The electric field of a point charge drops off as an inverse square law, but that of a dipole varies as $1/r^3$. Why does the field drop off more quickly for a dipole?
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Very simply, the field of the positive and negative elements of the dipole "almost" cancel out - but not quite. It is because they are some small distance away that there is a residual (third order) term. You can see this by taking two charges $+q$ and $-q$ at a distance $2d$, and look at the field a distance $r$ from the center of the two (on the same axis). We get
$$E \propto \frac{q}{(r+d)^2} - \frac{q}{(r-d)^2}\\ = \frac{q}{r^2}\left(\frac{1}{(1+\frac{d}{r})^2}-\frac{1}{(1-\frac{d}{r})^2}\right)$$
Doing Taylor expansion of the terms at the bottom, we get
$$E \approx \frac{q}{r^2}\left((1-\frac{2d}{r})-(1+\frac{2d}{r})\right)=-\frac{4qd}{r^3}$$
From which it follows that the electric field scales with the dipole moment $2qd$ and the inverse cube distance. There are higher order terms I omitted - but once you allow $d$ to go to zero (while maintaining the value $qd$ constant) you will see that those terms become negligible.
