Two identical dipoles are arranged in $\displaystyle x-y$ plane as shown in the figure. Find the magnitude and the direction of net electric field at the origin O .

Dipole moment due to dipole BA is $\displaystyle \vec{p_1}$ & dipole moment due to dipole DC is $\displaystyle \vec{p_2}$.Electric field $\displaystyle \vec{E_1}$ due to $\displaystyle \vec{p_1}$ is along OB.Electric field $\displaystyle \vec{E_2}$ due to $\displaystyle \vec{p_2}$ is along OD.Magnitude of resultant Electric field$\displaystyle \left|\vec{E}_{net}\right| = 2\sqrt{2}\,E$Since $\displaystyle \left|\vec{E_1}\right| = \left|\vec{E_2}\right| = E = \dfrac{1}{4\pi \varepsilon_0} \cdot \dfrac{q}{a^2}$$\displaystyle \vec{E}_{net} = \dfrac{1}{4\pi \varepsilon_0} \cdot \dfrac{2\sqrt{2}\; q}{a^2}$Direction of $\displaystyle \vec{E}_{net}$ is $\displaystyle 225$° to x-axis.Alternatively:Considering CB as dipole, electric field at O$\displaystyle E_1 = \dfrac{2kq \times (a/\sqrt{2})}{\left[\left(\dfrac{a}{\sqrt{2}}\right)^2 + \left(\dfrac{a}{\sqrt{2}}\right)^2\right]^{3/2}}$$\displaystyle E_1 = \dfrac{\sqrt{2}\,kq \times a}{a^3 \left(\dfrac{1}{2} + \dfrac{1}{2}\right)^{3/2}}$$\displaystyle E_1 = \dfrac{\sqrt{2}\,kq}{a^2}$Similarly considering AD as another dipole, electric field at O$\displaystyle E_2 = \dfrac{2kq \times (a/\sqrt{2})}{\left[\left(\dfrac{a}{\sqrt{2}}\right)^2 + \left(\dfrac{a}{\sqrt{2}}\right)^2\right]^{3/2}}$$\displaystyle E_2 = \dfrac{\sqrt{2}\,kq \times a}{a^3 \left(\dfrac{1}{2} + \dfrac{1}{2}\right)^{3/2}}$$\displaystyle E_2 = \dfrac{\sqrt{2}\,kq}{a^2}$$\displaystyle E_{net} = E_1 + E_2$$\displaystyle = \dfrac{\sqrt{2}\,kq}{a^2} + \dfrac{\sqrt{2}\,kq}{a^2}$$\displaystyle = \dfrac{2\sqrt{2}\,kq}{a^2}$