A charged cylindrical insulator of radius R and length L carries a uniform charge density λ as illustrated below. Determine the electric potential at a distance r from the center of the cylinder along its axis. Note that R << L. 

$V\left(r\right)=\frac{\lambda }{4\pi {\epsilon }_0}\ln \left(\frac{r+\frac{L}{2}}{r-\frac{L}{2}}\right)$

$V\left(r\right)=\frac{\lambda }{4\pi {\epsilon }_0}\ln \left(\frac{r-\frac{L}{2}}{r+\frac{L}{2}}\right)$

$V\left(r\right)=\frac{\lambda }{4\pi {\epsilon }_0}\ln \left(\frac{r+\frac{L}{2}}{r+\frac{L}{2}}\right)$

$V\left(r\right)=\frac{1}{4\pi {\epsilon }_0}\ln \left(\frac{r-\frac{L}{2}}{r+\frac{L}{2}}\right)$