Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
In general, we can assume that the attractive force between $+Q$ and $-q$ is greater than the repulsive force between $+Q$ and $+q$ since $-q$ is closer to $+Q$ than $+q$.
This means that there is a net force toward the charge $Q$.
Thus, the net force exerted on the dipole by $+Q$ in this case is toward the right.
$$\sum F_x=F_{Q\rightarrow(-q)}-F_{Q\rightarrow(+q)}$$
$$\sum F_x=\dfrac{kQq}{\left( r-\dfrac{s}{2} \right)^2}-\dfrac{kQq}{\left( r+\dfrac{s}{2} \right)^2}$$
Note that we ignored the charges' signs since we focused on the forces' directions.
$$\boxed{\sum F =kQq\left(\dfrac{1}{\left( r-\dfrac{s}{2} \right)^2}-\dfrac{1}{\left( r+\dfrac{s}{2} \right)^2}\right)\hat i}$$
$$\color{blue}{\bf [b]}$$
As we explained above, this net force is toward the charge $+Q$.
$$\color{blue}{\bf [c]}$$
Using the binomial approximation for $\left( r-\dfrac{s}{2}\right)^{-2}$ and $\left( r-\dfrac{s}{2}\right)^{-2}$ since $r\gt\gt s $.
$$\left( r-\dfrac{s}{2}\right)^{-2}=r^{-2} \left( 1-\dfrac{s}{2r}\right)^{-2} \approx r^{-2}\left(1+ \dfrac{2s}{2r} \right)\\
=\boxed{r^{-2}\left(1+ \dfrac{s}{r} \right)}$$
By the same approach,
$$\left( r+\dfrac{s}{2}\right)^{-2}=r^{-2} \left( 1+\dfrac{s}{2r}\right)^{-2} \approx r^{-2}\left(1- \dfrac{2s}{2r} \right) \\
=\boxed{r^{-2}\left(1- \dfrac{s}{r} \right)}\hat i$$
Plug these two boxed results into the boxed formula above,
$$ \sum F =\dfrac{kQq}{r^2}\left( 1+\dfrac{s}{r}-\left[ 1- \dfrac{s}{r} \right]\right) \hat i$$
$$ \sum F =\dfrac{kQq}{r^2}\left( 1+\dfrac{s}{r}- 1+ \dfrac{s}{r} \right) \hat i=\dfrac{kQq}{r^2}\left( \dfrac{2s}{r} \right) \hat i$$
$$\boxed{ \sum F =\dfrac{2kQqs}{r^3}\;\hat i}$$
$$\color{blue}{\bf [d]}$$
Coulomb’s law says that the electric force depends on the inverse square of the distance when we are dealing with the force between two point charges. Here we are dealing with force between a dipole.
