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Re: [Phys-l] field of a finite line of charge

On 02/27/2007 11:49 AM, pschoch@nac.net wrote:

I solved the E field problem of a finite line of charge on the x axis
from -L to 0, at a point (0,P).

One young man took an example from the textbook, tried to combine it with
my example, and got a solution that makes no sense and asked if I could do
it.

After trying, I also get something that doesn't seem right, and I need a
bit of help doing this with only freshman techniques. The problem is:

You have a line of charge from (-L,0) to (L,0), and a point at
(-L,P). Evaluate the E field at the point (L,P).

That's exactly the same problem. It's shifted horizontally and
rescaled, but the physics is the same. Draw the picture.

When I try to do this by Freshman physics methods and evaluate the
line integral to get the x and y components from the line of charge I get
the x component to cancel out (because of the integration from -L to +L). Now, this can't be correct, but I don't see my error.

1) My crystal ball is slightly hazy this morning, but the
first thing I would check for would be a missing "L" in
the integrand. When finding the field at (x,P), the
integrand is symmetric for x=0 and not otherwise.

There are lots of ways to arrive at this mistake:
1a) By using the integrand for the previous problem in this
problem

1b) A lot of textbooks solve for the field at (x,y) in the case
of an /infinite/ line charge, in which case they can get away
with leaving the x out of the integrand ... but plug-and-chug
misapplication of the textbook integrand will not work for
this version of the problem.

2) Or just shift and rescale, so as to reduce it to the problem
previously solved.

3) BTW the conventional approach is not to integrate the field
directly, but rather to integrate the potential and then
take the gradient. That way there's only one integral, and
the integral is doable by elementary methods. Hint:
http://www.google.com/search?q=trigonometric-substitution