In my book dipole moment is defined for two charges.

But I got a hw question to calculate dipole moment due to a ring.

enter image description here

I tried two different methods to solve this problem but I don't think that any of them is correct.

I sticked to the definition given in my book for 2 charges,"An electric dipole is a pair of equal and opposite point charges separated by a distance 2a."

My solution,

Small charge $dq$ = $λRdθ$.

Now by geometry distance between any two points, $2a$ = $2Rcosθ$ as shown in the image.

enter image description here

Now $dp$ = $2adq$

Integrated it from $-π/2$ to $π/2$ and got option A as the answer.

Net Dipole moment will clearly be in the $+x$ direction

Now second method,

Consider any two diametrically opposite points, by symmetry dipole moment will be in the $+x$ direction as it will be cancelled in the $y$ direction.

Now since they are diametrically opposite so $2a$ = $2R$

Hence, $dp$ = $2Rdqcosθ$ in the $+x$ direction.

But I don't think any of my method is correct as each charge will be under the influence of all the opposite charges so I can't actually decide what will $2a$ be.

If the points are diametrically opposite then it will be $2R$ but the distance can be anything as we can choose any two pair of opposite charge so the distance between the opposite charges can be anything from $0$ to $2R$.


Both the ways you did is right, and if you do the math properly you should get option A in both the case.

As for you latter question on what the distance should be,

As you have defined, you have to take distance between equal and opposite charges. In our situation, there are not many ways, but precisely two ways of choosing the equal and opposite charges.

  1. Diametrically opposite: For a charge at $\theta$ we have charge per unit length at that point to be $\lambda=\lambda_{0}\cos\theta$, and for a diametrically opposite point ($180^o +\theta$), we have $\lambda=\lambda_{0}\cos(180^o + \theta)=-\lambda_{0}\cos(\theta)$. Hence these two does form a dipole (equal and opposite) with dipole lenght $2R$.

  2. Mirror about y-axis: For a charge at $\theta$ we have charge per unit length at that point to be $\lambda=\lambda_{0}\cos\theta$, and for a point mirrored about y-axis ($180^o -\theta$), we have $\lambda=\lambda_{0}\cos(180^o - \theta)=-\lambda_{0}\cos(\theta)$. Hence these two does form a dipole (equal and opposite) with dipole length $2R \cos(\theta)$.

And, as you have already done, either way you should get the same result, as it should be.

The way you are thinking which is that each charge interacts with each other is the monopole way of going about it, which is perfectly valid. In fact on derives the formula for dipole ($p=\sum q_i d_i$) is using Coulombs Law on the system of charges. The dipolar, or for that matter any multipolar system is a very special kind of a system, in that it has some symmetry (in case of dipole its equal and opposite). Due to this very symmetry the Coulomb law takes a special form which is when you ignore the constants and call the numerator as a dipole charge.

Keeping it very simple and not going into details, what I mean to say is that your worry that each charge is interacting with all others is already taken care of by the symmetry of the system. For a general system, this particular way is not exactly valid and so we do a multipolar expansion.

  • 1
    $\begingroup$ Thankyou, I was really anxious about this problem from so many days but I got it now. Also, Could you please elaborate a little bit or share a link on monopole method? I've 3 books with me, University Physics, my school book and Concepts of Physics but nothing is mentioned about this monopole method in those books and neither any example is present to calculate dipole moment of continuous charge distribution. $\endgroup$ Oct 1 '21 at 20:31
  • $\begingroup$ I am not aware of many high school text, but any standard Electromagnetism book at undergraduate/graduate level should have a fairly detailed explanation. Why don't you look up the Wikipedia page here. The idea is straight forward. If you have just a single charge, its just standard Coulomb's at whichever scale you decides to look. But things get interesting when you have more than one. For ease of calculation as well as quite a nice physical interpretation, we prefer to break down the net effects to sum of smaller once. $\endgroup$ Oct 2 '21 at 4:31
  • 1
    $\begingroup$ Thankyou so much. $\endgroup$ Oct 2 '21 at 4:35

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