What is the physical significance of defining the dipole moment as a vector? I don't know why do we need to assign a direction to some quantity which is a product of the charge and the distance bewteen charges.
What is the physical reason of defining the dipole moment as a vector quantity ?
 A: The dipole moment is not just defined for two opposite charges. Any system of point charges with zero net charge can be assigned a dipole moment, which is defined by
$$
\mathbf d = \sum_j q_j \mathbf r_j,
$$
where $q_j$ and $\mathbf r_j$ are the charge and position of the $j$th particle. (Moreover, equivalent expressions hold for continuous distribution of charge.) This dipole moment is essential in understanding the electrostatic field produced by this charge distribution in the far field. And it should be obvious that there is no way to define it other than as a vector.

... and, that said, even for two charges, the separation between them still has a direction, and this direction is crucial in understanding the electric field produced by the pair, so why wouldn't it make sense to incorporate it into the dipole moment?
A: In chemistry, I've seen it with applications in polarization but I think the best way to understand it is via multipole expansion. The basic idea of the multipole expansion is that suppose you have a random charge distribution maybe a dipole ,maybe a a quadropole or whatever, there is one general formula which gives the potential of all of this. So you don't have to go about finding potential of each and then adding them up:
$$V(r)=\frac{1}{4\pi\epsilon_0r}∫_{V′}ρ(r′)(1− \frac{\hat{r} \cdot r'}{r^2})dV′+\text{stuff we don't care about in this example}$$

Math explanation: $\vec{r}$ is vector from our origin to the point at which we want to know potential $r'$ is vector from some point inside charge distribution to the point we want to know potential, the $V'$ denotes that we integrate over the volume containing the charge distribution, $\rho(r')$ is the charge density function
Ok, so, let us try to evaluate this integral incase of a dipole, where we have two opposite charges seperated by a distance $d$:
$$ V= \frac{1}{ 4 \pi \epsilon_o r} \int_{V'} \rho(r') dV  - \frac{1}{4 \pi \epsilon r} \int_{V'} \frac{ \hat{r} \cdot r'}{r^2} dV'$$
It is intuitive that since we have opposite and equal charges that the first integral vanishes, this leaves it with the second integral. Upon evaluating we would find that this lands in :$ \frac{k p \cos \theta}{r^2}$ which is what we get it from considering the fields and evaluating.
Tl;dr: Comes up naturally in a genera expression for potential of an arbitary charge distribution

Here are some helpful resources:

*

*Indetail explanation of multipole expansion

*Libretext explanation (also where picture is from)

*About density of point charge distribution (this is a bit complicated)

A: One way to look at the electric dipole moment is to look at the field produced by it,
which is arguably somewhat abstract.
Another aspect of the electric dipole moment is to look at how it responds to an external field... and this might be a better motivation for the

the electric dipole moment as a "directed quantity":
 $\vec p=Q\vec d$,
where $\vec d$ is the displacement vector from the negative-charge (-Q) to the positive charge (+Q).
Consider a bunch of seeds suspended in oil in the presence of an external electric field [take it to be uniform for simplicity].
https://youtu.be/Y6YdC2UoDYY?t=289

Seeds get electrically polarized and rotate to point along the direction of the local electric field, much like iron-filings
which act like little compass needles that point along the local external magnetic field.
The seeds are essentially electrostatic compass needles.

*

*Iron-filings, bar-magnets, and current-carrying coils have magnetic dipole moments (vectorial quantities) which POINT in the direction of the magnetic field.

*Similarly, seeds, opposite-charges-with-a-rigid-rod joining them, etc... have electric dipole moments (vectorial quantities) which POINT in the direction of the electric field.

Things that POINT (and not merely line-up with) remind me of vectors.
Additional comments.

*

*the displacement vector $\vec d$  is the vector difference between the two position-vectors that point from an origin to each charge. Note that $\vec d$ does not depend on the choice of origin used... and thus the separate position-vectors are not as physically meaningful as the displacement vector $\vec d$ from the negative charge to the positive charge.


*there is a notion of potential energy associated with
the orientation of the dipole moment vector in the external field.
The magnitude of the charges $|Q|$ and the separation distance $|\vec d|$
is not sufficient to describe the sign of this potential energy. The direction provided by the vector is needed.
(There is also a torque associated with the dipoles turning to tend to align with the field. The direction provided by the dipole moment determines if the turning is clockwise or counterclockwise.)


*One can also look at the field produced by the electric dipole moment.
Implied by the other comments, at a given observation point in space,
the magnitude and direction of the electric field due to that dipole
depends on more than just
...the position vector to the observation point from the center of the dipole,
...the magnitude of the charges $|Q|$
...and the separation distance $|\vec d|$.
It also depends on the direction of the dipole-moment vector
$\hat d$.... in particular, the angle between the direction of the position-vector to the observation point and the direction of the dipole moment.
A: Because if we have a system of charges whose net charge is 0 then we can replace that system of charges by an equivalent system of 2 charges that have same dipole moment as the original system.
•  Such that the electric field and potential by the two system is same.

(like here system 1 is equivalent to system 2)

*

*Now to find what the equivalent system(dipole) would be we can combine pairs of dipole in original system and place them in such ways that tail(negative charge) of one pair touches head(positive charge) of other.


*And now we can cancel the 2 charges place at a single point(as it will have no net electric field) and we would be left with only 2 charges.
The way above system is replaced is same as vector addition, hence diplole moment is defined as vector.
