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.

  • $\begingroup$ what is $\rho$? $\endgroup$
    – Sanya
    Jul 1 '16 at 20:40
  • $\begingroup$ ρ is the volumetric charge density $\endgroup$ 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 $\endgroup$ Jul 1 '16 at 21:13
  • $\begingroup$ Go here. What you are describing is called a linear electric quadrupole. $\endgroup$ Feb 28 '17 at 15:23

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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