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Electric Field for infinite sheet of uniform charge per unit area

  1. Oct 3, 2007 #1
    OK. In my book it says that for "any point" the Electric Field Magnitude for a Infinite sheet of charge with uniform charge per unit area is [tex] \frac{\sigma}{2\epsilon_{0}} [/tex], and for a wire of infinite length the Electric Field Magnitude is [tex] E = \frac{\lambda}{2\pi\epsilon_{0}r} [/tex] where r is the distance from the wire.

    Why is E for the infinite wire dependent on distance, but not the distance for the infinite sheet. Wouldn't the E get smaller the farther away from the infinite sheet??? This is not a HW question.
     
  2. jcsd
  3. Oct 3, 2007 #2
    Electric field lines cancel out from the sheet, yielding only those perpendicular, undiluted and extending out to infinity. In other terms, consider a "pillbox" intersecting the sheet's surface, where only perpendicular field lines emerge.

    For the wire the reinforcement is only partial. A cylindrical surface constructed in symmetry with the wire uses Gauss's law as

    2(pi)rlEr=(lambda)l/(epsilon)0

    so the wire's electric field is inversely proportional to the radial distance from the wire.
     
  4. Oct 3, 2007 #3
    I'm confused by what it means when you say " the reinforcement is only partial". And when you say a cylidrical surface constructed with symmetry with the wire, are the cylider and the wire perpindicar or paralell, I'm guessing maybe it doesn't matter but one is easier. And when your talking about the pill box with the plane down the middle, does that mean that it's 2EA because there's a field coming from both sides?
     
  5. Oct 3, 2007 #4
    By partial reinforcement I mean that electric field lines diverge. The Gaussian surface around the straight wire is a cylinder with ends perpendicular to the wire and curved surface centered on the wire.

    If the extended sheet is a conductor, the electric field within it is zero. You can construct the pillbox for the sheet with an end placed just inside the conducting sheet, and the other just outside it. The cylinder itself is perpendicular to the sheet. The only contribution here is due to the E-field through the outside end of the pillbox.
     
  6. Oct 3, 2007 #5

    Claude Bile

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    Look at the field lines in each case. For the single wire, the field lines clearly diverge as one moves away from the wire, thus E decreases with r. For the infinite plane case (think the mother of all capacitors), the field lines don't diverge, they all point normal to the surface, hence E will be constant everywhere.

    Claude.
     
  7. Oct 3, 2007 #6

    robphy

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    Another way to look at it is that the field of
    a point charge falls off as 1/r^2;
    an infinite line of charge (composed of an infinite line's worth of charges) falls of as r(1/r^2)=1/r;
    an infinite plane of charge (composed of an infinite plane's worth of charge) falls of as r^2(1/r^2)=constant.
     
  8. Oct 3, 2007 #7
    That makes much more sense, The Idea that the lines diverge from a circle but are constant in their spacing for a plane. It just seems strange to me that you can be for example galaxies away from this ideal plane and the E field would be the same if you were right next to it. Darn you physics! Like I actually overheard someone say in the hallway of our physics building today "that's why physicians are so weird" lol~
     
  9. Oct 8, 2007 #8

    Claude Bile

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    I agree, Physicians are weird.

    Us Physicists however are normal, balanced folk! :rolleyes:

    Sorry I couldn't resist.

    Claude.
     
  10. Oct 8, 2007 #9

    rbj

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    this has to do with the fundamental reason why there are inverse-square laws in a universe with 3 spatial dimensions. it has to do with the concept of "flux" or "flux density" of stuff and equating "field" to "flux density" (perhaps with a scaling constant that depends on our anthropocentric unit systems).

    check out

    http://en.wikipedia.org/wiki/Inverse-square
    http://en.wikipedia.org/wiki/Flux
    http://en.wikipedia.org/wiki/Gauss's_law

    you can go through the nasty double integrals, but think of it conceptually a little.

    take a look at a point charge as a little particle. as you move away from it, that particle appears smaller to you in two dimensions. it appears smaller both in height and in width.

    now take a look at an infinite line of charge. as you move away from it, that infinite line appears smaller (more precisely "thinner") to you in one dimension. it still looks like it's infinitely long, but it does appear thinner and thinner as you move away from it.

    now take a look at an infinite plane of charge. as you move away from it, how does it appear smaller? if you're in empty space, except for that infinite plane of charge, how can you tell that that infinite plane of charge is 1 kilometer away from you or 1 lightyear away from you?

    more properly, think of all 3 of these geometries as emitting "flux" whatever their "stuff" of interaction is. for illustration let's think of flux as emitted energy of some radiative source and "flux density" as the intensity of that radiation at some point in space. this intensity is the amount of energy flowing out that falls on a square meter (or whatever unit area you choose) that is oriented perpendicularly to the direction of energy flow.

    here at Earth, we get about 1300 watts of sunlight falling on every square meter, but you might expect it to be more when you're on Mercury or less when you're on Mars. knowing that we are about 150 billion meters from the Sun and that the surface area of a sphere (in our 3-D existance) is [itex] 4 \pi r^2 [/itex] (for radius r), we know there are about 2.8 x 1023 square meters in this sphere surrounding the Sun (where we, on Earth, are at or very near the surface of that sphere) and 1300 watts falls on every one of those square meters. so the Sun's output is about 3.7 x 1026 watts, a pretty sizable light bulb.

    so what would the intensity be if you got in a spaceship and doubled your distance from the Sun? the Sun's total output remains the same 3.7 x 1026 watts but now all of that energy is distributed on the surface area of a hypothetical sphere that is twice as big in radius and has 4 times the surface area. then the number of watts per square meter has to be around 1300/4 or 325 watts/m2.

    a point source emits a finite amount of total flux outward in all 3 spatial dimensions will have the flux density increase in proportion to the reciprocal of the square of the distance because the size of the encapsulating sphere is reduced directly in proportion to the square of the distance.

    now imagine an infinite line of charge. each meter of length of that infinite line of charge is emitting a fixed amout of flux. because of symmetry, the flow of flux must be perpendicular to the line and it flows out spreading in only 2 dimensions. for each meter of length, you can imagine a cylinder of the same length and radius r surrounding that segment of length and the surface area of that cylinder now increases proportional to r, not to r2 as did the sphere. so the same constant flux (for one finite segment of the infinite line) gets distributed on an area that increases proportional to r which means the flux density must be proportional to 1/r for all of the flux falling on that surface area to add up to the same constant total flux emitted by that finite segment of the infinite line.

    now imagine an infinite plane of charge. each square meter of area of that infinite plane of charge is emitting a fixed amout of flux. because of symmetry, the flow of flux must be perpendicular to the plane and it flows out in only one dimension and is not spreading. if you to place a square meter 1 kilometer from the infinite plane (and parallel to it, which is perpedicular to the direction of flow of the flux), the same amount of flux would flow through that sq. meter as it would if you were a lightyear away.

    a wordy, but i hope rigorous answer.
     
    Last edited: Oct 8, 2007
  11. May 13, 2010 #10
    can u pls explain what this divergence means? electric field is d force acting on a unit point charge so the flux lines from both line and sheet charges ought to converge at d point and d perpendicular components cancel leading to a net flux in d direction of d field.
     
  12. Feb 19, 2012 #11
    As we know we want to calculate elecrical intenisty at a point P mean we wanna to calculate (Force acting on point P per unit positive) so we Draw cylinderical Gaussain surface To calculate intenisty at point P so tell me why we Draw rare part of cylinder behind the plane because we don't want to claculate E behind the plane..............
     
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