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Charged object exerts a net attractive force on an electric dipole, permanent electric dipole consisting of charges +q and-q separated by the fixed distance s. Charge +Q is the distance r from the center of the dipole and $s<<r.$

The net force from the charge is coming from both the +ve and the-ve charges in the dipole. From the given figure, the distance fromq to Q is$r+s/2$, and so the force between these two charges is given by $\stackrel{鈫�}{F}\text{positive}=\frac{-k脳Q脳p}{(r+s/2{)}^2}\hat{i}$

here the force is negative because the two positive charges repel each other, it pushes the dipole away from it. Similarly, since the negative charge in the dipole is a distance $r-s/2$away fromQ, and is attractive, the force on -q is given by

$\stackrel{鈫�}{F}\text{negative}=\frac{+k脳Q脳p}{(r-s/2{)}^2}\hat{i}$

Then the net force is F_negative+F_positive

$$\begin{gathered} \stackrel{鈫�}{F}=\frac{-k脳Q脳p}{(r+s/2{)}^2}\hat{i}+\frac{+k脳Q脳p}{(r-s/2{)}^2}\hat{i} \\ \stackrel{鈫�}{F}=k脳Q脳p\left(\frac{1}{(r-s/2{)}^2}-\frac{1}{(r+s/2{)}^2}\right) \end{gathered}$$

Charged object exerts a net attractive force on an electric dipole, permanent electric dipole consisting of charges+q and -q separated by the fixed distance s. Charge+Q is the distancer from the center of the dipole and $s<<r$.

Because $r-s/2<r+s/2$, the first term is bigger, and so the force is to the right and it's an attractive force.

Charged object exerts a net attractive force on an electric dipole, permanent electric dipole consisting of charges +q and -q separated by the fixed distance s. Charge +Q is the distancer from the center of the dipole and $s<<r.$

Using a binomial expansion, ${\left(1+x\right)}^{-n}鈮�1-nx$, we can rewrite

$$\begin{gathered} \left(\frac{1}{{\left(r-s/2\right)}^2}-\frac{1}{{\left(r+s/2\right)}^2}\right)=\frac{1}{r^2}\left(\frac{1}{{\left(1-s/2r\right)}^2}-\frac{1}{{\left(1+s/2r\right)}^2}\right) \\ \left(\frac{1}{(r-s/2{)}^2}-\frac{1}{(r+s/2{)}^2}\right)=\frac{1}{r^2}\left(1+\frac{s}{r}-1+\frac{s}{r}\right) \\ \left(\frac{1}{(r-s/2{)}^2}-\frac{1}{(r+s/2{)}^2}\right)=\frac{2s}{r^2} \\ \text{Thismeansthat} \\ \stackrel{鈫�}{F}鈮�\frac{2k脳Q脳p脳s}{r^3}\hat{i} \end{gathered}$$

And this force is attractive.

Charged object exerts a net attractive force on an electric dipole, permanent electric dipole consisting of charges +q and -q separated by the fixed distance s. Charge +Q is the distancer from the center of the dipole and$s<<r.$

Coulomb's law applies between point charges. This situation is different- we have extended charge distributions. The negative charge is partially screened by the positive charge in the dipole. So, the force is not as strong. The force drops to zero because the dipole looks neutral at large distances; the two charges nearly cancel each other out.