Let's find the electric field of an electric dipole at a point on the axis joining the two charges. The Figure 1 shows that the centre of our coordinate system is the centre of the dipole. We are going to find the electric field at the point \(p\) shown in Figure 1.
It's quiet simple that you need to add the electric fields due to both charges at the point. Consider that the electric field due to positive charge is \(\vec E_1\) and the electric field due to negative charge is \(\vec E_2\). In an electric dipole the magnitude of both charges is the same say \(q\) and are separated by a distance \(d\). Therefore, the distance of positive charge from the point is \(y + d/2\) and the distance of negative charge from the point is \(y - d/2\). Note that the x-components of electric fields due to both charges is zero. The electric field vectors \(\vec E_1\) and \(\vec E_2\) are
\[{\vec E_1} = k\frac{q}{{{{\left( {y + \frac{d}{2}} \right)}^2}}} \hat j\]
\[{\vec E_2} = -k\frac{q}{{{{\left( {y - \frac{d}{2}} \right)}^2}}} \hat j\]
The unit vector \(\hat j\) gives the direction to the electric field vector which is along y-axis. If \(E_{1y}\) is the y-component of \(E_1\) and \(E_{2y}\) is the y-component of \(E_2\), then you know that \(E_1 = E_{1y}\) and \(E_2 = E_{2y}\) (there is no x-component of electric field at the point \(p\)). The net electric field which is \(\vec E = \vec E_y\) (the subscript y-represents the y-component) at the point \(p\) is
\[\begin{align*} \vec E &= k\left[ {\frac{q}{{{{\left( {y + \frac{d}{2}} \right)}^2}}} - \frac{q}{{{{\left( {y - \frac{d}{2}} \right)}^2}}}} \right]\widehat j\\ {\rm{or,}}\quad \vec E &= k\frac{q}{{{y^2}}}\left[ {{{\left( {1 + \frac{d}{{2y}}} \right)}^{ - 2}} - {{\left( {1 - \frac{d}{{2y}}} \right)}^{ - 2}}} \right]\widehat j \end{align*}\]
Note that in an approximation that \(y\) is much larger than \(d\), the term obviously \(\left| \frac{d}{2y} \right| < 1\). Now we use the binomial expansion to solve the terms \({\left( {1 - \frac{d}{{2y}}} \right)^{ - 2}}\) and \({\left( {1 + \frac{d}{{2y}}} \right)^{ - 2}}\). Note that the expression for the binomial expansion of \({(1 + x)^n}\) when \(\left| x \right|<1\) is \({{(1+x)}^{n}}=1+nx+n(n-1)\frac{{{x}^{2}}}{2}+...\). So,
\[\begin{align*} {\left( {1 - \frac{d}{{2y}}} \right)^{ - 2}} &= 1 + \frac{d}{y} + \frac{3}{4}\left( {\frac{{{d^2}}}{{{y^2}}}} \right) + ...\\ {\rm{and,}}\quad {\left( {1 + \frac{d}{{2y}}} \right)^{ - 2}} &= 1 - \frac{d}{y} + \frac{3}{4}\left( {\frac{{{d^2}}}{{{y^2}}}} \right) + ... \end{align*}\]
Now keeping only the first two terms neglecting the smaller terms we have \({\left( {1 - \frac{d}{{2y}}} \right)^{ - 2}} \cong 1 + \frac{d}{y}\) and \({\left( {1 + \frac{d}{{2y}}} \right)^{ - 2}} \cong 1 - \frac{d}{y}\). So the net electric field is,
\[\begin{align*} \vec E &= k\frac{q}{{{y^2}}}\left[ {\left( {1 - \frac{d}{y}} \right) - \left( {1 + \frac{d}{y}} \right)} \right]\hat j\\ &= - k\frac{{2qd}}{{{y^3}}}\hat j = - k\frac{{2p}}{{{y^3}}}\hat j = k\frac{{2\vec p}}{{{y^3}}} \tag{1} \label{1} \end{align*}\]
As you can see from the above expression of the net electric field that the electric field is proportional to \(\frac{1}{{{y^3}}}\) instead of \(\frac{1}{{{y}^{2}}}\). The above expression of net electric field tells us that the net electric field is along negative y-direction in our case shown in Figure 1. If you consider only the magnitude of the net electric field, it is
\[E = k\frac{2p}{y^3} \tag{2} \label{2}\]
