Chapter 22, Problem 75P

A thin rod of length L carries charge Q distributed uniformly over its length, (a) Show that the potential in the plane that perpendicularly bisects the rod is given by

$V\left(r\right)=\frac{2kQ}{L}\ln \left[\frac{L}{2r}+\sqrt{1+\frac{L^2}{4r^2}}\right]$

where r is the perpendicular distance from the rod center and where the zero of potential is taken at infinity, (b) Show that this expression reduces to an expected result when r >> L (Hint: See Appendix A for a series expansion of the logarithm.)