Question

For the dipole in Ex. 3.10, expand$1/r\pm$ to order ${\left(d/r\right)}^3$,and use this

to determine the quadrupole and octo-pole terms in the potential.

Short answer

The quadruple and octupletterms in the potential is and$\frac{{\mathbf{qd}}^{\mathbf{3}}}{\mathbf{32}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{4}}}\left(\mathbf{5}{\mathbf{cos}}^{\mathbf{3}}\mathbf{\theta }\mathbf{-}\mathbf{3}\mathbf{cos\theta }\right)$

Step-by-step solution

Define functions

Write the expression for the potential.

$V\left(r\right)=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q}{r_{+}}-\frac{q}{r_{-}}\right)$ …… (1)

Here,q is the charge and v is the potential.

As,

$$\begin{gathered} \frac{1}{r_{+}}=\frac{1}{r}\sum _{n=0}^{\infty }{\left(\frac{r^r}{r}\right)}^n{P}_n\left(\cos \theta \right) \\ \frac{1}{r_{-}}=\frac{1}{r}\sum _{n=0}^{\infty }{\left(\frac{r^r}{r}\right)}^n{P}_n\left(\cos \theta \right) \end{gathered}$$

Here,$P_n$ is the Legendre polynomial of order n .

Determine the quadruple and octuplet terms in the potential

Keep$r\text{'}=d/2$ in above equation.

$\frac{1}{r_{-}}=\frac{1}{r}\sum _{n=0}^{\infty }{\left(\frac{d}{2r}\right)}^n{P}_n\left(\cos \theta \right)$

For r take $\theta =180-\theta$, so$\cos \theta \to -\cos \theta$

Then,$\frac{1}{r_{-}}=\frac{1}{r}\sum _{n=0}^{\infty }{\left(\frac{d}{2r}\right)}^n{P}_n\left(-\cos \theta \right)$

As,

$\begin{array}{rcl}{P}_n\left(-n\right)& =& {\left(-1\right)}^n{P}_n\left(x\right)\\ P_n\left(-\cos \theta \right)& =& {\left(-1\right)}^n{P}_n\left(\cos \theta \right)\\ & & \end{array}$

Then,

$V\left(r\right)=\frac{q}{4\pi {\epsilon }_{0}r}\left[\sum _{n=0}^{\infty }{\left(\frac{d}{2r}\right)}^n{P}_n\left(\cos \theta \right)-\sum _{n=0}^{\infty }{\left(-1\right)}^n{\left(\frac{d}{2r}\right)}^n{P}_n\left(\cos \theta \right)\right]$…… (2)

Determine the quadruple and octuplet terms in the potential

Put $n=1$for dipole,

$$\begin{gathered} V_{dipole}=\frac{q}{4\pi {\epsilon }_{0}r}\left[P_1\left(\cos \theta \right)+P_1\left(\cos \theta \right)\right] \\ {V}_{dipole}=\frac{q}{4\pi {\epsilon }_{0}r}\frac{2qd\cos \theta }{2r^2} \\ {V}_{dipole}=\frac{q}{4\pi {\epsilon }_{0}r}\frac{qd\cos \theta }{r^2} \end{gathered}$$

$P_1\left(\cos \theta \right)=\cos \theta$

Now, put $n=2$for quadruple term,

$$\begin{gathered} V_{quad}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}{\left(\frac{d}{2r}\right)}^2\left[P_2\left(\cos \theta \right)-P_2\cos \theta \right] \\ =0 \end{gathered}$$

Now, put $n=3$for octuplet term,

$\begin{array}{rcl}{V}_{oct}& =& \frac{1}{4\pi {\epsilon }_0}\frac{q}{r}{\left(\frac{d}{2r}\right)}^3\left[P_3\left(\cos \theta \right)+P_3\left(\cos \theta \right)\right]\\ & =& \frac{2q}{4\pi {\epsilon }_{0}r}{\left(\frac{d}{2r}\right)}^3{P}_3\left(\cos \theta \right)\\ & =& \frac{q}{4\pi {\epsilon }_{0}r}{\left(\frac{d}{2r}\right)}^3\left(5{\cos }^3\theta -3\cos \theta \right)\\ & =& \frac{q{d}^3}{32\pi {\epsilon }_0{r}^4}\left(5{\cos }^3\theta -3\cos \theta \right)\end{array}$

Hence, the quadruple and octuplet terms in the potential are 0 and .

$\frac{q{d}^3}{32\pi {\epsilon }_0{r}^4}\left(5{\cos }^3\theta -3\cos \theta \right)$