# What exactly is electric dipole moment & how it works and why is electric dipole moment vector direction is from negative charge to positive charge?

What exactly is electric dipole moment & how it works/used? (don't go into deep mathematical calculations)

Please answer the question about the direction with good details. Don't just say that it's simply a convention because it's not. It surely has some deeper meanings attached to it.

I've seen a video on this recently. It said that permittivity of a medium is the resistance to build up an electric field inside that medium. The medium creates an opposing electric field when inside an external electric field reducing the neat External electric field (I knew these).

Then it said that this is very hard to work with these kind of fields that oppose each other like this. So, a trick was done & The scientists have defined polarization which is defined by vector 'P' and instead of going from positive to negative charge, Polarization goes from negative to positive charge. The opposite direction of what actually happens in the molecules. Then it said something about displacement vector (I don't understand much about this Vector)

So, my question -

1. Details about Electric Dipole Moment (significance) [I know the laws and Maths, you don't need to prove them, just tell me the significance of this & what it is anyway]

2. Is this polarization has something to do with the direction of The electric dipole moment? & if they are connected then please tell me the details about how & why they are connected? (but do not go into deep mathematical calculations)

3. My main question is Why the direction is like that? (details)

So, a good details about the whole electric dipole moment is what I need.

• You believe it or not.its all about convention. – Paul Jan 28 '15 at 10:26
• "instead of going from positive to negative charge, Polarization goes from negative to positive charge. The opposite direction of what actually happens in the molecules." What do you mean by "what actually happens"? – Steeven Aug 7 '16 at 10:07

A 'moment' here is an integral of co-ordinate over the elements of something. Momentum is the rate of change of moment of mass.

Suppose that you have a 6 kg at 5 metres. Its moment is 30 kg.m. If 3 seconds later its coordinate were at 7 metres, then its moment would be 42 kg m. In 3 seconds, it has changed by 12 kg m, and thus its momentum is 4 kg m/s.

If i were to change the origion of calculation, the 5 metres and 7 metres would become, say 10 metres and 12 metres, but the momentum would not change.

The 'dipole moment', is simply a measure of 'moment of charge', but the charges are equal and opposite. So if i have a charge of -2 C at 3 metres, and a +2 coulomb at 7 metres, the total moment of charge is -2*3 + 2*7 = 8 coulomb-metres. Because this does not change if i add 10 metres to the coordinate, it is a property of the charges themselves.

It's also a vector, so the vector of 8 C.m points in the direction of the positive charge, because that's what makes 2*7 bigger than 2*3. If i were to reverse the charge, it would be -8 C.m pointing to +infty, but 8 C.m pointing in the reversed direction.

The dipole moment is represented by $p$, and its density by $P$.

When you apply a field to a dielectric, the electrons in the atoms are moved more than the atoms themselves, setting up a dipole at atomic level. So for example, 8 C m is not produced by 2 C separated by 4 metres, but say, 20,000,000,000 C separated by 0.4 nm. In other words, the dielectric gets a very large amount of charge moving a very tiny distance.

The polarisation actually points the opposite direction to the way the charge has moved, because the dipole = negative charge * changed coordinate + positive charge * unchanged coordinate = - negative charged * changed coordinate.

When an electric field hits a dielectric, the flux of the original field is absorbed by the displaced charge. Inside the dielectric, you have lots of atoms acting like little dipoles to cancel out the previous atom's flux. So the total flux does not change.

What you do see is 'induced polarisation', which is the density of these atomic dipoles, given by $P_i + D = \epsilon E$, where $P_i$ is the induced density of polarisation (C.m/m³), and D is the flux outside of the diaelectric, (there is no 'new charge' created in the dielectric, so the flux is unchanged).

The ability for $D$ to produce $P_i$ is the susceptability $\chi_e = P_i/D = \epsilon_r -1$. That is, to produce the extra field, one requires something else than flux.