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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:

\[V(\mathbf{r})=\frac{1}{4\pi \epsilon _0 }\sum_{n=0}^{\infty } \frac{1}{r^{n+1}}\int (r')^n P_n(\cos \left( \alpha \right) )\rho (\mathbf{r}')d\tau ' \tag{Eq. 3.95, Page 150}\]

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

\[\rho (\mathbf{r}')d\tau '\rightarrow \lambda dl=\lambda Rd\phi ' \]

Where the prime distinguish the $\phi $ from position and the $\phi ' $ for the position of the charge configuration.

Our position vector, as usual, is:

\[\mathbf{r}=r\sin \left( \theta \right) \cos \left( \phi \right) \hat{\mathbf{x}} + r\sin \left( \theta \right) \sin \left( \phi \right) \hat{\mathbf{y}} + r\cos \left( \theta \right) \hat{\mathbf{z}} \]

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

\[\mathbf{r}'=R\cos \left( \phi ' \right) \hat{\mathbf{x}}+R \sin \left( \phi ' \right) \hat{\mathbf{y}} \]

With this in mind, we can compute:

\[\mathbf{r}\cdot \mathbf{r}'=rR\sin \left( \theta \right) \bigg( \cos \left( \phi \right) \cos \left( \phi ' \right) +\sin \left( \phi \right) \sin \left( \phi ' \right) \bigg) \]

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

\[\mathbf{r}\cdot \mathbf{r}'=|\mathbf{r}||\mathbf{r}'|\cos \left( \alpha \right) =rR\cos \left( \alpha \right) \]

So

\[\cos \left( \alpha \right) = \sin \left( \theta \right) \bigg( \cos \left( \phi \right) \cos \left( \phi ' \right) +\sin \left( \phi \right) \sin \left( \phi ' \right) \bigg)\] We now recall Table 3.1 of Legendre Polynomials (page 140): And compute each term separately. In all of them, $r' \rightarrow R$:

Monopole term: $n=0 $

\[P_0 (\cos \left( \alpha \right) )=1 \]

So

\begin{align*} V_0(r,\theta ,\phi ) &= \frac{1}{4\pi \epsilon _0 }\frac{1}{r}\int_{\phi '=0}^{2\pi } R^0 1 \lambda R d\phi ' \\ &= \frac{\lambda R}{2\epsilon _0 r} \end{align*}

Dipole term: $n=1 $

\begin{align*} P_1(\cos \left( \alpha \right)) &= \cos \left( \alpha \right) \\ &= \sin \left( \theta \right) \bigg( \cos \left( \phi \right) \cos \left( \phi ' \right) +\sin \left( \phi \right) \sin \left( \phi ' \right) \bigg) \end{align*}

So

\begin{align*} V_1(r,\theta ,\phi ) &= \frac{1}{4\pi \epsilon _0 }\frac{1}{r^2 }\int_{\phi '=0}^{2\pi } R^1 \sin \left( \theta \right) \bigg( \cos \left( \phi \right) \cos \left( \phi ' \right) +\sin \left( \phi \right) \sin \left( \phi ' \right) \bigg) \lambda R d\phi ' \\ &= 0 \end{align*}

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

Quadrupole term: $n=2 $

\begin{align*} P_2 (\cos \left( \alpha \right) ) &= \frac{1}{2}(3\cos ^2 \alpha -1) \\ &= \frac{1}{2} \bigg( 3 \bigg[ \sin \left( \theta \right) \bigg( \cos \left( \phi \right) \cos \left( \phi ' \right) +\sin \left( \phi \right) \sin \left( \phi ' \right) \bigg) \bigg] ^2 -1 \bigg) \\ &= \frac{1}{2} \bigg( 3 \sin ^2 \theta \left( \cos ^2 \phi \cos ^2 \phi ' + \sin ^2 \phi \sin ^2 \phi ' + 2 \cos \phi \sin \phi \cos \phi ' \sin \phi ' \right) -1 \bigg) \end{align*}

So

\begin{align*} V_2(r,\theta ,\phi ) &= \frac{1}{4\pi \epsilon _0 }\frac{1}{r^3}\int_{\phi '=0}^{2\pi } R^2 \frac{1}{2} \bigg( 3 \sin ^2 \theta \left( \cos ^2 \phi \cos ^2 \phi ' + \sin ^2 \phi \sin ^2 \phi ' + 2 \cos \phi \sin \phi \cos \phi ' \sin \phi ' \right) -1 \bigg) \lambda R d\phi ' \\ &= -\frac{R^3 \lambda (1+3\cos \left( 2\theta \right) )}{16r^3 \epsilon _0 } \end{align*}

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