Find the magnitude of the electric field at the point P in the configuration shown in figure for $\mathrm{d}>>\mathrm{a}$. Take 2qa = p.

The given situation is similar to the dipole at a point on an equatorial line (as shown in the following figure).From the figure, we can see that sin$\mathrm{\theta }$ components cancel out each other and cos$\mathrm{\theta }$ components add up. That is,
${\mathrm{E}}_{\text{net }}=2\mathrm{E}(\cos \mathrm{\theta })=2\times \frac{\mathrm{kq}}{{\mathrm{x}}^2}(\cos \mathrm{\theta })=2\times \frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{\mathrm{q}}{{\left({\mathrm{a}}^2+{\mathrm{d}}^2\right)}^{3/2}}\right]\mathrm{a}$
The dipole moment is given as p = 2qa. Substituting this value of p in the above relation, we get the magnitude of the net electric field at the point P (in the given configuration) due to dipole= $\frac{p}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{1}{{\left(a^2+d^2\right)}^{3/2}}$

Now net electric field= $$\begin{gathered} \sqrt{\frac{k^2{q}^2}{d^4}+\frac{k^2{p}^2}{{\left(a^2+d^2\right)}^3}} \\ = \sqrt{\frac{k^2}{d^4}(q^2+\frac{p^2}{d^2}}) \\ =\sqrt{\frac{k^2}{d^6}(q^2{d}^2+p^2)} \\ =\frac{k}{d^3}\sqrt{q^2{d}^2+p^2} \end{gathered}$$

as ${\mathrm{E}}_{\text{net }}=\frac{1}{4{\mathrm{\pi \epsilon }}_0{\mathrm{d}}^3}\sqrt{{\mathrm{q}}^2{\mathrm{d}}^2+{\mathrm{p}}^2}$