Limit example in Zangwill "Modern Electrodynamics"

Zangwill shows that the potential of a finite line segment going from $-L$ to $L$ on the $z$-axis with constant line charge density $\lambda$ is:

$$\phi(z,\rho) = \frac{\lambda}{4\pi\epsilon_0}\ln\left(\frac{\sqrt{(L-z)^2 + \rho^2} + L - z}{\sqrt{(L+z)^2 + \rho^2} - L - z}\right)$$

where $\rho$ is defined as the distance in the picture and $R = \sqrt{\rho^2 + z^2}$ He goes on to take limits of this potential:

If z >> L then

$\phi(z,\rho) \approx \frac{\lambda}{4\pi\epsilon_0} \ln\left(\frac{1 - z/R + L/R}{1 - z/R - L/R}\right)$

and if z << L and $\rho$ << L then

$\phi(z,\rho) = \frac{\lambda}{4\pi\epsilon_0}\ln\left(\frac{\sqrt{(L)^2 + \rho^2} + L}{\sqrt{(L)^2 + \rho^2} - L}\right)$

My confusion is with the first limit. In the second limit it seems like Zangwill just said "z is small compared to L so I will drop it from the $(L - z)^2$ term and the $(L - z)$ term". But in the first limit, he only drops the L in the $(L - z)^2$ term and not the $(L-z)$ term even though z >> L. Why is this? Surely, in the limit of z >> L, L should be dropped from both terms.

I think the main reason for keeping me the $L$ even though it is sub-leading is that if you set it to zero immediately, then the numerator and denominator cancel and you get zero potential for all $z$ which is trivial.

The proper way to to take the limit is by making a Taylor expansion in $L/R$, since $R>z\gg L$.

We may then write the two fractions as \begin{eqnarray*} f_{\pm} & = & \sqrt{\left(L-z\right)^{2}+\rho^{2}}-z\pm L\\ & = & \sqrt{L^{2}-2Lz+R^{2}}-z\pm L \end{eqnarray*} whre now $R,z\gg L$.

Dividing by a factor of $R$ we have \begin{eqnarray*} \frac{f_{\pm}}{R} & = & \sqrt{1-\frac{2Lz}{R^{2}}+\frac{L^{2}}{R^{2}}}-\frac{z}{R}\pm\frac{L}{R}\\ & \approx & 1-\frac{Lz}{R^{2}}+\frac{L^{2}}{2R^{2}}-\frac{z}{R}\pm\frac{L}{R}\\ & \approx & 1-\left(\frac{L}{R}\cdot\frac{z}{R}\right)-\frac{z}{R}\pm\frac{L}{R} \end{eqnarray*} where we may drop the term $\frac{L^{2}}{R^{2}}$ as it is quadratically vanishing.

In my opition we should also keep the term $\frac{L}{R}\cdot\frac{z}{R}$ as it is of the same order as $\frac{L}{R}$ giving

$$\frac{f_{\pm}}{R} = 1-\frac{z}{R}\pm\frac{L}{R}\left(1\mp\frac{z}{R}\right)$$

$$\frac{f_{\pm}}{R} = 1-\frac{z}{R}\pm\frac{L}{R}$$
• @user373763 I may have expressed myself badly, then i say you "don't get what you want", i mean that you have removed all $z$ and $\rho$ dependence in the answer, and this is unphysical. I have eddited my asnwer to take this into account. Feb 1 '17 at 13:04