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.
![](/images/user/7/uploads/fig11.12.png)
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
\[E = k\frac{2p}{y^3} \tag{2} \label{2}\]