# Electric field generated by a uniform rod

When you calculate the electric field of a uniform rod of length $L$ and charge density $\lambda$ at a distance $d$ on its axis you can remove the effect of the ends considering $L$ is big enough.
However if you take other point, not on the axis (for example, at a distance $a$ of one end), you should consider the effect of the end, how do you do it?
I think, this effect only adds another component to the electric field, but I'm not sure. Can someone tell me if this is correct?

A uniformly charged rod or needle with finite length ($L=2c$) is the limiting case for a conducting prolate spheroid.

For the charge distributes at $-c<z<c$ and $x=y=0$, the equipotential surface is

$$\frac{x^2}{s}+\frac{y^2}{s}+\frac{z^2}{c^2+s}=1$$

where $s>0$.

The electrostatic potential is

\begin{align} \phi (x,y,z) &= \frac{\lambda}{4\pi \varepsilon_0} \ln \frac{z+c+\sqrt{x^2+y^2+(z+c)^2}} {z-c+\sqrt{x^2+y^2+(z-c)^2}} \\ &= \frac{\lambda}{4\pi \varepsilon_0} \left( \sinh^{-1} \frac{z+c}{\sqrt{x^2+y^2}}- \sinh^{-1} \frac{z-c}{\sqrt{x^2+y^2}} \right) \\ &= \frac{\lambda}{2\pi \varepsilon_0} \sinh^{-1} \frac{c}{\sqrt{s}} \end{align}

The electric field in cylindrical coordinates $(\rho, \phi, z)$ is

\begin{align} \mathbf{E} &= -\nabla \phi \\ &= \frac{\lambda}{4\pi \varepsilon_0} \begin{pmatrix} \dfrac{z+c}{\sqrt{\rho^2+(z+c)^2}}- \dfrac{z-c}{\sqrt{\rho^2+(z-c)^2}} \\ 0 \\ \dfrac{1}{\sqrt{\rho^2+(z-c)^2}}- \dfrac{1}{\sqrt{\rho^2+(z+c)^2}} \\ \end{pmatrix} \end{align}