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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.

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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 $L\ll z$: $$E=\frac{1}{4 \pi \varepsilon_{0}}\frac{2 \lambda L}{z\sqrt{z^{2}+L^{2}}}\approx\frac{1}{4 \pi \varepsilon_{0}}\frac{2 \lambda L}{z\sqrt{z^{2}}}=\frac{1}{4 \pi \varepsilon_{0}}\frac{2 \lambda L}{z^2}$$

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