Question

A thin insulating rod, running from z =-a to z=+a ,carries the

indicated line charges. In each case, find the leading term in the multi-pole expansion of the potential: $\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}\mathbf{\lambda }\mathbf{=}\mathbf{k}\mathbf{}\mathbf{cos}\left(\mathrm{\pi z}/2a\right)\mathbf{,}\mathbf{}\left(b\right)\mathbf{}\mathbf{\lambda }\mathbf{=}\mathbf{k}\mathbf{}\mathbf{sin}\left(\mathrm{\pi z}/a\right)\mathbf{,}\mathbf{}\left(c\right)\mathbf{}\mathbf{\lambda }\mathbf{=}\mathbf{k}\mathbf{}\mathbf{cos}\mathbf{}\left(\mathrm{\pi z}/a\right)\mathbf{,}\mathbf{}\mathbf{where}\mathbf{}\mathbf{k}\mathbf{}\mathbf{is}\mathbf{}\mathbf{a}\mathbf{}\mathbf{constant}\mathbf{.}$

Short answer

$$\begin{gathered} \mathrm{a})\mathrm{The}\mathrm{potential}\mathrm{due}\mathrm{to}\mathrm{monopole}\mathrm{is}\mathrm{V}\left(\mathrm{r},\mathrm{\theta }\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{4\mathrm{aK}}{\mathrm{\pi }}\right)\frac{1}{\mathrm{r}}. \\ \mathrm{b})\mathrm{The}\mathrm{potential}\mathrm{due}\mathrm{to}\mathrm{a}\mathrm{dipoleis}\mathrm{V}\left(\mathrm{r},\mathrm{\theta }\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{2{\mathrm{a}}^2\mathrm{K}}{\mathrm{\pi }}\right)\frac{1}{{\mathrm{r}}^2}\cos \mathrm{\theta }. \\ \mathrm{c})\mathrm{The}\mathrm{potential}\mathrm{due}\mathrm{to}\mathrm{Quadra}\mathrm{pole}\mathrm{is}\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{{\mathrm{r}}^3}\left(\frac{3{\cos }^2\mathrm{\theta }-1}{2}\right)\left(\frac{-4{\mathrm{a}}^3\mathrm{K}}{{\mathrm{\pi }}^2}\right) \end{gathered}$$

Step-by-step solution

Define function

Write the expression for potential at a distance r,

$\mathbf{V}\left(r,\theta \right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{P}}_{\mathbf{n}}\left(\mathrm{cos\theta }\right)}{{\mathbf{r}}^{\mathbf{n}\mathbf{+}\mathbf{1}}}{\mathbf{\int }}_{\mathbf{-}\mathbf{a}}^{\mathbf{+}\mathbf{a}}{\mathbf{z}}^{\mathbf{n}}\mathbf{\lambda }\left(z\right)\mathbf{dZ}$ …… (1)

This is for an insulating rod running from z=-a to z=+a.

Determine λ=K cos(πz/2a)

a)

Consider the integral, from equation (1),

$$\begin{gathered} {\mathrm{I}}_{\mathrm{n}}={\int }_{-\mathrm{a}}^{+\mathrm{a}}{\mathrm{Z}}^{\mathrm{n}}\mathrm{\lambda }\left(\mathrm{z}\right)\mathrm{dZ} \\ ={\int }_{-\mathrm{a}}^{+\mathrm{a}}{\mathrm{Z}}^{\mathrm{n}}\mathrm{K}\cos \left(\mathrm{\pi z}/2\mathrm{a}\right)\mathrm{dZ} \\ \mathrm{If}\mathrm{n}=0,\mathrm{then} \\ {\mathrm{I}}_0=\mathrm{K}{\int }_{-\mathrm{a}}^{+\mathrm{a}}{\mathrm{Z}}^{\left(0\right)}\mathrm{K}\cos \left(\mathrm{\pi z}/2\mathrm{a}\right)\mathrm{dZ} \\ {\mathrm{I}}_0={\left[\frac{2\mathrm{a}}{\mathrm{\pi }}\left(\sin \frac{\mathrm{\pi z}}{2\mathrm{a}}\right)\right]}_{-\mathrm{a}}^{+\mathrm{a}} \\ =\frac{2\mathrm{aK}}{\mathrm{\pi }}\left[\sin \left(\frac{\mathrm{\pi }}{2}\right)-\sin \left(\frac{-\mathrm{\pi }}{2}\right)\right] \\ =\frac{4\mathrm{aK}}{\mathrm{\pi }} \\ \mathrm{If}\mathrm{put}\mathrm{n}=0,\mathrm{then}\mathrm{potential}\mathrm{due}\mathrm{to}\mathrm{a}\mathrm{monopole}, \\ \mathrm{v}\left(\mathrm{r},\mathrm{\theta }\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{P}}_0\left(\mathrm{cos\theta }\right)}{{\mathrm{r}}^{0+1}}\left(\frac{4\mathrm{aK}}{\mathrm{\pi }}\right) \\ \mathrm{Thus},\mathrm{potential}\mathrm{due}\mathrm{to}\mathrm{monopole}, \\ \mathrm{v}\left(\mathrm{r},0\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{4\mathrm{aK}}{\mathrm{\pi }}\right)\frac{1}{\mathrm{r}} \end{gathered}$$

Determine λ=k sin (πz/a)

$$\begin{gathered} \mathrm{b}) \\ \mathrm{Consider}\mathrm{the}\mathrm{integral}\mathrm{part}, \\ {\mathrm{I}}_{\mathrm{n}}={\int }_{-\mathrm{a}}^{+\mathrm{a}}{\mathrm{z}}^{\mathrm{n}}\mathrm{\lambda }\left(\mathrm{z}\right)\mathrm{dz} \\ {\mathrm{I}}_{\mathrm{n}}={\int }_{-\mathrm{a}}^{+\mathrm{a}}{\mathrm{z}}^{\mathrm{n}}\mathrm{K}\sin \left(\frac{\mathrm{\pi z}}{\mathrm{a}}\right)\mathrm{dZ} \\ \mathrm{If}\mathrm{n}=0,\mathrm{then} \\ {\mathrm{I}}_0={\int }_{-\mathrm{a}}^{+\mathrm{a}}{\mathrm{z}}^{\left(0\right)}\mathrm{K}\sin \left(\frac{\mathrm{\pi z}}{\mathrm{a}}\right)\mathrm{dZ} \\ =-\frac{\mathrm{a}}{\mathrm{\pi }}{\left(\cos \frac{\mathrm{\pi z}}{\mathrm{a}}\right)}_{-\mathrm{a}}^{+\mathrm{a}} \\ =0 \\ \mathrm{If}\mathrm{n}=1,\mathrm{then} \\ {\mathrm{I}}_1={\int }_{-\mathrm{a}}^{+\mathrm{a}}{\mathrm{z}}^{\left(1\right)}\mathrm{K}\sin \left(\frac{\mathrm{\pi z}}{\mathrm{a}}\right)\mathrm{dZ} \\ \mathrm{Using}\mathrm{integration}\mathrm{by}\mathrm{parts}\mathrm{method}, \\ {\mathrm{I}}_1=\mathrm{K}{\left\{{\left(\frac{\mathrm{a}}{\mathrm{\pi }}\right)}^2\sin \left(\frac{\mathrm{\pi }2}{\mathrm{a}}\right)-\frac{\mathrm{aZ}}{\mathrm{\pi }}\cos \left(\frac{\mathrm{\pi Z}}{\mathrm{a}}\right)\right\}}_{-\mathrm{a}}^{+\mathrm{a}} \\ =\mathrm{K}\left\{\left[{\left(\frac{\mathrm{a}}{\mathrm{\pi }}\right)}^2\sin \left(\mathrm{\pi }\right)-\sin \left(-\mathrm{\pi }\right)\right]\right\} \\ =\mathrm{K}\left\{\left[{\left(\frac{\mathrm{a}}{\mathrm{\pi }}\right)}^2\sin \left(\mathrm{\pi }\right)-\sin \left(-\mathrm{\pi }\right)\right]-\frac{{\mathrm{a}}^2}{\mathrm{\pi }}\mathrm{cos\pi }-\frac{{\mathrm{a}}^2}{\mathrm{\pi }}\cos \left(-\mathrm{\pi }\right)\right\} \\ =\mathrm{K}\left[0+\frac{{\mathrm{a}}^2}{\mathrm{\pi }}+\frac{{\mathrm{a}}^2}{\mathrm{\pi }}\right] \\ {\mathrm{I}}_1=2\mathrm{k}\frac{{\mathrm{a}}^2}{\mathrm{\pi }} \\ \mathrm{If}\mathrm{n}=1,\mathrm{potential}\mathrm{due}\mathrm{to}\mathrm{a}\mathrm{dipole}\mathrm{then}, \\ \mathrm{V}\left(\mathrm{r},\mathrm{\theta }\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{2{\mathrm{a}}^2\mathrm{K}}{\mathrm{\pi }}\frac{1}{{\mathrm{r}}^2}\mathrm{cos\theta } \end{gathered}$$

Determine λ=K cos(πz/a)

c)

Put n=0 in potential function,

$$\begin{gathered} {\mathrm{I}}_0={\int }_{-\mathrm{a}}^{+\mathrm{a}}{\mathrm{Z}}^{\left(0\right)}\mathrm{K}\cos \left(\frac{\mathrm{\pi z}}{\mathrm{a}}\right)\mathrm{dZ} \\ {\mathrm{I}}_0=\mathrm{K}\frac{\mathrm{a}}{\mathrm{\pi }}{\left(\sin \frac{\mathrm{\pi z}}{\mathrm{a}}\right)}_{-\mathrm{a}}^{+\mathrm{a}} \\ =\frac{\mathrm{Ka}}{\mathrm{\pi }}\left[\mathrm{sin\pi }-\sin \left(-\mathrm{\pi }\right)\right] \\ =0 \end{gathered}$$

Put n=1 ,

${\mathrm{I}}_1={\int }_{-\mathrm{a}}^{+\mathrm{a}}\mathrm{zcos}\left(\frac{\mathrm{\pi z}}{\mathrm{a}}\right)\mathrm{dZ}$

By integration by parts,

$$\begin{gathered} I_1={\left[\mathrm{Z}\left(\frac{\mathrm{a}}{\mathrm{\pi }}\right)\sin \frac{\mathrm{\pi Z}}{\mathrm{a}}\right]}_{-\mathrm{a}}^{+\mathrm{a}}-\frac{\mathrm{a}}{\mathrm{\pi }}{\int }_{-\mathrm{a}}^{+\mathrm{a}}\sin \left(\frac{\mathrm{\pi Z}}{\mathrm{a}}\right)\mathrm{dZ} \\ =\left[\frac{{\mathrm{a}}^2}{\mathrm{\pi }}\mathrm{sin\pi }+\frac{{\mathrm{a}}^2}{\mathrm{\pi }}\sin \left(-\mathrm{\pi }\right)\right]+{\left(\frac{\mathrm{a}}{\mathrm{\pi }}\right)}^2{\left(\cos \frac{\mathrm{\pi z}}{\mathrm{a}}\right)}_{-\mathrm{a}}^{+\mathrm{a}} \\ =0+{\left(\frac{\mathrm{a}}{\mathrm{\pi }}\right)}^2\left[\mathrm{cos\pi }-\cos \left(-\mathrm{\pi }\right)\right] \\ =0 \\ \mathrm{put}\mathrm{n}=2, \\ {\mathrm{I}}_2=\mathrm{K}{\int }_{-\mathrm{a}}^{+\mathrm{a}}{\mathrm{Z}}^2\cos \left(\frac{\mathrm{\pi Z}}{\mathrm{a}}\right)\mathrm{dZ} \\ =\mathrm{K}{\left\{\frac{2\mathrm{z}\cos \left(\mathrm{\pi z}/\mathrm{a}\right)}{\left(\mathrm{\pi }/{\mathrm{a}}^2\right)}+\frac{\left(\mathrm{\pi z}/{\mathrm{a}}^2\right)-2\sin \left(\mathrm{\pi z}/\mathrm{a}\right)}{{\left(\mathrm{\pi }/\mathrm{a}\right)}^3}\right\}}_{-\mathrm{a}}^{+\mathrm{a}} \\ =2\mathrm{K}{\left(\frac{\mathrm{a}}{\mathrm{\pi }}\right)}^2\mathrm{a}\left(\cos \mathrm{\pi }+\cos \left(-\mathrm{\pi }\right)\right) \\ {\mathrm{I}}_2=\frac{-4{\mathrm{Ka}}^3}{{\mathrm{\pi }}^2} \end{gathered}$$

Then put n=2 , get the potential due to Quadra pole

$$\begin{gathered} \mathrm{V}\left(\mathrm{r},\mathrm{\theta }\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{{\mathrm{r}}^3}{\mathrm{P}}_2\left(\mathrm{cos\theta }\right){\mathrm{I}}_2 \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{{\mathrm{r}}^3}\left(\frac{3{\cos }^2\mathrm{\theta }-1}{2}\right)\left(\frac{-4{\mathrm{a}}^3\mathrm{K}}{{\mathrm{\pi }}^2}\right) \\ \mathrm{Therefore},\mathrm{The}\mathrm{potential}\mathrm{due}\mathrm{to}\mathrm{Quadra}\mathrm{pole}\mathrm{is}\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{{\mathrm{r}}^3}\left(\frac{3{\cos }^2\mathrm{\theta }-1}{2}\right)\left(\frac{-4{\mathrm{a}}^3\mathrm{K}}{{\mathrm{\pi }}^2}\right) \end{gathered}$$