Solution:
Solution
$$E_{D}=\large\frac{2 k\;q }{((\large\frac{d}{2})^{2}+x^2)^{\large\frac{3}{2}}}\times \large\frac{d}{2}$$
This is because only horizontal components will add vertical cancel each other being equal and opposite
$$E_{D} \approx \large\frac{2 \;k \;q\; d}{2x^3}$$
$$E_{D} \approx \large\frac{k \; P}{x^3}$$
From the diagram it is clear the ED & P are in opposite direction
Therefore
$$\overrightarrow{E}_{D}=-\large\frac{k\;\overrightarrow{P}}{x^3}\;.$$