Two identical thin rods each of length 2a carry equal charges +Q that is distributed uniformly along their lengths. The rods lie along the x –axis with their centers separated by a distance b > 2a.

Field due to left rod at a distance x from the right end is $\Rightarrow \mathrm{E}=\frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_0\mathrm{x}\left(\mathrm{x}+2\mathrm{a}\right)}$

Force on the right rod due to the left rod is $\Rightarrow \mathrm{F}=\frac{{\mathrm{Q}}^2}{16{\mathrm{\pi \epsilon }}_0{\mathrm{a}}^2}{\log }_{\mathrm{e}}\left(\frac{{\mathrm{b}}^2}{{\mathrm{b}}^2-4{\mathrm{a}}^2}\right)$

Field due to left rod at a distance x from the right end is $\Rightarrow \mathrm{E}=\frac{\mathrm{Q}}{2{\mathrm{\pi \epsilon }}_0\mathrm{x}\left(\mathrm{x}+2\mathrm{a}\right)}$

Force on the right rod due to the left rod is $\Rightarrow \mathrm{F}=\frac{{\mathrm{Q}}^2}{8\sqrt{2}{\mathrm{\pi \epsilon }}_0{\mathrm{a}}^2}{\log }_{\mathrm{e}}\left(\frac{{\mathrm{b}}^2}{{\mathrm{b}}^2-4{\mathrm{a}}^2}\right)$