σ cos(θ)/z² (again, z is NOT a distance perpendicular to the plate, see image)

    dE = k σ r dr dθ/(r² + x² - 2 r x Cos(θ))

    http://img442.imageshack.us/img442/7708/asidbfhisabdfg.png [Broken]

    How can we find a function σ(r) that satisfies this and makes the energy be minimal
    Last edited by a moderator: May 6, 2017
  9. Apr 17, 2013 #8

    Simon Bridge

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    Now I'm lost ...

    ... you say the field everywhere in the plane should be zero then you say "we have electric field in the plane".

    ...you say the plane is "metallic" (which is a technical term in chemistry btw) - what do you mean by this: that it is a conductor? But later you talk about an insulator.

    z is some distance in the plane and then it isn't and then it is?

    I am afraid I will have to insist on a clear question if I am to help you.
    What is the problem you are trying to solve?
    Tell me what you are trying to achieve?
  10. Apr 17, 2013 #9

    Simon Bridge

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    If you know what the field is, and you want to get the charge disrtibution, then you may find Poisson's equation or the differential form of Gauss' Law easier to use. You plug in the known ##\vec{E}## and get charge distribution out. If you only know the field at one point - then the problem is under-specified - it has infinite solutions.

    Off post #7 - you want the least-energy charge distribution that gives a zero field at point P. That's easy - ##\sigma=0##
  11. Apr 18, 2013 #10
    No. I said that if the statement (I) was right then we would have to have E field not zero inside the plate, which turns to be an absurd, so (I) should be wrong.

    As metallic I mean conductor (sorry by that)

    The insulator was an analogy: The charge distribution at the insulator I've mentioned was identical to the metallic one. So the E field and electric potential at any point should be the same too. I did that because we KNOW how to calculate the electric potential at the rim and at the center of a very long disc of charge (actually I got this result from the book Solved problems in Physics). We know there is a potential difference and that would be an absurd if the plate was metallic instead of insulated. So (I) have to be wrong. A metallic plate is not similar to THIS insulated plate I've mentioned, the metallic one could not have σ constant.

    As I said, look at the image. The circle is the metallic disc. All the points and lines are at the disc plane. z is QP distance. Sorry I used z here, but it has nothing to do with the z AXE. Neither do x. I don't know why I used these letters, sorry.

    I don't know what is the electric field outside the disc (if I did I could use E = σ/ ε0 to find σ of course).
    I only know the electric field inside should be zero, what gives me that equation.
    I want to find a possible function σ(r) that satisfies that and also satisfies the minimal energy condition.

    Again: ∫ σ r dr = Q/2π
    If we had σ=0,
    ∫ σ r dr would be zero
    Last edited: Apr 18, 2013
  12. Apr 18, 2013 #11


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    That's fine for the field in the PQ direction, but you only want the component in the OQ direction, yes? I see you multiplied by cos(θ) next, but that would only gives the desired component if dE were in the OP direction.
  13. Apr 18, 2013 #12
    Sorry, now I saw what I did
    You are right, I have to multiply by Cos(180 - PGO) :/
    Let me integrate again...
    Last edited: Apr 18, 2013
  14. Apr 19, 2013 #13
    It seems the answer is not that easy

    The first Integral (by wolfram) is

    (2 ((a - d) EllipticE[2/(1 + a)] - (-1 + a) EllipticK[2/(
    1 + a)]))/((-1 + a) Sqrt[1 + a])

    where a = (r^2 + x^2)/2r x and d = x/r
  15. Apr 19, 2013 #14
    Actually I was not hoping to find an easy answer, but to plot a graphic

    Does anyone know how this graphic would look like? Surface Charge Density vs distance from center
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