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Homework Help: Gauss's Law applied to an infinite sheet of charge

  1. Feb 10, 2014 #1
    1. The problem statement, all variables and given/known data
    The problem that inspired the upcoming question is: Find the magnitude of the electric field due to an infinite sheet of charge with uniform charge density σ using gauss's law.

    I have in fact arrived at the character answer σ/2ε (epsilon nought), but I don't understand this solution. To elaborate, I'm unsure as to why I was able to solve this problem by using a cylinder of area A that only encloses a small portion of the sheet of charge (similar to every example online of this classic problem). Why does this small Area A represent the E field for the entire sheet? I'm also wondering why I cannot use a sphere and symmetry to solve this problem; Is the E field not constant?

    2. Relevant equations

    electric flux = e * da;
    flux = charge enclosed/epsilon nought
    area of ends of a cylinder 2*pi*r^2

    3. The attempt at a solution

    I have found the answer using a cylinder, but I don't see why this works, or why I can ignore part of the charge of the sheet; Or in other words, why charge enclosed in the above equation need not be equal to the charge of the sheet, but that of a tiny cross sectional area.
  2. jcsd
  3. Feb 10, 2014 #2
    You can use just about any type of surface for this problem. A cylinder just works well. The crux of the problem is that the electric field is constant for the sheet, and to find it you examine just a small portion of area and a small portion of charge enclosed within that area, and figure out the field produced by that area. Then, because the field is constant, the value of the field you find there applies to the whole sheet.
  4. Feb 10, 2014 #3


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    Homework Helper

    The electric field reflects the symmetry of the charge distribution. An infinite sheet of charge is laterally homogeneous, the charge /unit area is the same everywhere: so must be the electric field. That means, the field can depend only on the distance from the sheet.
    The charge distribution does not change if you rotate the sheet about an axis normal to the sheet: The electric field is also invariant for rotation about the same axis. It does not depend on direction. That means the electric field is normal to the sheet.

    If there is nothing else but the sheet then the sheet is a mirror plane. The electric field at one side of the sheet is the mirror image of the electric field at the other side.

    Knowing all these you can choose a Gaussian surface, which also has the symmetry of the situation: A right cylinder or right prism with opposite faces parallel with the sheet, the other face(s) normal to it. For such surface, the net flux is equal to the sum of fluxes across the parallel faces.

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