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My physics knowledge is (sadly) very infinitesimal and I have a question about a calculation which I think is wrong, but want to be sure.

I tried to calculate the potential of a finite line charge. According to my understanding, this is simply done by integrating point charges along the line. When I do that, I get something that looks like this (lines are the equipotential lines)

line charge

This plot makes intuitive sense to me since the potential computation is basically convolving a $1/\|x\|$ function with a "wide" delta pulse which I expect to look like this.

The integration result is

$$\frac{\lambda}{4\pi\varepsilon_{0}}\log\left(\frac{\sqrt{x_{2}^{2}+\left(a-x_{1}\right)^{2}}+x_{1}-a}{\sqrt{x_{2}^{2}+\left(b-x_{1}\right)^{2}}+x_{1}-b}\right)$$ where $(x_1,x_2)$ are the coordinates where the potenial is computed and (a,b) are the limits of the line charge that lives on the x-axis.

However, many calculations that I found on the web (e.g. this one) obtain a result that is independent of the abscissa. First I thought that their result might only hold for x-values between the ends of the line, but even then the potential should not be independent of the abscissa.

Am I getting something important wrong here? Thanks in advance!

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Your method of summing up points charges is qualitatively correct, and your image looks good as well. Can you share the functional form of your potential? –  DJBunk Jun 25 '12 at 15:28
    
Thanks for your comment. I included the result. –  fabee Jun 25 '12 at 15:36
    
Hi fabee, so not to be elusive but you are on the right track and I think it would be best if I gave you a hint for now. I would draw very careful pictures of both scenarios and label each parameter {x1,x2,a,b}, {a,b,d} and think very carefully about what each means for the relevant case. Right away we can see that there are a different number of parameters between the 2 formulas - what does this mean? –  DJBunk Jun 25 '12 at 16:05
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So what is the analogue of x1 for the potential you linked to? –  DJBunk Jun 25 '12 at 16:29
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Your problem seems to be that their formula doesn't use abscissa variable, equivalent to your $x_1$. This is because their $a$ and $b$ variables are defined differently. You may notice that $b$ only ever occurs as $b-x_1$. Replace that expression with $b$, you'll eliminate $x_1$, and it's likely you will get the formulas to correspond. –  Alan Rominger Jun 25 '12 at 16:35

1 Answer 1

up vote 1 down vote accepted

The coordinate system in the example you cited is kind of strange, since it defines the observation point as x=0. As the observation point changes, both a and b will change (so the variation with x is "hidden"). Your coordinate system looks more rational to me.

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Yes, I just realized that this is the problem. They are basically computing the potential for all points for which $x_1=0$. For any other point, their calculation is not correct. Thanks for the hint (also to Alan and DJBunk). –  fabee Jun 25 '12 at 16:49
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@fabee, you're welcome. Btw, beautiful graphic. What tool did you use? –  Art Brown Jun 25 '12 at 17:42
    
Thanks. I used python's contour plot form matplotlib. –  fabee Jun 25 '12 at 20:53
    
@fabee, Thanks. –  Art Brown Jun 25 '12 at 21:42

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