1
$\begingroup$

When I read about multipole expansion in Griffith book, i saw that "if total charge is zero, dipole moment is independent of origin." I have some doubts about this.

  1. I learned that the dipoles are two equal and opposite charges separated by distance. So, can we form dipole in which charges are of different magnitude?

  2. In multipole expansion, I read that if charge shifted from origin, there will form a dipole moment also. Here, there is only one charge, right? Then how it form a dipole moment? Or my understanding of dipole moment is wrong? Or is this usual dipole moment and multipole expansion is different cases?

Reference: Griffith electrodynamics,page no:151,origin of coordinates in multipole expansion

$\endgroup$

3 Answers 3

3
$\begingroup$

The key point is to make a distinction between the notions of "dipole" and "dipole moment". The latter is not the same thing as the presence of a dipole. In fact, given what is said regarding the dependence of dipole moment on the origin, it turns out you can have a nonzero dipole moment when there is only one point charge present - just take the origin as anywhere but the exact location of that point charge.

A dipole is, by definition, two separate and equal but opposite point charges. So no, you cannot have a dipole with unequal charges, by definition.

But the dipole moment refers to a specific term in an asymptotic expansion of the electrical potential at large distance, i.e. in mathematician's terms, the "expansion at infinity". It is called as such because a dipole always has a nonzero dipole moment (which is, as said in the book, also independent of the choice of origin) and, moreover, a "pure" dipole moment - i.e. where this term only is taken in the expansion, the potential described is that of an "ideal" dipole, meaning one in which we take the mathematical limit as the charge separation goes to zero while the charge magnitude goes to infinity, with both of these balanced against each other "just so" to end up in a finite and nonzero electric field configuration. Thus the term measures "how much 'ideal dipoleness'" there is in the field configuration in question. The more it dominates other terms, the more the field looks like an ideal dipole, at least very far away from the center.

To see that there is no need at all to have a dipole to have a dipole moment, simply note that the dipole moment of $N$ point charges is just

$$\mathbf{p} := \sum_{i=1}^{N} q_i \mathbf{r}_i$$

where $\mathbf{r}_i$ are the charge positions. For a solitary point charge, this is just

$$\mathbf{p} = q\mathbf{r}$$

and thus one charge anywhere but the origin, i.e. $\mathbf{r} \ne \mathbf{0}$, has a nonzero dipole moment, yet is quite obviously about as far from being a "dipole", as just defined, as you can possibly get. The reason that a dipole - i.e. without "moment" after it - has a dipole moment independent of the origin is that if you take the above formula with two charges, i.e. $N = 2$, and equal and opposite, i.e. $q_1 = q_2 = q$, you get that the dipole moment becomes $q(\mathbf{r}_1 - \mathbf{r}_2)$, and while a position vector depends on the choice of origin, the difference of two such does not, because effectively, it is measuring the directed "distance" between two points, and intuitively, simply moving your reference point doesn't make things get closer or further apart from each other! If the two charges are slightly unbalanced, i.e. the scenario you are thinking of, you can think of the resulting dipole moment as this term for the mean charge, plus some unbalancing terms that depend on the position vectors individually, and so upon the origin. The dipole moment will never be outright zero in this case, and it will have a minimum, but it will also not be constant with respect to changes in the origin. Moreover, with regard to what we said about the asymptotic expansion, there will be other terms than just the dipole moment present in these cases - in fact, even for a "non-ideal" dipole - which account for the differences from the ideal dipole situation and, given the strong differences in field, these terms will be comparable if not dominant contributors in terms of their associated coefficients.

$\endgroup$
0
$\begingroup$

The dipole moment definition is ${\bf{d}}=q\bf{r}$ for one charge. For many charges you add them together.

$\endgroup$
0
$\begingroup$

1) A dipole is by definition two equal and opposite charges held a fixed distance apart. What's wrong in your understanding is the following. Any charge distribution ( or system of charges) can have a dipole moment. In addition to the potential due to the dipole moment , an arbitrary charge distribution also has potential contributions due to other terms in the multipole expansion( like the monopole, quadruapole etc.) . Even a dipole system has potential contributions due to these terms except when the dipole is a perfect dipole ( which si the artificial situation that the separation between the two charges becomes 0 and the charges become infinte), in which case only the dipole potential exists and the other terms( in the multipole expansion) become zero.
2) A system of a single point charge away from the origin not only has a dipole moment ( and thus a potential due to the dipole moment) but also all the other terms in the multipole expansion as said earlier. So, a charge distribution need not be a dipole to have have a dipole moment nor does it need to be a quadrupole to have a potential contribution from the quadrapole term.

$\endgroup$
2
  • $\begingroup$ So,Can a single charge form dipole?it need two charges ..right?,then how a single charge distent from origin will get dipole moment,(because P=qd,d is distemce between two charge and here is only one charge..then what is d ? $\endgroup$
    – Sajeer
    Commented Jan 25, 2019 at 9:11
  • $\begingroup$ A single charge has a dipole moment given by the formula d=qr ( which becomes zero if the charge is at the origin as r=0 then) but it is not called a dipole. A charge distribution need not be a dipole to have a dipole moment. The nomenclature is a bit confusing but that's how it is. $\endgroup$
    – D.Mason
    Commented Jan 25, 2019 at 9:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.