# Why does the electric polarisation (polarisation density) have this direction?

It can be inferred that the amount polarisation $P$ is dependent upon and proportional to the magnitude of the electric field $E$ applied:

$$P \propto E.$$

This makes sense from what we know about the nature of electric polarisation and is also supported by the well-known relationship that

$$\mathbf{P} = \epsilon_0\chi_e \mathbf{E}.\ \ \ \ \ \ \ (1)$$

However, the electric field will push positive charges in the direction of the field and negative charges against the field direction.

Convention dictates that vectors regarding electric charge must point from the positive charge to the negative, implying that

$$\hat{\mathbf{P}} \propto - \hat{\mathbf{E}},$$

which we know to not be true due to equation 1 (both $\epsilon_0$ and $\chi_e$ are always greater than 0.)

What is the justification for the vector direction that allows $\hat{\mathbf{P}}=\hat{\mathbf{E}}$ by equation 1 when the arrangement of charge suggests the opposite?

• It is simply a convention. Equation (1) BTW is only true in LIH media. The polarisation field is the amount of electric dipole moment per unit volume and is in the same direction as the dipole moments of those dipoles. It points in the same direction as and is proportional to the electric field only in linear media. – Rob Jeffries Apr 15 '16 at 16:50

The reason is simple: the convention is that the polarization vector goes from negative to positive, not the other way around.

• I also came to a similar conclusion but found it frustratingly simple. What is the physical reason for the change of convention? – Harry Smith Aug 3 '15 at 6:45
• @HarrySmith: it's not a change of convention. As far as I know, a dipole vector always goes from negative to positive. – Javier Aug 3 '15 at 11:50
• That seems odd considering that the convention of both electric current and electric field is positive to negative. Magnetic fields also follow the same idea. Do you have any sources that explicity confirm this? – Harry Smith Aug 3 '15 at 12:05
• Okay I have found sources that agree with what you say. It just seems odd to me that we'd abandon the usual positive-to-negative deal when it comes to dipoles. Is this just for convenience? – Harry Smith Aug 3 '15 at 12:07
• @HarrySmith: If you mean the fact that the electric field lines go from positive to negative, that's not a convention. That's, in a sense, the definition of positive and negative; you can't escape it. I admit that I don't know why $\mathbf{P}$ goes from negative to positive, but I guess that $\mathbf{P}$ and $\mathbf{E}$ going in the same direction is a good reason. – Javier Aug 3 '15 at 16:52

Imagine two opposite charges very very close together

In the tiny region between the charges the field goes from the positive charge to the negative charge. However if you come towards the positive charge from the opposite direction as the negative charge the field points in the opposite direction.

And if you come towards the negative charge from the opposite direction as the positive charge the field points in the opposite direction.

Why? Because both charges exert forces, but the $q/r^2$ gets weaker the farther away you are so you still point away from the positive charge and towards the negative charge.

Now if you imagine the distance between the charges as getting smaller eventually it looks like the field just comes into the dipole over where the negative charge was and jumps out the other side where the positive charge was.

That vector is the polarization vector. Check out the gif at wikipedia https://en.m.wikipedia.org/wiki/Electric_dipole_moment#/media/File:VFPt_dipole_animation_electric.gif

Now a real electric dipole has that field pointing the opposite direction in between you can approximate it with an ideal dipole which doesn't have that space between the charges.

If it helps, you can think that the vector pointing from negative to positive is your reminder that you are approximating it as if the charges were on top of each other (closer than they are really) and you changed how strong the charges were. Not because it was accurate, but because that was pretty close to the correct field when you are not very close.

You always want to know when you are approximating, so the polarization vector pointing that way can tell you that you are approximating as if there were ideal dipoles there.

• In what way does the closer charge exert a stronger force? It seems strange to me that the force should be subjective to the relative position of the observer! It might help to clarify when you're referring to the external $E$-field vector and when you're referring to the induced polarisation vector. – Harry Smith Aug 3 '15 at 6:43