Calculate the electric field due to a uniformly charged rod of length $L$at a point $P$ at distance $r$from one end of the rod. Assume total charge on the rod to be $Q$.

For this we consider an infinitesimal element of length $d x$ on rod as shown in figure. Charge on the infinitesimal elemental length $d x$ is

$dq=\frac{Q}{L}dx$

 This $d q$ (an infinitesimal element) care a point charge, hence the electric field element at point P is $d E$given by $dE=\frac{dq}{4\pi {\epsilon }_0{x}^2}$

 This$d q$(an infinitesimal element) can be regarded as a point charge, hence the electric field $d E$due to this

$\Rightarrow dE=\frac{\left(\frac{Q}{L}\right)dx}{4\pi {\epsilon }_0{x}^2}$

The net electric field strength at point P can be given by integrating this expression over the whole length of rod as

$E_P=\int dE={\int }_r^{r+L}\frac{Qdx}{4\pi {\epsilon }_{0}L{x}^2}$

$$\begin{gathered} \Rightarrow E_P=\frac{Q}{4\pi {\epsilon }_{0}L}{\int }_r^{r+L}{x}^{-2}dx \\ \Rightarrow E_P=\frac{Q}{4\pi {\epsilon }_{0}L}{\left[-\frac{1}{x}\right]}_r^{r+L} \\ \Rightarrow E_P=\frac{Q}{4\pi {\epsilon }_{0}L}\left[\frac{1}{r}-\frac{1}{r+L}\right] \\ \Rightarrow E_P=\frac{Q}{4\pi {\epsilon }_{0}r(r+L)} \end{gathered}$$