A charge is distributed with a linear density $\lambda$, over a rod of the length $L$ placed along radius vector drawn from the point where a point charge $q$ is located. The distance between $q$ and the nearest point on linear charge is $R$. The electrical force experienced by the linear charge due to $q$ is

As coulomb's force is an action reaction pair, so force experienced by the linear charge is equal and opposite to the force experienced by point charge $q$. Here, we are computing electric force experienced by $q$ due to the charge.

Consider a small element on line charge as shown, then force experienced by $q$ due to this element is,

$dF=\frac{q\lambda dr}{4\pi {\epsilon }_0{r}^2}$

By integrating both side, we get

$F={\int }_R^{R+L} \frac{q\lambda dr}{4\pi {\epsilon }_0{r}^2}=\frac{q\lambda L}{4\pi {\epsilon }_{0}R\left(R+L\right)}$