Answer
$E_{net} = \frac{qd}{4~\pi~\epsilon_0~r^3}$
Work Step by Step
The electric field due to the positive charge is directed away from the positive charge, and the electric field due to the negative charge is directed toward the negative charge
Note that since the magnitude of the charges are equal, the horizontal components of the electric field due to each charge cancels out, and the net electric field is the sum of the vertical components.
We can find the magnitude of $E_{net}$:
$E_{net} = 2\times \frac{1}{4~\pi~\epsilon_0}~\frac{\vert q \vert}{R^2}~sin~\theta$
$E_{net} = \frac{1}{2~\pi~\epsilon_0}~\frac{q}{(\sqrt{r^2+(d/2)^2})^2}~(\frac{d/2}{\sqrt{r^2+(d/2)^2}})$
$E_{net} = \frac{qd}{4~\pi~\epsilon_0~[r^2+(d/2)^2](\sqrt{r^2+(d/2)^2})}$
$E_{net} = \frac{qd}{4~\pi~\epsilon_0~[r^2+(d/2)^2]^{3/2}}$
$E_{net} \approx \frac{qd}{4~\pi~\epsilon_0~(r^2)^{3/2}}$
$E_{net} \approx \frac{qd}{4~\pi~\epsilon_0~r^3}$
