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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 3.30

Let's start by recalling from Section 3.4.1: "Approximate Potentials at Large Distances" the Multipole expansion:

$$\begin{array}{}\text{(Eq. 3.95, Page 150)}& V(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{r^{n+1}}\int (r^{\prime }{)}^n{P}_n(\cos \left(\alpha \right))\rho ({\mathbf{r}}^{\prime })d{\tau }^{\prime }\end{array}$$

We need to polish this formula a bit to suit our problem better. Firstly, the charge configuration is 1-dimensional (with parameter $\varphi$):

$$\rho ({\mathbf{r}}^{\prime })d{\tau }^{\prime }\to \lambda dl=\lambda Rd{\varphi }^{\prime }$$

Where the prime distinguish the $\varphi$ from position and the ${\varphi }^{\prime }$ for the position of the charge configuration.

Our position vector, as usual, is:

$$\mathbf{r}=r\sin \left(\theta \right)\cos \left(\varphi \right)\hat{\mathbf{x}}+r\sin \left(\theta \right)\sin \left(\varphi \right)\hat{\mathbf{y}}+r\cos \left(\theta \right)\hat{\mathbf{z}}$$

The vector to the charge distribution, ${\mathbf{r}}^{\prime }$, is:

$${\mathbf{r}}^{\prime }=R\cos \left({\varphi }^{\prime }\right)\hat{\mathbf{x}}+R\sin \left({\varphi }^{\prime }\right)\hat{\mathbf{y}}$$

With this in mind, we can compute:

$$\mathbf{r}\cdot {\mathbf{r}}^{\prime }=rR\sin \left(\theta \right)(\cos \left(\varphi \right)\cos \left({\varphi }^{\prime }\right)+\sin \left(\varphi \right)\sin \left({\varphi }^{\prime }\right))$$

And $\alpha$ is defined as the angle between those two vectors, meaning that we can define it through the dot product:

$$\mathbf{r}\cdot {\mathbf{r}}^{\prime }=|\mathbf{r}||{\mathbf{r}}^{\prime }|\cos \left(\alpha \right)=rR\cos \left(\alpha \right)$$

So

$$\cos \left(\alpha \right)=\sin \left(\theta \right)(\cos \left(\varphi \right)\cos \left({\varphi }^{\prime }\right)+\sin \left(\varphi \right)\sin \left({\varphi }^{\prime }\right))$$ We now recall Table 3.1 of Legendre Polynomials (page 140): And compute each term separately. In all of them, $r^{\prime }\to R$:

Monopole term: $n=0$

$$P_0(\cos \left(\alpha \right))=1$$

So

$$\begin{array}{rl}{V}_0(r,\theta ,\varphi )& =\frac{1}{4\pi {ϵ}_0}\frac{1}{r}{\int }_{{\varphi }^{\prime }=0}^{2\pi }{R}^{0}1\lambda Rd{\varphi }^{\prime }\\ & =\frac{\lambda R}{2{ϵ}_{0}r}\end{array}$$

Dipole term: $n=1$

$$\begin{array}{rl}{P}_1(\cos \left(\alpha \right))& =\cos \left(\alpha \right)\\ & =\sin \left(\theta \right)(\cos \left(\varphi \right)\cos \left({\varphi }^{\prime }\right)+\sin \left(\varphi \right)\sin \left({\varphi }^{\prime }\right))\end{array}$$

So

$$\begin{array}{rl}{V}_1(r,\theta ,\varphi )& =\frac{1}{4\pi {ϵ}_0}\frac{1}{r^2}{\int }_{{\varphi }^{\prime }=0}^{2\pi }{R}^1\sin \left(\theta \right)(\cos \left(\varphi \right)\cos \left({\varphi }^{\prime }\right)+\sin \left(\varphi \right)\sin \left({\varphi }^{\prime }\right))\lambda Rd{\varphi }^{\prime }\\ & =0\end{array}$$

Which is not too surprising! Our system doesn't have the type of geometry a dipole has...

Quadrupole term: $n=2$

$$\begin{array}{rl}{P}_2(\cos \left(\alpha \right))& =\frac{1}{2}(3{\cos }^2\alpha -1)\\ & =\frac{1}{2}(3[\sin \left(\theta \right)(\cos \left(\varphi \right)\cos \left({\varphi }^{\prime }\right)+\sin \left(\varphi \right)\sin \left({\varphi }^{\prime }\right)){]}^2-1)\\ & =\frac{1}{2}(3{\sin }^2\theta ({\cos }^2\varphi {\cos }^2{\varphi }^{\prime }+{\sin }^2\varphi {\sin }^2{\varphi }^{\prime }+2\cos \varphi \sin \varphi \cos {\varphi }^{\prime }\sin {\varphi }^{\prime })-1)\end{array}$$

So

$$\begin{array}{rl}{V}_2(r,\theta ,\varphi )& =\frac{1}{4\pi {ϵ}_0}\frac{1}{r^3}{\int }_{{\varphi }^{\prime }=0}^{2\pi }{R}^2\frac{1}{2}(3{\sin }^2\theta ({\cos }^2\varphi {\cos }^2{\varphi }^{\prime }+{\sin }^2\varphi {\sin }^2{\varphi }^{\prime }+2\cos \varphi \sin \varphi \cos {\varphi }^{\prime }\sin {\varphi }^{\prime })-1)\lambda Rd{\varphi }^{\prime }\\ & =-\frac{R^3\lambda (1+3\cos \left(2\theta \right))}{16r^3{ϵ}_0}\end{array}$$

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