Find the force of interaction-3

Clearly, as $R<<L$ , the electric field(radial from axis) due to the cylinder is given by :

$E(r) = \begin{cases} \dfrac{\rho r}{2 \epsilon{0}} , & \text{if }r \leq R \\ \dfrac{\rho R^2}{2 r \epsilon_{0}}, & \text{if } r>R , r<< L \end{cases}$

Now, consider an arc of thickness of radius $r$ , and $dr$ one one of the cylinder, whose center coincides with the center of other cylinder as shown :

Now , clearly, the horizontal force on a small element on the arc corresponding to angle $\theta$ , where $\theta$ varies from $- \phi$ to $\phi$ is $E(r) \rho r L \cos \theta d \theta dr$

Hence , the net force of interaction is given by:

$F = \displaystyle \int_{0}^{2R} \rho L r E(r) \bigg(\int_{- \phi}^{\phi}\cos \theta d \theta\bigg) dr$

= $\displaystyle \int_{0}^{2R} 2\rho L r E(r) \sin \phi dr$

Now, clearly, by cosine rule,

$\cos \phi = \dfrac{R^2+r^2-R^2}{2Rr} = \dfrac{r}{2R}$

Hence, $\sin \phi = \sqrt{1 - \bigg(\dfrac{r}{2R}\bigg)^2}$

Thus ,

$F = \displaystyle \int_{0}^{2R} 2\rho L r E(r) \sqrt{1 - \bigg(\dfrac{r}{2R}\bigg)^2} \ dr$

= $\displaystyle \int_{0}^{R}2\rho L r \dfrac{\rho r}{2 \epsilon_{0}} \sqrt{1 - \bigg(\dfrac{r}{2R}\bigg)^2} \ dr + \displaystyle \int_{R}^{2R} 2 \rho r L \dfrac{\rho R^2}{2 r \epsilon_{0} }\sqrt{1 - \bigg(\dfrac{r}{2R}\bigg)^2 } \ dr$

Now, after the substitution $r = 2R \sin t$ ,

$F =\displaystyle \dfrac{\rho^2 R^3 L}{ \epsilon_{0}} \bigg( \int_{0}^{\pi/6} 8 \sin^2 t \cos^2 t dt + \int_{\pi/6}^{\pi/2} 2 \cos^2 t dt \bigg)$

$= \boxed{\dfrac{\rho^2 R^3 L}{ \epsilon_{0}} \bigg(\dfrac{\pi}{2} - \dfrac{3 \sqrt{3}}{8} \bigg) }$