EM: Electric field, Two thin rods uniform line charge

[RJLiberator]

Homework Statement

Two identical thin rods of length L carry the same uniform line charge distribution (charge per unit length) of . If the two rods are collinear (aligned along the same line), with a distance of d between their nearest ends, calculate the Coulomb force (magnitude and direction) each rod exerts on the other. Check that your answer makes sense in the limit d >> L.

Homework Equations

Electric field of a line charge
[tex]E(r)=\frac{1}{4 \pi \epsilon_0} \int \frac{\lambda (r')}{χ^2}\hat χ dl'[/tex]

The Attempt at a Solution

I've set up a diagram of the problem, but this can't be posted here. From the diagram the electric field generated by each line charge points in opposite directions. The electric field is only found in the ##\hat i## direction.

If we calculate the electric field that the first rod from enacts on the second rod we have bounds of integration that go from 0 to L as my origin is placed on (0,0) and the first line charge goes from (0,L).

There is a separation then from (L,L+d) of d. The next rod goes from (L+d, 2L+d).

The problem I am having with this question is understanding what χ represents.
Right now, I have χ = (L+d)-x.
I believe χ to be (the point that we are calculating the electric field at) - (the location of the electric field). ##\hat χ = \hat i##.The integral then sets up as follows

[tex]E_{1 on 2} = \int dE = -\frac{1}{4 \pi \epsilon_0} \frac{Q}{L} \int_0^L \frac{dx}{(L+d-x)^2} \hat i[/tex]

The integral comes out to be ##\frac{1}{d} - \frac{1}{d-x}##.

What am I doing wrong in this calculation? I feel like my χ is off, but I don't logically understand how it would be.

 

[NFuller]

The problem I am having with this question is understanding what χ represents.

##\chi## is the vector pointing from a piece of the rod ##dl^{\prime}## to the point in space where you are calculating the electric field ##\mathbf{E}##.

The integral then sets up as follows

[tex]E_{1 on 2} = \int dE = -\frac{1}{4 \pi \epsilon_0} \frac{Q}{L} \int_0^L \frac{dx}{(L+d-x)^2} \hat i[/tex]

Careful, there is no ##\mathbf{E}_{1on2}##. You can only calculate ##\mathbf{E}## at a point on the second rod. The total force on the second rod is the integral of ##\mathbf{E}## generated by the first rod, acting on each charge element ##dq## of the second rod. This problem requires two different integrals to be done.