# Help using Coulomb's Law for a semi-infinite plane

1. Apr 1, 2014

### FallenLeibniz

Alright, so I have tried my hand at this problem but keep hitting a wall.
(NOTE: Every time I have tried to use $\LaTeX$ syntax in this post, it has not worked. As a result of this, and in the interest of making it easier for those who read this post to help me, I have attached a LaTex formatted pdf and referenced the equations in that pdf here. Please note however that it is my first time trying to use LaTex, so forgive me if I miss any subtle formatting points.
Actually, to those who read this post, would it be easier if I just wrote my full question as a pdf LaTex doc and then just attached it to whatever posts I do in the future?)

The Problem
A long, uniformly charged ribbon is located in the xz plane, parallel to
the z axis occupying the region -\infty<=z<=\infty and -a/2<=x<=a/2. The charge per
unit of area on the ribbon is \sigma.

a)Determine E(r-r') at (x,0,0) where x is greater than a/2.

My attempt

I decided that it would be easiest to attempt this problem using Cartesian
Coordinates. My seperation vector is given by eq1 and its magnitude by eq2. The field vector I have come up with for
the problem is given by eq 3.

Now I start off by trying to solve the integrals for the ith component of
the field. I first do the integration with respect to x'. The results of this integration are given by eq 4 and eq 5.

Since the next integration involves integrating over the result of eq 5 with respect to z' and both parts of the difference take the same form, I decided to just substitute b for x-a/2 and x+a/2 and use the result of eq 6 to get my final answer.

Thus here is my dilemna. The value in the log expression goes to zero and
thus "blows up" the field in the process. I've been through my integration three
times over and still can not find anything wrong with it. I could really use
any assistance or insight that anyone can give me.

#### Attached Files:

• ###### PF_Question1.pdf
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Last edited: Apr 1, 2014
2. Apr 1, 2014

### BvU

Hello bruised Gottfried,

Re $\TeX$: just quote a post that has some $\TeX$ markup to see how it's done.

$\sigma$ suggests a charge per unit of area ($\lambda$ is usually used for charge / unit of length).

I find it hard to think of the meaning of $\vec E(\vec r - \vec r')$ and then ask for that at (x,0,0).
My simple picture is that $\vec E(\vec r) = \iiint \vec E(\vec r - \vec r') \, d^3\vec r'\$ would be a more logical thing to ask (in this case in 2d).

Then there is only one non-zero component due to symmetry.

And the final answer can only depend on $\sigma$, a and x. Maybe you need to introduce an x0 if E doesn't fall off fast enough to infinity.

3. Apr 1, 2014

### TSny

Hello FallenLeibniz.

I believe you have an overall sign error in your result of integration (5) in your pdf.

Integration of the result of (5) with respect to z' will give two logarithm terms. Combine the two logarithms before taking the limit of z going to plus or minus infinity.

I believe the whole problem would be easier if you integrate over z' first and then integrate over x'.

4. Apr 1, 2014

### FallenLeibniz

Thank you for the tip in regards to the Tex. Do you think it could be that I tried to use LaTex over just straight Tex?

I'm not sure what you're indicating here, as the charged object in question is a surface. Did I write something wrong (please note I am not being sarcastic in asking this, but would like to know so if something is wrong, I can fix it before it turns away any other potential eyes to the probelm)?

5. Apr 1, 2014

### FallenLeibniz

Ok the idea of combining the log terms did help. Once I did that the expression came out to be ln|(z'+√(z'2+(x+a/2)2))/(z'+√(z'2+(a/2-x)2))| to be evaluated at the limits -∞ and ∞. Using L'Hospitals rule on the innards of the ln, I got the inner expression to be (x+a/2)2/(a/2-x)2 ((z'2+(a/2-x)2)/(z'2+(x+a/2)2))3/2. And since z'→∞ and -∞, the z' ratio goes to one. Thus leaving only an x dependence and so ln|(x+a/2)2/(a/2-x)2|

6. Apr 1, 2014

### FallenLeibniz

Thank you very much. To both of you.