a) There's a radially symmetric charge density $\rho(r)$ centered around the origin. Determine the dipolemoment of that charge density.

b) Let $\rho(r)$ be an arbitrary charge density now. Under what circumstances does the dipole moment of the displaced charge density $\rho '(\vec{r}) = \rho (\vec{r}-\vec{b})$ differ from the one not displaced at all.

Here were my ideas so far:

a) Just thinking about the situation it has to be zero, right? I mean, since there's no real dipole. But how do I show that mathematically?

I was thinking of just going like this (it may be wrong):

Let the charge density be $\rho (r)=kr$, then we can get the charge q by integrating:

$$q=4\pi \int_0^R kr\cdot r^2dr=\pi k R^4$$

I'm looking at the charge distribution as a spherical electron cloud with radius $R$.

Then, since $p=qd$ and $d$ is zero because there are no two different charges the dipole moment is zero. Is that sufficient as an answer?

b) I don't know how to approach this one. My guess is that its dipole moment is also zero because we're only looking at a displacement here.

Anyone got any idea? I would appreciate any advice on this.


closed as off-topic by ACuriousMind, user36790, Gert, Kyle Kanos, HDE 226868 Nov 8 '15 at 22:24

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Remember, the dipole is a vector. So, its not simply $p = qd$.

For a general charge distribution $\rho(\mathbf r')$, you need the multipole expansion of potential in spherical coordinates, for powers of $1/r$. Meaning, take the potential, make expansion of $1/r$ powers. The $1/r$ term is the monopole. The $1/r^2$ term is the dipole. The $1/r^3$ term is the quadrupole. And so on.

From this you will get the real dipole definition: $$ \mathbf p = \iiint_V\mathbf r'\rho(\mathbf r')dV $$

Check if your calculations arrives at this answer. Now, for the problem (b), its simple once you have this expression. Just make a translation in the coordinate axis in the expression of the dipole moment, and it will give you the answer without effort. I hope this was helpful. Any questions, let me know.

  • $\begingroup$ Sorry for my late response. There is construction work in my neighborhood and for some reason my internet connection has been off for a couple of days. I have tried solving the integral you have given (in spherical coordinates it's $\iiint r'\rho(r') r'^2 \sin (\theta) drd\theta d\phi$) but I don't get zero. I have never dealt with dipoles in such a manner before so I don't know how to do it properly. I tried finding an example of the multipole expansion in my workbook to work by it but there wasn't one. $\endgroup$ – Rafa Fafa Nov 8 '15 at 7:53

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