In electrodynamics, for a line segment of length 2L with a uniform line charge $\lambda$, the electric field at a distance z above the midpoint of this straight line segment is
$E\left(\vec{r}\right)=\frac{1}{4 \pi \varepsilon_{0}}\frac{2 \lambda L}{z\sqrt{z^{2}+L^{2}}}\hat{z}$
For a field point z > > L:
We may 'translate' L to the 'zero' point so that
$E\approx \frac{1}{4 \pi \varepsilon_{0}}\frac{2 \lambda L}{z^{2}}$
and
for the limit $L\rightarrow \infty$:
$E = \frac{1}{4 \pi \varepsilon_{0}}\frac{2 \lambda}{z}$
In the case of the second, I have forgotten how the derivation was done.
It seems to involve Taylor series expansion if I remember.
Any jolt to my memory would be appreciated.
In the square root term, either $z\ll L$ or $L\ll z$. Just keep the larger one, and completely ignore the other.
For $z\ll L$: $$E=\frac{1}{4 \pi \varepsilon_{0}}\frac{2 \lambda L}{z\sqrt{z^{2}+L^{2}}}\approx\frac{1}{4 \pi \varepsilon_{0}}\frac{2 \lambda L}{z\sqrt{L^{2}}}=\frac{1}{4 \pi \varepsilon_{0}}\frac{2 \lambda L}{zL}=\frac{1}{4 \pi \varepsilon_{0}}\frac{2 \lambda }{z}$$
For $L\ll z$: $$E=\frac{1}{4 \pi \varepsilon_{0}}\frac{2 \lambda L}{z\sqrt{z^{2}+L^{2}}}\approx\frac{1}{4 \pi \varepsilon_{0}}\frac{2 \lambda L}{z\sqrt{z^{2}}}=\frac{1}{4 \pi \varepsilon_{0}}\frac{2 \lambda L}{z^2}$$