Why is the direction of the dipole moment always taken from -q to +q?

[ananthu]

Homework Statement

Can anyone explain why the direction of the dipole moment of an electric dipole is always taken as "from -q to +q" but not "from +q to -q"? In fact when we draw the electric lines of force we are only drawing in such a way that they start from +q and terminate at -q.Then why this contradiction? What is the correct explanation for this convention?

Homework Equations

The Attempt at a Solution

 

[Doc Al]

Can anyone explain why the direction of the dipole moment of an electric dipole is always taken as "from -q to +q" but not "from +q to -q"?

It's just a convention.

In fact when we draw the electric lines of force we are only drawing in such a way that they start from +q and terminate at -q.Then why this contradiction?

What does that have to do with the definition of dipole moment? There's no contradiction.

 

[tiny-tim]

But why is the convention for a dipole the opposite to the other convention?

 

[Doc Al]

But why is the convention for a dipole the opposite to the other convention?

Ah... now I understand the question. The "natural" definition of dipole moment (the first moment of the charge distribution) is:

\vec{p} = q_1\vec{r}_1 + q_2\vec{r}_2

That will give the direction of the dipole moment as minus to plus.

 

[rl.bhat]

When you keep a dipole in an electric field, it acquires the stable equilibrium position with positive charge toward the electric field. Potential energy for a dipole is given by
U = - p.E
It has minimum value = -pE at the stable equilibrium position. It is possible only when p is parallel to E, i.e. p is from -q to +q.

 

[Doc Al]

When you keep a dipole in an electric field, it acquires the stable equilibrium position with positive charge toward the electric field. Potential energy for a dipole is given by
U = - p.E
It has minimum value = -pE at the stable equilibrium position. It is possible only when p is parallel to E, i.e. p is from -q to +q.

But if you defined the dipole moment with the opposite convention, U = p.E. And the minimum value would be when p is anti-parallel to E. The physics wouldn't change. (Not that I'm suggesting one flout convention. )

 

[ananthu]

Ah... now I understand the question. The "natural" definition of dipole moment (the first moment of the charge distribution) is:

\vec{p} = q_1\vec{r}_1 + q_2\vec{r}_2

That will give the direction of the dipole moment as minus to plus.

Will you please elaborate this point?

 

[Doc Al]

Will you please elaborate this point?

I'll try. Let q1 = +q and q2 = -q, then:

\vec{p} = q_1\vec{r}_1 + q_2\vec{r}_2 = q\vec{r}_1 - q\vec{r}_2 = q(\vec{r}_1 - \vec{r}_2)

The vectors r1 and r2 are the position vectors of +q and -q. Thus the vector r1 - r2 points from -q to +q.

 

[ananthu]

I'll try. Let q1 = +q and q2 = -q, then:

\vec{p} = q_1\vec{r}_1 + q_2\vec{r}_2 = q\vec{r}_1 - q\vec{r}_2 = q(\vec{r}_1 - \vec{r}_2)

The vectors r1 and r2 are the position vectors of +q and -q. Thus the vector r1 - r2 points from -q to +q.

Thank you. Now it is clear.