Question

A circular ring in the$\mathit{x}\mathit{y}$ plane (radius R , centered at the origin) carries a uniform line charge $\mathit{\lambda }$. Find the first three terms$\left(n=0,1,2\right)$ in the multi pole expansion for $\mathit{V}\left(r,\theta \right)$.

Short answer

The first term of the multi-pole expansion is $\frac{\mathbf{\lambda R}}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{r}}$.

The second term of the multi-pole expansion is $\mathbf{0}$.

The third term of the multi-pole expansion is $\mathbf{-}\frac{{\mathbf{\lambda R}}^{\mathbf{3}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}\left(P_2\mathrm{cos\theta }\right)$.

Step-by-step solution

Define function

Write the expression for the multi pole potential.

$V_n=\frac{1}{4\pi {\epsilon }_0}\sum _{n=0}^n\frac{1}{r^{n+1}}\int {\left(r\text{'}\right)}^n{P}_{n}cos\alpha \rho \left(r\text{'}\right)d\tau \text{'}$ …… (1)

Here, ${\epsilon }_0$is the permittivity of free space, $V_n$is the multi-pole potential, r is the constant, $r\text{'}$is the radius, $P_n$is the Legendre polynomial and $\rho$is the density.

The below diagram shows the circular ring placed in $xy$plane.

Here,$\theta$is the angle between r and z axis and$\varphi$is the angle between radius vector and y axis.

Determine expression for multi pole 

Write the expression for the surface charge density.

$\rho \left(r\text{'}\right)d\tau \text{'}=\lambda dr\text{'}d\varphi \text{'}$ …… (2)

Here,$\lambda$ is the line charge density.

Write the expression for the constant .

$r=r\sin \theta \cos \varphi \hat{x}+r\sin \theta \sin \varphi \hat{y}+r\cos \theta \hat{z}$ …… (3)

Now, write the expression for radius of the sphere.

$r\text{'}=R\cos \varphi \text{'}\hat{x}+R\sin \varphi \text{'}\hat{y}$ …… (4)

Calculate the dot product of r and $r\text{'}$.

$\begin{array}{rcl}r·r\text{'}& =& \left(r\sin \theta \cos \varphi \hat{x}+r\sin \theta \sin \varphi \hat{y}+r\cos \theta \hat{z}\right)·\left(R\cos \varphi \text{'}\hat{x}+R\sin \varphi \text{'}\hat{y}\right)\\ & =& Rr\cos \varphi \text{'}\sin \theta \cos \varphi +Rr\sin \varphi \text{'}\sin \theta \sin \varphi \\ & =& Rr\sin \theta \left(\cos \varphi \text{'}\cos \varphi +\sin \varphi \text{'}\sin \varphi \right)\\ & =& Rr\cos \alpha \\ & & \end{array}$ ……. (5)

Determine multi-pole for first term 

Now, suppose$\sin \theta \left(\cos \varphi \text{'}\cos \varphi +\sin \varphi \text{'}\sin \varphi \right)=\cos \alpha$ .

Thus, the multi-pole potential for$n=0$ i.e. the monopole is written as by substituting

for in equation (1).

$$\begin{gathered} V_n=\frac{1}{4\pi {\epsilon }_0}\sum _{n=0}^n\frac{1}{r^{n+1}}\int {\left(r\text{'}\right)}^n{P}_n\cos \alpha \rho \left(r\text{'}\right)d\tau \text{'} \\ \therefore V_0=\frac{1}{4\pi {\epsilon }_0}\int \rho \left(r\text{'}\right)d\tau \text{'} \end{gathered}$$ ……. (6)

Now substitute $\lambda dr\text{'}d\varphi \text{'}$for $\rho \left(r\text{'}\right)d\tau \text{'}$

$\begin{array}{rcl}{V}_0& =& \frac{1}{4\pi {\epsilon }_0}\int \rho \left(r\text{'}\right)d\tau \text{'}\\ & =& \frac{\lambda }{4\pi {\epsilon }_{0}r}\int dr\text{'}d\varphi \text{'}\\ & & \end{array}$

Integrate the above equation,

$\begin{array}{rcl}{V}_0& =& \frac{\lambda }{4\pi {\epsilon }_{0}r}\underset{0}{\overset{R}{\int }}dr\text{'}\underset{0}{\overset{2\pi }{\int }}d\varphi \text{'}\\ & =& \frac{\lambda }{4\pi {\epsilon }_{0}r}\left(R-0\right)\left(2\pi -0\right)\\ & =& \frac{\lambda R}{2{\epsilon }_{0}r}\\ & & \end{array}$

Thus, the first term of the multi-pole expansion is $\frac{\lambda R}{2{\epsilon }_{0}r}$.

Determine multi-pole for second term 

The multi-pole potential for $n=1$, that is quadrat-pole is written by substituting 1 for in the equation (1).

$$\begin{gathered} V_n=\frac{1}{4\pi {\epsilon }_0}\sum _{n=0}^n\frac{1}{r^{n+1}}\int {\left(r\text{'}\right)}^n{P}_n\cos \alpha \rho \left(r\text{'}\right)d\tau \text{'} \\ \therefore V_1=\frac{1}{4\pi {\epsilon }_0}\int R\cos \alpha \rho \left(r\text{'}\right)d\tau \text{'} \end{gathered}$$

Now, substitute $\lambda dr\text{'}d\varphi \text{'}$ for $\rho \left(r\text{'}\right)d\tau \text{'}$

$$\begin{gathered} V_1=\frac{1}{4\pi {\epsilon }_0}\int R\cos \alpha \rho \left(r\text{'}\right)d\tau \text{'} \\ =\frac{R\lambda }{4\pi {\epsilon }_0{r}^2}\int \cos \alpha dr\text{'}d\varphi \text{'} \end{gathered}$$

Substitute $\sin \theta \left(\cos \varphi \text{'}\cos \varphi +\sin \varphi \text{'}\sin \varphi \right)$for$\cos \alpha$in above equation.

$V_1=\frac{R\lambda }{4\pi {\epsilon }_0{r}^2}\int \sin \theta \left(\cos \varphi \text{'}\cos \varphi +\sin \varphi \text{'}\sin \varphi \right)dr\text{'}d\varphi \text{'}$

Integrate the above the equation.

$\begin{array}{rcl}{V}_1& =& \frac{R\lambda }{4\pi {\epsilon }_0{r}^2}\underset{0}{\overset{R}{\int }}dr\text{'}\underset{0}{\overset{2\pi }{\int }}\sin \theta \left(\cos \varphi \text{'}\cos \varphi +\sin \varphi \text{'}\sin \varphi \right)dr\text{'}d\varphi \text{'}\\ & =& \frac{R\lambda \sin \theta }{4\pi {\epsilon }_0{r}^2}\left(R\right)\underset{0}{\overset{2\pi }{\int }}\cos \left(\varphi \text{'}-\varphi \right)d\varphi \text{'}\\ & =& 0\\ & & \end{array}$

Here, the trigonometric identity is used, i.e$\left(\cos A\cos B+\sin A\sin B\right)=\cos \left(A-B\right)$

Thus, the second term of the multi-pole expansion is 0 .

Determine multi-pole for third term

The multi-pole potential for $n=2$, that is quadrat-pole is written by substituting 2 for in the equation (1).

$$\begin{gathered} V_n=\frac{1}{4\pi {\epsilon }_0}\sum _{n=0}^n\frac{1}{r^{n+1}}\int {\left(r\text{'}\right)}^n{P}_n\cos \alpha \rho \left(r\text{'}\right)d\tau \text{'} \\ {V}_2=\frac{1}{4\pi {\epsilon }_0{r}^3}\int R^2\left(\frac{3}{2}{\cos }^2\alpha -\frac{1}{2}\right)\rho \left(r\text{'}\right)d\tau \text{'} \end{gathered}$$

Now, substitute $\lambda dr\text{'}d\varphi \text{'}$for$\rho \left(r\text{'}\right)d\tau \text{'}$

$$\begin{gathered} V_2=\frac{1}{4\pi {\epsilon }_0{r}^3}\int R^2\left(\frac{3}{2}{\cos }^2\alpha -\frac{1}{2}\right)\rho \left(r\text{'}\right)d\tau \text{'} \\ {V}_2=\frac{1}{4\pi {\epsilon }_0{r}^3}\int R^2\left(\frac{3}{2}{\cos }^2\alpha -\frac{1}{2}\right)\lambda dr\text{'}d\varphi \text{'} \\ =\frac{\lambda }{4\pi {\epsilon }_0{r}^3}\left(\frac{R^2}{2}\right)\int \left(3{\cos }^2\alpha -1\right)dr\text{'}d\varphi \text{'} \end{gathered}$$

Substitute$\sin \theta \left(\cos \varphi \text{'}\cos \varphi +\sin \varphi \text{'}\sin \varphi \right)$ for$\cos \alpha$in above equation.

$$\begin{gathered} V_2=\frac{\lambda }{4\pi {\epsilon }_0{r}^3}\left(\frac{R^2}{2}\right)\int \left(3{\cos }^2\alpha -1\right)dr\text{'} \\ {V}_2=\frac{\lambda }{4\pi {\epsilon }_0{r}^3}\left(\frac{R^2}{2}\right)\int \left(3\left(\sin \theta {\left(\cos \varphi \text{'}\cos \varphi +\sin \varphi \text{'}\sin \varphi \right)}^2\right)-1\right)dr\text{'}d\varphi \text{'} \end{gathered}$$

Now, integrate equation,

$$\begin{gathered} \begin{array}{rcl}{V}_2& =& \frac{\lambda }{4\pi {\epsilon }_0{r}^3}\left(\frac{R^2}{2}\right)\int \left(3\left(\sin \theta {\left(\cos \varphi \text{'}\cos \varphi +\sin \varphi \text{'}\sin \varphi \right)}^2\right)-1\right)dr\text{'}d\varphi \text{'}\\ V_2& =& \frac{\lambda }{4\pi {\epsilon }_0{r}^3}\left(\frac{R^2}{2}\right)\underset{0}{\overset{R}{\int }}dr\text{'}\underset{0}{\overset{2\pi }{\int }}\int \left(3\left(\sin \theta {\left(\cos \varphi \text{'}\cos \varphi +\sin \varphi \text{'}\sin \varphi \right)}^2\right)-1\right)d\varphi \text{'}\\ & =& \frac{\lambda }{4\pi {\epsilon }_0{r}^3}\left(\frac{R^2}{2}\right){\left(r\text{'}\right)}_0^R\left[\backslash \mathrm{begingathered}3{\sin }^2\theta \left({\cos }^2\varphi \right)\underset{0}{\overset{2\pi }{\int }}{\cos }^2\varphi \text{'}d\varphi \text{'}+\backslash \mathrm{hfill} \\ {\sin }^2\varphi \underset{0}{\overset{2\pi }{\int }}{\sin }^2\varphi \text{'}d\varphi \text{'}+2\sin \varphi \cos \varphi \underset{0}{\overset{2\pi }{\int }}\left(\cos \varphi \text{'}\sin \varphi \text{'}d\varphi \text{'}\right)\backslash \mathrm{hfill} \\ \backslash \mathrm{endgathered}\right]-\underset{0}{\overset{2\pi }{\int }}d\varphi \text{'}\\ & =& \frac{\lambda R^2}{4\pi {\epsilon }_0{r}^3}\left(R\right)3{\sin }^2\theta \left(\pi {\cos }^2\varphi +\pi {\sin }^2\varphi \right)-{\left(\varphi \text{'}\right)}_0^{2\pi }\\ & =& \frac{\lambda R^3}{4\pi {\epsilon }_0{r}^3}\left(3{\sin }^2\theta \left({\cos }^2\varphi +{\sin }^2\varphi \right)-2\right)\\ & =& \frac{\lambda R^3}{4\pi {\epsilon }_0{r}^3}\left(3{\sin }^2\theta -2\right)\\ & & \end{array} \end{gathered}$$

Here, trigonometry identity is used. $\left({\cos }^{2}A+{\sin }^{2}A\right)=1$

Solve as further,

$\begin{array}{rcl}{V}_2& =& \frac{\lambda R^3}{8\pi {\epsilon }_0{r}^3}\left(3\left(1-{\cos }^2\theta \right)-2\right)\\ & =& \frac{\lambda R^3}{8\pi {\epsilon }_0{r}^3}\left(1-3{\cos }^2\theta \right)\\ & & \end{array}$ ……. (7)

Now, from the Legendre polynomial,

$\frac{1}{3}\left(4P_2\cos \theta -P_0\cos \theta \right)=\cos 2\theta$

Solve the equation,

$2P_2\cos \theta =3{\cos }^2\theta -1$

Put this equation in equation (7).

$\begin{array}{rcl}{V}_2& =& \frac{\lambda R^3}{8\pi {\epsilon }_0{r}^3}\left(1-3{\cos }^2\theta \right)\\ & =& -\frac{\lambda R^3}{8\pi {\epsilon }_0{r}^3}\left(2P_2\cos \theta \right)\\ & =& -\frac{\lambda R^3}{4\pi {\epsilon }_0{r}^3}\left(P_2\cos \theta \right)\\ & & \end{array}$

Therefore, the third term of the multi-pole expansion is $-\frac{\lambda R^3}{4\pi {\epsilon }_0{r}^3}\left(P_2\cos \theta \right)$.