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Electric Fields using Gauss' law

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data

    http://i.minus.com/jbxXiQZmmiOQtS.png [Broken]

    2. Relevant equations

    Elecric flux = Integral of closed surface of E dotted into dA = qenclosed / ε0
    Rho = 50 x 10^-6 abs(z)

    3. The attempt at a solution

    Integral of closed surface of E dotted into dA will end up being EA when E is constant.

    as such,

    EA =qenclosed / ε0


    I believe I have to use the function of the Rho in the integral
    qenclosed = Intergal Rho * dV

    As given we know to find dV we use the area (A) multiplied by dz.

    Such as that

    A * 50 x 10^-6 * Integral abs(z) dz

    It here I believe that defining the limits tells us rather we are taking from within inside or outside. In my notes we are using a cylinder bar and Gaussian surface of a ring and going from 0 to R (R radius being that of cylinder bar) for inside and 0 to r (r being distance from center of cylinder bar to ring).

    I have no idea what to put for my limits to get the inside and outside.

    Moderator note: Moved to the introductory physics forum.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 31, 2012 #2
    gauss's law basically says that net flux through a surface is proportional to the enclosed charge.

    gauss's law is for symmetrical charge configurations (i.e. you want the electric field to be constant across the Gaussian surface.

    now your charge distribution is symmetric on the xy plane (z=0).

    when z=0, you have an equal amount of charge above and below you, so E=0

    if you took a really small box, centered at the origin, and increased it height (z direction), the total enclosed charge would be the integral of rho, over the area of this box. your limits of integration are the bottom and top of this box (hence dV = A*change in height(dz).

    as you integrate inside the slab, more of the q is enclosed so your electric field gets stronger, once you gaussian box is outside the slab, no additional charge is inclosed so your limits of integration for Qenc is just the thickness of the slab
     
  4. Jan 31, 2012 #3
    That is what I originally thought, given L = 0.20m I would do:

    Limits from -L/2 to L/2 , or in other words -0.10m to 0.10m. Taking into consideration I am integrating the absolute value of z dz I am left with L * |L| / 4 or 0.01.

    So,

    EA = (50 x 10 ^-6)(A)(0.01)/enought
    E = (50 x 10 ^-6)(0.01)/enought

    However, this no longer gives me a "function of z" so I am getting confused here. Also how would the equations differ from inside and outside the slab? I think my issue is that I didn't make a Gaussian surface so I have no "r", that is a measurement from the origin to the distance "r" to the Gaussian surface.
     
    Last edited: Jan 31, 2012
  5. Feb 1, 2012 #4
    ∫E.da = μ[itex]_{o}[/itex]∫ρdv

    the value of E at points located on the surface equals
    the integration of ρ (charge/unit volume) over any
    charge distribution that is inclosed by the surface where we are finding E

    so when we find E inside the slab, the gaussian box extends to a point z (-a/2<z<a/2)

    and the integration of ρ extends to the surface of this box.

    you integrate both integrals to plus / minus z, and this gives you E as a function of z
     
  6. Feb 1, 2012 #5
    So you are saying it should be:

    [itex]{50}*10^{-6} * A * \int\limits_{-z}^{z} |z| \, \mathrm{d} z.[/itex]

    But that integral doesn't evaluate correctly
     
  7. Feb 1, 2012 #6
    Upon looking at the picture again I realize the slab is not location in the middle but starting at 0.

    I know that the outside electric field would encompass all charges enclosed and thus would be:

    EA = [itex]\frac {50 * 10 ^{-6} * A *\int_{0}^{0.20} |z| dz}{ε0}[/itex]

    EA = [itex]\frac {50 * 10 ^{-6} * A * (0.02) }{ε0}[/itex]

    EA = [itex]\frac {1* 10 ^{-6} * A}{ε0}[/itex]

    E = [itex]\frac {1* 10 ^{-6}}{ε0}[/itex]


    So how would I do this for the inside? Would I just makeup a variable and have it be between 0 and 0.20? let's use the variable r (for radius) as I am use to cylinder problems of this nature (where r vs R, but in this case r vs 0.20)

    Would it then be

    EA = [itex]\frac {50 * 10 ^{-6} * A *\int_{0}^{r} |z| dz}{ε0}[/itex]
     
  8. Feb 1, 2012 #7
    the graph of rho vs z makes it seem the slab is centered at the origin. regardless, you want the middle of your gaussian surface to be aigned with the middle of the slab. this way, as the height of the box is increased equal amounts in +/_ z direction, there is an equal amount of charge in the upper and lower halves of the box, allowing the vaue of E to be constant at the surface of the box.

    for inside the slab, the E . da is zero on the walls of the box, and equals magn. of E on the top and bottom surface.

    Ienc = integral of rho from (middle of slab - z) to (middle of slab +z)

    after integrating,

    E*(area integral) = enclosed charge as a function of the height(z) of the gaussian surface.

    when you evaluate the enclosed charge integral, your limits are a func of z, so your evaluated integral will be a function of z

    "In my notes we are using a cylinder bar and Gaussian surface of a ring and going from 0 to R (R radius being that of cylinder bar) for inside and 0 to r (r being distance from center of cylinder bar to ring)."

    so i think you ment your gaussian surface was a pipe, or the surface of a cylinder.

    here you go from 0 to r b/c the charge distribution is cylindrically symetric, i.e. you use cylindrical coordinates.

    for the slab, the is no cylindrical symmetry, rather "planar" symmetry or whatever you want to call it. In this case, it is more convenient to use cartesian coordinates.

    likewise, if the charge dist. was a sphere, your gauss surface is also a sphere and you use shpereical coordinates.

    **the whole point is that the strength of the electric field is constant on the surface of your gauusian surface
     
  9. Feb 1, 2012 #8
    After speaking to the professor the answer turns out to be:

    Outside:



    [itex]E * A = \frac{50 μC * A * 2\int_{0}^{0.10} z dz}{ε0}[/itex]

    [itex]E = \frac{50 μC * 0.01}{ε0}[/itex]

    [itex]E = \frac{0.5 μC}{ε0}[/itex]


    Inside:

    [itex]E * A = \frac{50 μC * A * 2\int_{0}^{z} z dz}{ε0}[/itex]

    [itex]E * A = \frac{50 μC * A * 2 \frac{z^2}{2}}{ε0}[/itex]

    [itex]E = \frac{50 μC * z^2}{ε0}[/itex]




    It was the trick of getting of of the absolute value by just doing [itex]2\int_{0}^{z} z dz[/itex] instead of [itex]\int_{-0.10}^{z} |z| dz[/itex], The z being 0.10 when outside

    As thought, she was using a variable (z) to symbolize any value between 0 and 0.10. Using z as one of the limits in [itex]\int_{0}^{z} z dz[/itex] threw me off as my ti-89 said it was an incorrect integral.
     
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