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Line charge integral (2013)

Page history last edited by Matt Harrington 7 years, 10 months ago Saved with comment

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The integral of the field near to a very long (approximated by infinitely long) straight line of uniform charge gives a nice example of how to add up contributions by integration.  Let's suppose we have a very long straight line of charge that we are going to model for the sake of simplifying the math as infinitely long. The situation looks like the figure below:

(We'll use D for the distance instead of d to avoid confusion with the "d" in the "dx" of our integral.) The line charge is drawn in blue and is intended to go on forever on both sides. We'll put the origin of our coordinate system right underneath the point at which we are trying to find the E field (shown as a pink dot).


Let's consider a little bit of the line, labeled dx, a distance x out from the origin. It has a little bit of charge, dq = λ dx and it's a distance



away from our pink point. It therefore creates a bit of E field there, labeled dE. By Coulomb's law we know that the magnitude of dE is given by



We can use our equations to express both dq and r in terms of x.



We can't quite add this all up (do the integral) to get the final result since this dE is a vector. And from our discussion in A simple electric model: a line charge, we know that the horizontal component of this vector is going to be cancelled by an opposite component from a bit of charge on the other side. So we ONLY want to add up the vertical components of this vector. 


As we see from the inset diagram (just dE made larger), what we need to do to get the vertical component of dE is to multiply it by cos(θ). But from looking at our main diagram, we can see that




So what we really have to add up is (really the y component of E but we don't need to specify it since the x component vanishes):

This looks like and awful mess, but it's not really so bad. It's just a bunch of constants out front times an integral that is beginning to look like a standard integral from a calculus class. We can make it look more like that by changing integration variable (getting rid of the units) by defining u = x/D. If we change to this variable our integral becomes



That integral can be found in a table or online. You can put it into Wolfram|alpha as integral(1/(1+u^2)^(1.5),-infinity,infinity) and discover that the value of the integral is simply 2. (Or if you are gung-ho you can do a trigonometric substitution and work it out for yourself.) When we put all of this together we get the simple result:

as claimed in "A simple electric model: a line charge".


Now if you are really gung-ho, you can do that integral of u numerically (say on a spreadsheet) and see how far out you have to go to get, say, 1.99. (That, of course, will tell you how far in u you have to go. To get x, you have to multiply that result by D.) That way you can see what part of the line near to where we are measuring the field really matters.


Note that we already knew the result had to be proportional to kCλ/D by dimensional analysis. All the mathematical heavy lifting was required just to get the factor of 2 in front.


Joe Redish 2/15/12

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