December 24, 2017

Multipole Expansion of a Potential

Q. An insulating rod running from (z=-a to z=a) carries indicated line charges. Determine the leading term in the multipole expansion of the potential is given by $$ \lambda=kcos(\frac{\pi z}{2a})$$ where k is a constant.

Solution:
Potential due to multipole expansion is given by $$ V(\vec{r})=\frac{1}{4\pi\epsilon_0}\sum_{0}^{\inf} \frac{P_n(cos\theta)}{r^{n+1}}I_n$$ where (I_n=\int_{-a}^{+a}z^n\lambda(z)dz)
Inserting the value of charge density in the integral to obtain: $$I_0=k\int_{-a}^{a} cos(\frac{\pi z}{2a})dz=\frac{2ak}{\pi}sin(\frac{\pi z}{2a}$$
Applying limits (z=-a to z=a) $$ I_0= \frac{2ak}{\pi}[sin (\pi/2)+sin(\pi/2)]=\frac{4ak}{\pi}$$
The potential $$ V(r,\theta)=\frac{1}{4\pi\epsilon_0}(4ak/pi)$$
.

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com