# north tower is one pillar the south is the other pillar they are mag poles the 6 / 9 is 2 opposing of opp polarity vortex,s colliding creating an explosion

Imagine you have a box, and you aren’t allowed to examine the contents of the box, but you are told the contents are stationary, and you can look at the electric field outside. If you see an inverse square field, you know there’s a net charge inside the box (which you can work out using Gauss’s law). If all the charges in the box cancel each other out, there won’t be an inverse square field; the field will have to fall faster than 1/r^2. You can actually expand the field in an infinite series in powers of r. If you have a dipole made up of cancelling charges, you’ll see a 1/r^3 field at leading order, and the strength of that field will be proportional to the magnitude of the dipole moment inside the box. So, in fact, we can just define the dipole moment of the contents of the box to be the proportionality factor that gives the observed strength of the 1/r^3 field.

Therefore the answer to your question is “by definition”. If something has an electric field that falls of as an inverse cube, it is automatically a dipole, by definition.

If the dipole moment is also zero, you may get an electric field that falls off as 1/r^4—that 1/r^4 behaviour is then controlled by the quadrupole moment. If the quadrupole moment is zero, there may be an octopole moment, which gives a 1/r^5 field. And so on, ad infinitum. When you define these higher moments, you don’t define them by putting a bunch of charges together the way you do for a dipole, because the formulas get messy then. You define them in terms of the field they produce.