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

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