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Here is a question I came across in a book:

Three point charges $-q$,$-q$ and $2q$ are placed on the vertices of an equilateral triangle of side length $d$ units.What is the dipole moment of the combination?

I had learnt that the dipole moment is defined for 2 point charges only with equal magnitudes but opposite character.I am in a bit of a fix regarding what the author means by that phrase in bold.Thank you in advance!

Edit:I just noticed theoretical physics has been merged here;I sincerely apologise if the question is unbearably stupid.

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2 Answers

up vote 6 down vote accepted

Dipole moment is a vector and can be calculated using formula

$$\vec{p} = \sum_i q_i \vec{r}_i.$$

It can be shown easily using the formula above that in case of two charges separated by distance $d$

$$\vec{p} = q \vec{d},$$

where vector $\vec{d}$ goes starts at negative ends at positive charge.

http://en.wikipedia.org/wiki/Electric_dipole_moment#Dipole_moment_density_and_polarization_density

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I had learnt that the dipole moment is defined for 2 point charges only with equal magnitudes but opposite character.

Actually, that is not the case. You can calculate a dipole moment for any charge distribution.

In fact, the dipole moment is just one of a whole series of multipole moments which can be used to describe a charge distribution.

  • The monopole moment $$q = \sum_i q_i$$
  • The dipole moment $$\vec{p} = \sum_i q_i \vec{r}_i$$
  • The quadrupole moment $$\overset{\leftrightarrow}{Q} = \sum_i q_i(3 \vec{r}_i \otimes \vec{r}_i - r_i^2)$$

and so on. These moments are kind of analogous to the coefficients of a Taylor series expansion.

It turns out that if you take a positive point charge and a negative point charge of equal magnitude and bring them infinitesimally close together, then the resulting charge distribution (called a dipole) has only a dipole moment. All the other multipole moments are zero. That's where the dipole moment gets its name. You've been taught that you can calculate a dipole moment for a system of two charges because that is the simplest way to create a charge distribution with a dipole moment. But it is not the only way.

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