Trouble proving equivalence at limit

[boris.rarden]

Homework Statement

I'm trying to prove that Electric field away from a line segment of charge, is equivalent to the field away from a point charge, provided I observe from far enough.

Homework Equations

Ignoring all the constants:

potential_line = log( (sqrt(r^2 + a^2) + a) / (sqrt(r^2 + b^2) - b) )

potential_charge = 1/r

Here a+b is the length of the line segment, such that a and b are the parts of the line segment 'above' and 'below' the line of sight of the observer, assuming the line is vertical one. 'r' is the distance to the line along the line of sight.

Trying to show that the two equations become equivalent (close) when r is much bigger than a+b.

The Attempt at a Solution

I graphed with WorlframAlpha both formulas and the graphs look the same. Here are the two links. I took a=2, b=1 for an experiment.

http://www.wolframalpha.com/input/?i=y=log((sqrt(x^2+4)+2)/(sqrt(x^2+1)-1))++from+1+to+100

http://www.wolframalpha.com/input/?i=3/x+from+1+to+100

The graph looks very close, which is good. But how do I show this algebraically ? I tried to simplify that sqrt(r^2 + a^2) = sqrt(r^2 + b^2) = r, when r >> a+b. The graphs continue to look similar. I tried to simplify the numerator, and got this:

log( 1 + (a+b)/r + ab/r^2 )

 

[Redbelly98]

Welcome to PF.

Looks like an interesting problem!

Normally, I try to rewrite functions in the form f(1+ε), where ε is a small quantity compared to 1. Then replace the expression f(1+ε) with a series expansion to 1st or 2nd order in ε.

So, for example, if we had an expression \sqrt{x^2 + 9}, where x is large compared to 9, we can rewrite this as

\begin{align}<br /> \sqrt{x^2 + 9} &amp; = \sqrt{x^2 (1 + 9/x^2)} \\<br /> &amp;= \sqrt{x^2} \cdot \sqrt{1 + 9/x^2} \\<br /> &amp;\approx x \cdot (1 + \frac{1}{2}\frac{9}{x^2}) \\<br /> &amp;= x + \frac{9}{2x}<br /> \end{align}

This works because the quantity 9/x² is small when x is large. Note the use of the approximation \sqrt{1+\epsilon} =(1+\epsilon)^{1/2} \approx 1+\frac{1}{2} \epsilon.

See if you can apply that technique to your √-expressions and eventually to the logarithm as well.

 

[boris.rarden]

Actually, is there an equivalent approximation trick for log ? I'm stuck because I have to consolidate 1/r and log( f(r) ). Should I use Taylor series ?

 

[Redbelly98]

Actually, is there an equivalent approximation trick for log ? I'm stuck because I have to consolidate 1/r and log( f(r) ). Should I use Taylor series ?

Yes, for the log use the Taylor series about the point x=1. I.e.,

log(1+ε) ≈ ?​

 

[boris.rarden]

thanks, looks like i was able to show it after all following your ideas and the log expansion.