# Multipole expansion sphere

I saw the solution on griffiths, I didnt understand why in the monopole term they integrate from 0 to r, and then in the dipole they integrate on theta from 0 to pi and in the quadrupole the integration is on phi. why integrate separately, i would have integrate r,theta and phi in each term? i dont understand the contribuitions from each term in the case of a sphere.

• Do you know how to calculate the monopole, dipole, and quadrupole moment for a general charge distribution? – probably_someone Apr 29 '20 at 2:19
• yes, i just dont get why in solution of problem, why the monopole term is only integrated from 0 to r, for example? why not integrate the term in theta and phi too? and the same for the dipole, only the quadrupole is integrated in r, phi and theta. – Ana Branco Apr 29 '20 at 2:25

I am assuming that you do know how $$d\tau$$ became $$\displaystyle r^2sin\theta \;dr\;d \theta \; d\phi$$.

And also the order of the integration. (If not check the section 1.4 Curvilinear Coordinates)

Proceeding to your question, the problem says there is a sphere,centered at origin,carries charge density $$\rho (r,\theta)= k \frac{R}{r^2}(R-2r) sin \theta$$ See here you don't know how the potential due to this term acts at a specific distance, whether it's a dipole or monopole or something else. So we use the formula provided to us in the previous page, whose sole purpose is to calculate the potential for whatsoever charge density.

$$V(r)=\displaystyle \frac{1}{4 \pi \epsilon_0} \sum^\infty_{n=0}\frac{1}{r^{n+1}}\int (r')^n P_n (cos \theta')\rho(r')d \tau'$$

Expansion

\begin{aligned} V(r) &=\frac{1}{4 \pi \varepsilon_{0}} \sum_{n=0}^{\infty} \frac{1}{r^{(n+1)}} \int\left(r^{\prime}\right)^{n} P_{n}(\cos \theta) \rho d \tau \\ &=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{1}{r} \int \rho d \tau+\frac{1}{r^{2}} \int r^{\prime} \cos \theta \rho d \tau+\frac{1}{r^{3}} \int\left(r^{\prime}\right)^{2}\left(\frac{3}{2} \cos \theta-\frac{1}{2}\right) \rho d \tau+\cdots\right] \end{aligned}

Each next term in the expansion of above equation represents what would have happened if the distribution was having one more pole.

You can check yourself by separating each term and comparing them with formulas for the monopole,dipole and quadropole in earlier classes.

Coming to the question , you don't know what "pole" your charge density represents so you put it in the equation above.

You check if it was monopole by considering only first term but it turns out while computing $$dr$$ part the whole integral goes zero. Dipole contribution to the potential is zero. So you move on to next term in the expansion considering what your charge distribution has as dipole contribution.

But again you get the integral going to zero.

Then you calculate the third term in the integral for Quadropole. You get an answer and stick to it.

Mostly we go on to two-three terms at most. It is because at larger distances we assume the distribution in focus by it's net charge and the difference in values we get by calculating only three terms and four terms is not that significant for now(college level).

Like Griffiths says successive terms tell us how to improve the approximation if greater precision is required