Force between two uniformly charged identical rods

[j0n]

Homework Statement

Identical thin rods of length 2a carry equal charges, +Q, uniformly distributed along their lengths. The rods lie along the x-axis with their centers separated by a distance b > 2a. Find the force, F, exerted by the left rod on the right one.

Left rod's center @ (0,0) → rod stretches from (-a,0) to (a,0)
Right rod's center @ (b,0) → rod stretches from (b-a, 0) to (b+a, 0)

Homework Equations

They give that

$$F = \frac{k Q^2}{4 a^2} \ln \frac{b^2}{b^2 -4a^2}$$

Also we know that $$λ = \frac{Q}{2a} = \frac{dQ}{dx} , \space and \space Q = Q_1 = Q_2$$

The Attempt at a Solution

Electric field produced by rod 1 at a point x > 2a:

$$E_1(x) = \int \frac{k}{χ^2}dQ = \int_{x-a}^{x+a} \frac{k}{χ^2} λ dχ = \int_{x-a}^{x+a} \frac{k}{χ^2} \frac{Q_1}{2a} dχ = \frac{k Q_1}{2a} \int_{x-a}^{x+a} \frac{dχ}{χ^2}$$
$$= \frac{k Q_1}{x^2 - a^2}$$

Now I understand that F_TOT = Ʃ q_iE₁(x_i),
where q_i are the infinitesimal contributions of Q2 (= Q).

So shouldn't I have $$F = \int_{x-a}^{x+a} E_1(χ) λ dχ ?$$

If so then I have:

$$F = \int_{x-a}^{x+a} \frac{k Q_1}{χ^2 - a^2} \frac{Q_2}{2a} dχ$$
$$= \frac{k Q^2}{2a}\int_{x-a}^{x+a} \frac {dχ}{χ^2 - a^2}$$

And I'm stuck.

 

[j0n]

By the way I used chi (χ) as a dummy variable when integrating since I wanted to obtain E as a function of position along the x-axis.

Thus, $$E_1(x= b) = \frac{k Q_1}{b^2-a^2}$$

Edit:

I think I might've solve it now. Instead of cancelling 2a out when I calculated E₁(x), I just left it.

So then $$E_1(x) = \frac{k Q_1}{2a} (\frac{1}{x-a} - \frac{1}{x+a})$$

Then in finding the Force on rod 2 my range of integration should specifically be from b-a to b+a (not x-a to x+a).

So, $$F = \frac{k Q^2}{4a^2}\int_{b-a}^{b+a} (\frac{1}{χ-a} - \frac{1}{χ+a}) dχ$$