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We know that due to expansion, the Quadrupole potential equals $$1/4\pi \epsilon r^3 . \int (r^\prime)^2(\frac32. \cos^2\theta^\prime-\frac12)\rho(r^\prime)d\tau$$

but what is the equation for point charges? for example, we have two charge $q$ and one charge $-2q$ with defined locations.

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  • $\begingroup$ what is $\rho$? $\endgroup$ – Sanya Jul 1 '16 at 20:40
  • $\begingroup$ ρ is the volumetric charge density $\endgroup$ – M.H.Muhamadi Jul 1 '16 at 20:50
  • $\begingroup$ and how does $\rho$ look like for your example? $\endgroup$ – Sanya Jul 1 '16 at 20:56
  • $\begingroup$ imagine the the 2q is at "0", one of the q charges is at "+l" and the other q is at "-l" and they're all at one line $\endgroup$ – M.H.Muhamadi Jul 1 '16 at 21:13
  • $\begingroup$ Go here. What you are describing is called a linear electric quadrupole. $\endgroup$ – Prasad Mani Feb 28 '17 at 15:23
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You're trying to understand how a point charge is represented in the charge density $\rho(r)$.

The concept you need is the Dirac delta function $\delta(r)$, which can describe the density of a finite amount of stuff packed into an infinitesimal point:

$$\delta(r) = \left\{ ^{\infty \text{ if } r=0 }_{0 \text{ if } r\ne0} \right\} $$

$$\int_{-\infty}^{+\infty} \delta(x) dx = 1$$

And it has the following really useful property:

$$\int_{-\infty}^{+\infty} f(x)\delta(x-a) dx = f(a)$$

And in your case the the charge density could be written

$$\rho(r') = q\delta(r'-r_1) + q\delta(r'-r_2) - 2q\delta(r'-r_3)$$

Where $r_i$ is the location of the $i$th charge. The useful property above makes it easy to calculate your integral.

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