Cheenta
How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
Learn More

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 $$ V(r,\theta)=\frac{1}{4\pi\epsilon_0}(4ak/pi)$$
.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

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