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Let $\sigma$ be the areal charge density $\frac{Q}{\pi R^2}$ of a disk of radius $R$; then the electric field on the line pependicular to the disk and passing through is centre, if we use its direction as the $x$ axis, is$$\frac{\sigma x}{2\varepsilon_0}\Bigg(\frac{1}{|x|}-\frac{1}{\sqrt{x^2+R^2}}\Bigg).$$ I read that, if $|x|\gg R$, it can be approximated by $\frac{Qx}{4\pi\varepsilon_0}$, but I cannot prove it to myself... I tried to write the expression in several ways, including with the use of the Taylor polynomial for $\frac{1}{\sqrt{x^2+R^2}}=\frac{1}{|x|}\big( 1-\frac{1}{2}\big(\frac{R}{x}\big)^2+o\big(\big(\frac{R}{x}\big)^2 \big)\big)$, $\frac{R}{x}\to 0$, but I get nothing useful... I thank you all for any answer!

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    $\begingroup$ Substitute your Taylor expansion into the expression, and the charge density. $\endgroup$ – krvolok Sep 1 '15 at 13:18
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    $\begingroup$ Also, the expression is $\frac{Q}{4\pi \epsilon_0 x^2}$, just like a point charge. Your textbook may have a typo! $\endgroup$ – krvolok Sep 1 '15 at 13:18
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    $\begingroup$ Maybe this will help: physics.stackexchange.com/questions/169976/… $\endgroup$ – Dor Sep 1 '15 at 13:34
  • $\begingroup$ @krvolok Well, a typo would explain everything. I thank you all! $\endgroup$ – Self-teaching worker Sep 1 '15 at 15:48
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$$\frac1{|x|}-\frac1{\sqrt{x^2+R^2}}=\frac{\sqrt{x^2+R^2}-|x|}{|x|\sqrt{x^2+R^2}}=\frac{x^2+R^2-x^2}{|x|\sqrt{x^2+R^2}(\sqrt{x^2+R^2}+|x|)}.$$

For $|x|\gg R$, $\sqrt{x^2+R^2}\approx|x|$ and the expression simplifies to $$\frac{R^2}{2|x|^3}.$$

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  • $\begingroup$ ...which gives us the expression given by krvolok in the comment. Thank you very much!!! $\endgroup$ – Self-teaching worker Sep 1 '15 at 15:49
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    $\begingroup$ @Self-teachingDavide: absolutely. An electric field that grows with distance would be a nonsense. $\endgroup$ – Yves Daoust Sep 1 '15 at 16:01

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