Answer
\begin{align*}\text{Electric field, E}=\frac{1}{4\,\pi\,\epsilon_{0}}\,\frac{qd}{[z^2+(d/2)]^{3/2}}\,\hat{x}\end{align*}
Work Step by Step
Here, $E_1$ is the field due to the positive charge alone, and $E_2$ is the field due to the negative charge alone (in figure).
The vertical components of the field cancel each other out, leaving a net horizontal component along the positive x-axis -
\begin{align*}
E_x = 2\,\frac{1}{4\,\pi\,\epsilon_{0}}\,\frac{q}{r^2}\,\sin\theta\tag{1}
\end{align*}
where the symbols are as described in the figure.
Also, from figure $r=\sqrt (z^2+(d/2)^2)$ and $\sin\theta=d/2r$.
Thus, substituting in equation (1) the electric field is
\begin{align*}
\text{E}=\frac{1}{4\,\pi\,\epsilon_{0}}\,\frac{qd}{[z^2+(d/2)]^{3/2}}\,\hat{x}\end{align*}
