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Electric field for a semi infinitely charged plane (again)

  1. Nov 24, 2014 #1
    Mod note: Below is a revised version of the problem.
    1. The problem statement, all variables and given/known data

    Determine E at (x,0,0) where x>a/2. What is the asymptotic field on the x-axis?
    2. Relevant equations
    1)Coulomb's law for charged surface of density Sigma
    ##\vec E(\vec x-\vec x')=\frac{\sigma}{4\pi\epsilon_0} (\iint \frac{(\vec x-\vec x')}{|(\vec x-\vec x')|^3} \, dA')##
    2)Field of an infinitely long charged wire
    ## E(s)= \frac{\lambda}{2\pi\epsilon_0 s}##
    3. The attempt at a solution
    Ok. So I first started with tailoring equation 1 to the problem. I Then had the following setup.
    ##E_x=\int_{-\infty}^{\infty}\int_\frac{-a}{2}^\frac{a}{2} \frac{x-x'}{(z'^2+(x-x')^2)^\frac{3}{2}} \, dx' \, dz'##
    ##E_z=\int_{-\infty}^{\infty}\int_\frac{-a}{2}^\frac{a}{2} \frac{z'}{(z'^2+(x-x')^2)^\frac{3}{2}} \, dx' \, dz'##
    Now, since I know from the geometry of the problem that the z-axis component will be zero for the field, so I concentrated on calculating the x-axis component.
    Now, the idea occurred to me that I could actually approach this problem by building up the plane with an infinite sum of infinite line charges. I had an example in my book which already worked out the field at the mid-plane for an infinite wire as what I list above in equation 2. I then worked out that the analog here for lambda would be that:
    ##\lambda=\sigma \, dx'##
    Thus ##\, dE_x=\frac{(\sigma \, dx')}{(2\pi\epsilon_0 (x-x'))}##
    Thus for my first try at the answer I get
    ##E_x=\frac{\sigma}{2\pi\epsilon_0}\ln|\frac{x-\frac{a}{2}}{x+\frac{a}{2}}|##
    I only thought of this method after starting in first though with Couloumb's Law. I integrate the x-component of the field and get the following integral after the first integration:
    ## E_x=\frac{\sigma}{4\pi\epsilon_0} \int_{-\infty}^{+\infty} \frac{1}{(z'^2+(x-\frac{a}{2})^2)^\frac{1}{2}}-\frac{1}{(z'^2+(x+\frac{a}{2})^2)^\frac{1}{2}} \,dz'##
    Now. Every way shape, and form I try to solve this integral, I always end up getting zero for the answer. I've been looking at my work for days and can't find anything. I am hoping for a hint of where I'm going wrong since as this is a semi-infinite plane, the fringing effects should give an x-component.
    The following is the intermediate result of what I got when I tried to evaluate the above integral
    ## E_x=\frac{\sigma}{4\pi\epsilon_0} \ln \Big|\frac{(x+\frac{a}{2})(z'+(z'^2+(x-\frac{a}{2})^2)^\frac{1}{2})}{(x-\frac{a}{2})(z'+(z'^2+(x+\frac{a}{2})^2)^\frac{1}{2})}\Big|\Big|_{-\infty}^{\infty} ##
     
    Last edited by a moderator: Nov 25, 2014
  2. jcsd
  3. Nov 24, 2014 #2

    mfb

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    What is "a", where is the plane?

    The integral via the lines should give some non-zero value for x != 0, how did you calculate it?
     
  4. Nov 24, 2014 #3
    a is the distance that the non-infinite part of the plane. The plane is situated such that it occupies the space from x=-a/2 to x=a/2 and from z=##{-\infty}## to z=##{\infty}##.

    As for how I solved, I got from the first ##E_x## integral to the second using U-Substitution with ##u=x-\frac{a}{2}##. To get from that integral to the log expression I listed last, I used trigonometric substitution. I then proceeded to try and evaluate the natural log expression at the infinite limits using L'Hospital's rule. After three applications of the rule, I got that the limit (regardless of whether it was going to negative or positive infinity) of the expression inside the argument for the ##\ln## at the end is ##\frac{x-\frac{a}{2}}{x+\frac{a}{2}}## leading to a final result of zero once this result is applied to both of the infinite limits.

    I also have just seen that I have made a huge mistake in how I've written the original post. The "a/2"'s in the original integral setup where ##x-\frac{a}{2}## is in the expression that is raised to the 3/2 should be x'. I will see what I can do to edit the post now.
     
    Last edited: Nov 24, 2014
  5. Nov 24, 2014 #4
    IMPORTANT UPDATE: I also have just seen that I have made a huge mistake in how I've written the original post. The "a/2"'s in the original integral setup where ##x−\frac{a}{2}## is in the expression that is raised to the 3/2 should be x'. I will see what I can do to edit the post now.

    Edit from Mark44: I made changes to the OP, so this should now be fixed.
     
    Last edited by a moderator: Nov 25, 2014
  6. Nov 25, 2014 #5

    BvU

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    Try a few approximations: ##\displaystyle {1-a/2x\over 1+a/2x} = 1-{a\over x}## and ## \ln(1+\epsilon) = \epsilon##
     
  7. Nov 25, 2014 #6
    Although the limit is to be taken with respect to z' and not x, would the same idea apply just instead doing the division by z'?

    If I divide the argument in the ln by z' and try to take the limit (note the limit can be then taken by inspection then), the problem still lies that you get a constant minus the same constant.

    Also to note, if you try the symmetrical approach after doing the first integral with respect to x' in the double integral and then making the assumption that the field is symmetric (i.e. replace the ##{-\infty}## with a zero and multiple all by two), you do get a reasonable non-zero answer as the ln evaluated at the lower limit is actually zero. In fact it comes out to what I got via building the plane using an infinite amount of infinite line charges as I described in the post (the non-Couloumb's law method).
     
    Last edited: Nov 25, 2014
  8. Nov 26, 2014 #7

    BvU

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    Integration over z' gives you ##\ln\left ({\frac{x-\frac{a}{2}}{x+\frac{a}{2}}}\right )##.

    The limit I suggested show that for large x you get the same field as for a wire
     
  9. Nov 27, 2014 #8
    Anytime I try to integrate over z' and go to evaluate the limits (after integrating the integrand of what remains of the double integral), I get a constant minus the same constant after applying L'hospitals rule for z'->##{-\infty}## and z'->##{\infty}##. How do I account for the fact that the limit is the same no matter "which" infinity I take the limit over?
     
  10. Nov 28, 2014 #9

    BvU

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    Ah, I start to think you have difficulty with
    $$
    E_x=\frac{\sigma}{4\pi\epsilon_0} \ln \left |\frac{(x+\frac{a}{2})(z'+(z'^2+(x-\frac{a}{2})^2)^\frac{1}{2})}{(x-\frac{a}{2})(z'+(z'^2+(x+\frac{a}{2})^2)^\frac{1}{2})}\right | \ \ \Big|_{-\infty}^{\infty}\ \ ?$$
    Divide numerator and denominator by z'and then take the limits...
     
  11. Dec 2, 2014 #10
    If you divide that expression in the numerator and the denominator by z' then take the limit of the quotient at infinity, it produces the same result as when you take the limit to negative infinity (i.e. ##\frac{x+\frac{a}{2}}{x-\frac{a}{2}}##) is the argument of the ln in both cases). When you subtract both logs you get zero.
     
  12. Dec 4, 2014 #11

    BvU

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    Negative infinity limit does something different. I should have typed "divide numerator and denominator by |z'| " to make that clearer...
     
  13. Dec 31, 2014 #12
    I took a bit of a leap and switched the order of integration. This gave me something that allowed me to evaluate the limiting behavior as the result went to infinity and negative infinity (I did it more by looking at the graphical behavior of the first integration result as direct evaluation proved a little more difficult). I think that helped to work it out as it did give a result that I was expecting.
     
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