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I've a charge distribution, whose potential will be given as $$\begin{aligned} V(\mathbf{r})=& \frac{1}{4 \pi \epsilon_{0}}\left[\frac{1}{r} \int \rho\left(\mathbf{r}^{\prime}\right) d \tau^{\prime}+\frac{1}{r^{2}} \int r^{\prime} \cos \alpha \rho\left(\mathbf{r}^{\prime}\right) d \tau^{\prime}\right.\\ &+\frac{1}{r^{3}} \int\left(r^{\prime}\right)^{2}\left(\frac{3}{2} \cos ^{2} \alpha-\frac{1}{2}\right) \rho\left(\mathbf{r}^{\prime}\right) d \tau^{\prime}+\ldots \end{aligned}$$

If the above charge distribution is replaced by a dipole having two discrete charges distanced by infinitesimal length the potential reduces in the infinitesimal length limit to $$V_{\mathrm{dip}}(\mathbf{r})=\frac{1}{4 \pi \epsilon_{0}} \frac{\mathbf{p} \cdot \hat{\mathbf{r}}}{r^{2}}$$

Is there a way in which I can have a charge distribution however small, whose potential also reduces to the above limit? Can an infinitesimal volume of charge ( neutral overall so that the monople term is zero) also have the same potential given by the above equation by arranging the distribution such that all the higher terms in the multipole expansion vanish and all that's left is the dipole term?

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  • $\begingroup$ Isn't what you are asking for exactly the ideal dipole? (two opposite charges infinitesimally close to each other?) In that case, it is impossible to experimentally realize, but there is no real theoretical problem with it (only as much as comes with the Dirac delta function). $\endgroup$ – Rahul Arvind Apr 18 at 9:38
  • $\begingroup$ I'm asking that can an infinitesimal blobs' potential have only the dipole term as its multipole expansion ? Even if we let it's volume go to zero it will not be like the ideal point dipole. $\endgroup$ – Kashmiri Apr 19 at 7:09
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Your first equation describes a multipole expansion. The dipole term is the second term (with the $1/r^2$ factor) in this equation. You get this term without the others (i.e. your second equation) only if you have two equal charges of opposite polarity (that is an electric dipole). If your neutral charge distribution consists of more charges you would get the higher orders in your first equation as well, e.g. quadrupole, octupole terms and so on. So your second equation would then not apply anymore. It only holds for two charges. See this link for more details.

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  • $\begingroup$ Cannot we arrange so that the higher terms vanish? $\endgroup$ – Kashmiri Apr 19 at 7:02
  • $\begingroup$ @Kashmiri You could in principle arrange it this way: you have one dipole ( 1positive+ 1 negative charge) and you arrange the other charges spherically symmetrically around it. From the outside you would then only see the dipole. But you would need a continuous charge distribution for this, which does not exist in reality. So you will always have some small higher terms for any finite number of charges. $\endgroup$ – Thomas Apr 23 at 19:50

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