1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gauss Law and Electrostatic Field

  1. Jul 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Use Gauss' law and symmetry to find electrostatic as a function of position for an infinite plane of charge. Let the charge lie in the yz-plane and denote the charge per unit area by [tex]\rho=\alpha*e^{-abs(x/b)}[/tex]

    2. Relevant equations

    Q=triple integral of density

    3. The attempt at a solution

    First compute the Q. After the integration I plan to apply Gauss law to get the electric field.

    I first tried to convert the density to cylindrical coordinates and the tried to integrate the density using right circular cylinder. The triple integral was

    [tex]\int\int\int\alpha*e^{-abs(r*cos(O)/b}*r*dr*dO*dz[/tex]
    The integration limits were r = [0,r], dO = [0,2*pi] and dz = [0,1]

    However the integration becomes very difficult. Is there an easier way?
     
    Last edited: Jul 27, 2009
  2. jcsd
  3. Jul 27, 2009 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi Sami Lakka,

    Here you are converting from Cartesian (x,y,z) coordinates to cylindrical (r,theta,z) coordinates. But there is no reason why the cylindrical z-axis has to be the same as the Cartesian z-axis.

    In this case, the plane is in the y-z (cartesian) directions, and it changes in the x-direction, so try making the new z-axis the same direction as the original x-axis.
     
  4. Jul 27, 2009 #3

    tms

    User Avatar

     
  5. Jul 27, 2009 #4

    djeitnstine

    User Avatar
    Gold Member

     
  6. Jul 27, 2009 #5

    turin

    User Avatar
    Homework Helper

    tms has a valid point. The expression for charge density (not field lines) is given as an x-dependent expression, which contradicts a previous statement in the problem that the charge lays in the yz-plane. Furthermore, the field lines will curve somewhat in the axial direction as well if the charge density is not uniform in the plane. Sami Lakka needs to rephrase the problem statement more precisely.
     
  7. Jul 28, 2009 #6
     
  8. Jul 28, 2009 #7

    turin

    User Avatar
    Homework Helper

    Sami Lakka, you still need to correct the expression for the charge density. Right now, it depends on x.
     
  9. Jul 28, 2009 #8

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If it is a slab, then it must have a thickness from some lower limit of x to some higher limit of x, say -d < x < +d. Is that the case? Otherwise the problem does not make sense as stated.
     
  10. Jul 28, 2009 #9

    turin

    User Avatar
    Homework Helper

    A slab of charge with zero thickness makes no less sense than a point charge, which we use in physics all the time. A slab of charge can be modeled as a plane of charge with zero thickness, depending on the problem. This is not the issue that makes the problem nonsensical.
     
  11. Jul 28, 2009 #10

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I agree, but the problem implies that the slab has finite thickness in the x-direction. If this were not the case and the slab is infinitely thin, then the correct volume charge density should be written in terms of a Dirac delta function, [tex]\rho(x)=[/tex]a*e-abs(x/b)*[tex]\delta(x)[/tex]. In either case, one can formally integrate over the volume to find the enclosed charge or whatever else is needed.

    What does not make sense to me is what the problem wants. Find the electrostatic ... what?
     
  12. Jul 28, 2009 #11

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    [/URL]

    The blue and red statements both imply an infinitesimally thin sheet of charge. But then the x-dependent equation contradicts that.

    We need an exact, word-for-word statement of the problem before any useful help can be given.
     
    Last edited by a moderator: May 4, 2017
  13. Jul 29, 2009 #12
    Here is the problem statement from word to word. The problem is in Div, Grad, Curl and all that, p. 55, problem II-11

    a) Use Gauss' law and symmetry to find the electrostatic field as a function of position for an infinite uniform plane of charge. Let the charge lie in the yz-plane and denote the charge per unit area by [tex]\sigma[/tex]

    b) Repeat part (a) for an infinite slab of charge parallel to the yz-plane whose density is given by
    [tex]
    \rho(x)=\alpha[/tex] when -b < x < b
    [tex] \rho(x)=0 [/tex] when abs(x) >= b


    where [tex]\alpha[/tex] and b are constants.

    c) Repeat part (b) with [tex]
    \rho(x)=\alpha*e^{-abs(x/b)}
    [/tex]


    I'm doing the c) part. The correct answer is [tex]E=\alpha*b/\epsilon_{0}*(1-e^{-abs(x)/b})\textbf{i}[/tex] for x > 0 and -E for x < 0

    Yesterday I actually managed to get the right result by changing to cylindrical coordinates where the cylinder runs through x-axis instead of the normal z-axis (thanks alphysicist!).
    Then the triple integral is [tex]Q=\int\int\int\alpha*e^{-abs(x/b)}*r*dr*dO*dx [/tex]
    I used following integration limits x=[0,x], r=[0,r], O=[0,2*pi]. Then I multiplied the integral by two (because of the integration limit for x). The flux for that cylinder is 2*E(x)*pi*r2 as only the cylinder's ends contribute to the flux. However I'm not sure about my reasoning.

    Sorry for not posting the complete problem definition right from the start!
     
  14. Jul 29, 2009 #13

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Triple integrals aren't necessary here. Just integrate over x, using some fixed hypothetical area A that remains parallel to the yz-plane.
     
  15. Jul 29, 2009 #14

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    There are two regions where the electric field is different and you need to give expressions for both (call them EI(x) and EII(x): (I) outside the slab (x > b) and (II) inside the slab (x < b). If x is inside the slab, to find the enclosed charge you add all the contributions from zero to x as you have already done. If x is outside the slab, there is no more charge density beyond x = b, so what does this say about your limits of integration in x?
     
  16. Jul 29, 2009 #15

    turin

    User Avatar
    Homework Helper

    Sami Lakka is working part (c). There is no "outside the slab" for part (c), because the charge distribution fills all of space. For |x|>b, the charge distribution is ρ<αe, but not zero.
     
  17. Jul 29, 2009 #16

    turin

    User Avatar
    Homework Helper

    Well, what is your reasoning? I don't disagree with your conclusion; we just can't help you with your reasoning unless you tell us what it is.

    Do you understand what direction the electric field is pointing through each of the distict, well-defined bounding surfaces of the integration region? The reasoning is easier to understand in Cartesian coordinates (as djeitnstine somehow realized before we even knew what the problem statement was).
     
  18. Jul 29, 2009 #17

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Interesting ... The instructions in part (c) ask us to repeat part (b) which mentions a slab. Having fixed the idea of a slab in my head from part (b), I interpreted the instructions in (c) to mean "replace the non-zero constant charge density with the exponential form and leave the zero density alone". To clinch the "no slab" interpretation in part (c), the problem's author should have added "for all x" after the exponential form.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Gauss Law and Electrostatic Field
Loading...