This question is about the equation describing the net dipole moment for an arbitrary number of point charges.

Here's what I understand:

For two point charges $q$ and $-q$ with positions $\mathbf{r_1}$ and $\mathbf{r_2}$, the dipole moment is $$\mathbf{p} = q\mathbf{r_1} +- q\mathbf{r_2} = q\mathbf{d}$$ where we define $\mathbf{d}$ to point from the negative charge to the positive charge.

It would seem to me, incorrectly, that to find the net dipole moment of a set of $N$ point charges you just add up the individual dipole moments between every pair of them. I'll call this $\mathbf{p_{wrong}}$ for obvious reasons.

$$\mathbf{p_{wrong}} = \sum_{j=1}^{N-1}\left(\sum_{k=j+1}^{N} q_j\mathbf{r_j} + q_k\mathbf{r_k} \right) $$

which,for 3 charges would be

$(q_1\mathbf{r_1} + q_2\mathbf{r_2}) + (q_1\mathbf{r_1} + q_3\mathbf{r_3}) + (q_2\mathbf{r_2} + q_3\mathbf{r_3})$

My textbook and all sources I have consulted list the below as the correct answer with very little explanation.

$$\mathbf{p_{right}} = \sum_i^N q_i\mathbf{d_i} $$

I can't for the life of me see how this is right. In this context, I don't even understand what $\mathbf{d_i}$ means. Is my expression somewhow equivalent, is $\mathbf{d_i}$ just the absolute position vector of each point charge, or am I completely out in the weeds here?

  • 1
    $\begingroup$ Look back at center of mass to get an idea. $\endgroup$
    – AHusain
    Commented Jun 23, 2016 at 2:06
  • $\begingroup$ center of mass, if my memory serves is $$\mathbf{r_{cm}} = \frac{1}{M_{tot}} \sum(m_i\mathbf{r_i})$$. So then it seems you're saying it would be $$\mathbf{p} = \sum(q_i\mathbf{r_i}$$ $\endgroup$
    – MmmQuarks
    Commented Jun 23, 2016 at 20:26

2 Answers 2


Yes, ${\bf d}_i$ simply means the absolute position vector for the $i$th point charge. It turns out that if the total charge is zero (i.e. $\sum_i q_i = 0$), then the value of this vector quantity does not depend on your choice of origin. However, this fact is not necessarily immediately obvious to a beginning student, so your textbooks should have explained why it is true.

It's better to memorize and think about that the general formula ${\bf p} := \sum_i^N q_i {\bf d}_i$, and to consider the case of two opposite point charges as a special case, than to start from the special case and try to generalize to an arbitrary number of charges.


In the expression $$\mathbf{p_{right}} = \sum_i^N q_i\mathbf{d_i} $$ you are summing dipoles and assuming that the net charge is zero so the sum is independent of the choice of reference point.
So $q_i$ is the charge on the dipole and $\mathbf{d_i} $ is the position vector of the positive charge relative to the negative charge.

If you have two charges $+q$ and $-q$ at positions $\mathbf{r_+}$ and $\mathbf{r_-}$ relative to a reference point then the dipole moment is $+q\mathbf{r_+} - q \mathbf{r_-} = +q(\mathbf{r_+} - \mathbf{r_-}) = q\mathbf{d} $, thus independent of the reference point.

The Wikipedia article Electric Dipole Moment is possibly worth consulting as is the answer to this question and the link therein provided by @EmilioPisanty where monopoles and multipoles are mentioned.

  • $\begingroup$ Here's my confusion. This is supposed to hold for an arbitrary number of point charges so when you say: "$q_I$ is the charge on the dipole and $\mathbf{d_i}$ is the position vector of the positive charge relative to the negative charge" how do I know which pair to use to calculate $\mathbf{d_i}$? Also what if there are no two charges $q$ and $q\prime$ for which $q = -q\prime$ ? What do I use as $q_i$ then? Does $\mathbf{p_{right}}$ only apply to systems of even number of point charges where every point charge has an equal and opposite point charge? $\endgroup$
    – MmmQuarks
    Commented Jun 23, 2016 at 20:22

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