- #1
FallenLeibniz
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Mod note: Below is a revised version of the problem.
1. Homework Statement
Determine E at (x,0,0) where x>a/2. What is the asymptotic field on the x-axis?
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}##
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} ##
1. Homework Statement
Determine E at (x,0,0) where x>a/2. What is the asymptotic field on the x-axis?
Homework 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}##
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} ##
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