We then simplify the above expression using $x\gg d$ and the Taylor expansion ${\left(1+\varepsilon \right)}^{\alpha }\approx 1+\alpha \varepsilon $ which is valid when $\varepsilon \ll 1$. Factoring $x$ out of the denominator of each fraction yields the following

\begin{aligned} E_{net}&=\frac{kq}{\displaystyle{{\left(x-\frac{d}{2}\right)}^2}}-\frac{kq}{\displaystyle{{\left(x+\frac{d}{2}\right)}^2}} \\ \\ &=\frac{kq}{\displaystyle{x^2{\left(1-\frac{d}{2x}\right)}^2}}-\frac{kq}{\displaystyle{x^2{\left(1+\frac{d}{2x}\right)}^2}} \\ \\ &=\frac{kq}{x^2}\left[\frac{1}{\displaystyle{{\left(1-\frac{d}{2x}\right)}^2}}-\frac{1}{\displaystyle{{\left(1+\frac{d}{2x}\right)}^2}}\right] \end{aligned}

Noticing that $d/2x\ \ll 1$, we apply the Taylor expansion above and derive the following expression

\begin{aligned} E_{net}&=\frac{kq}{x^2}\left[\frac{1}{\displaystyle{{\left(1-\frac{d}{2x}\right)}^2}}-\frac{1}{\displaystyle{{\left(1+\frac{d}{2x}\right)}^2}}\right] \\ \\ &=\frac{kq}{x^2}\left[{\left(1-\frac{d}{2x}\right)}^{-2}-{\left(1+\frac{d}{2x}\right)}^{-2}\right] \\ \\ &\approx \frac{kq}{x^2}\left[1+\frac{2d}{2x}-\left(1-\frac{2d}{2x}\right)\right] \\ \\ &\approx \frac{kq}{x^2}\cdot \frac{2d}{x} \\ \\ &\approx \frac{2kqd}{x^3} \end{aligned}

Thus, the far field generated by the electric dipole at a point $x\gg d$ along its axis is given by

\boxed{{\overrightarrow{E}}_{net}\approx\frac{2kqd}{x^3}\ \ \hat{x}\approx \frac{2kp}{x^3}\ \ \hat{x}}

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