Consider an electric quadrupole of length $2l$. Then, for a point $P$ on perpendicular bisector of quadrupole at a distance $r$ from the centre.

Here r is much much greater than l.

 

At $P$, vertical components of positive charges $q$ are in same direction and horizontal components cancels out each other,

So, $F_{\text{net }}=2E_1\cos \theta -E_2$

where, $E_1=$ field of positive charge

and $E_2=$field of negative charge

Now, $E_1=\frac{kq}{{\left(\sqrt{r^2+l^2}\right)}^2}\frac{r}{\sqrt{r^2+l^2}}$

and $E_2=\frac{k\left(2q\right)}{r^2}$

$$\begin{gathered} \therefore E_{\text{net }}=\frac{2kqr}{{\left(r^2+l^2\right)}^{3/2}}-\frac{2kq}{r^2} \\ \Rightarrow E_{\text{net }}=2kq\left\{\frac{r}{{\left(r^2+l^2\right)}^{3/2}}-\frac{1}{r^2}\right\} \end{gathered}$$

For $r>>l,{\left(r^2+l^2\right)}^{3/2}\simeq r^3$

$$\begin{gathered} \therefore E_{\text{net }}=2kq\left\{\frac{r^3-{\left(r^2+l^2\right)}^{3/2}}{r^2{\left(r^2+l^2\right)}^{3/2}}\right\} \\ =2kq{r}^3\left\{\frac{1-{\left(1+l^2/r^2\right)}^{3/2}}{r^5{\left(1+l^2/r^2\right)}^{3/2}}\right\} \\ =\frac{2kg{r}^3}{r^5}·\left\{\frac{1-\left(1+\frac{3}{2}·\frac{l^2}{r^2}\right)}{1+\frac{3}{2}·l^2/r^2}\right\} \\ =\frac{2kq{r}^3\left(-\frac{3}{2}\frac{I^2}{r^2}\right)}{r^5}=\left(-\frac{3}{2}\right)\frac{2kg{l}^2}{r^4} \\ =-\frac{3}{4}\frac{k\left(q(2l{)}^2\right)}{r^4}=-\frac{3}{4}\frac{kp}{r^4} \\ \therefore E\propto \frac{1}{r^4} \end{gathered}$$

and it points towards negative charge.