There are two identical rods of length 2a lying along the x-axis with their centers separated by a distance b (b > a). Each rod carries an equal charge of +Q uniformly distributed along their lengths. Please calculate the magnitude of the force exerted by the left rod on the right one.
[latexpage]
Let the variable $x_1$ be the position placing on the left rod and the variable $x_2$ placing on the right rod. Here $dx_1$ ($dx_2$) is the infinitesimal displacement on the left (right) rod. The linear charge density is $\lambda=Q/2a$ and the infinitesimal charge $dq$ can be expressed as $dq=\lambda dx_1$. The force exerted on the left rod with an infinitesimal displacement $dx_1$ and an infinitesimal charge $dq_1=\lambda dx_1$ by an infinitesimal charge $dq_2=\lambda dx_2$ is:
\[
dF=\frac{kdq_1dq_2}{(x_2-x_1)^2}=\frac{k\lambda^2dx_1dx_2}{(x_2-x_1)^2}
\]
The total force is obtained through the integration:
\[
F=k\lambda^2\int_{b-a}^{b+a}\left[\int_{-a}^a\frac{dx_1}{(x_1-x_2)^2}\right]dx_2
\]
In the inner integration, the $x_1$ is a variable and the $x_2$ is a constant so we have the equivalent relation $dx_1=d(x_1-x_2)$. The above equation leads to:
\begin{eqnarray*}
F&=&k\lambda^2\int_{b-a}^{b+a}\left[\int_{-a}^a\frac{d(x_1-x_2)}{(x_1-x_2)^2}\right]dx_2 \\
&=&k\lambda^2\int_{b-a}^{b+a}\left[-\frac{1}{x_1-x_2}\right]_{x_1=-a}^adx_2 \\
&=&k\lambda^2\int_{b-a}^{b+a}\left(-\frac{1}{a-x_2}+\frac{1}{-a-x_2}\right)dx_2 \\
&=&k\lambda^2\int_{b-a}^{b+a}\left(\frac{1}{x_2-a}-\frac{1}{x_2+a}\right)dx_2 \\
&=&k\lambda^2\int_{b-a}^{b+a}\frac{1}{x_2-a}d(x_2-a)-k\lambda^2\int_{b-a}^{b+a}\frac{1}{x_2+a}d(x_2+a) \\
&=&k\lambda^2\left[\ln(x_2-a)\right]_{b-a}^{b+a}-k\lambda^2\left[\ln(x_2+a)\right]_{b-a}^{b+a} \\
&=&k\lambda^2\ln(\frac{b}{b-2a})-k\lambda^2\ln(\frac{b+2a}{b}) \\
&=&k\lambda^2\ln(\frac{b^2}{(b-2a)(b+2a)})
\end{eqnarray*}
