Question

Using the formulas of Chapter 12, Section 5, sum the series in (8.20) to get (8.21).

Short answer

The sum series is derived given in (8.20).

$V=\frac{q}{\sqrt{r^2-2ar\cos \theta +a^2}}-\frac{\left(\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right)q}{\sqrt{r^2+{\left(\raisebox{1ex}{$R^2$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right)}^2-2r\left(\raisebox{1ex}{$R^2$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right)\cos \theta }}$

Step-by-step solution

Given Information:

It has been asked to use the formula given in section 5 of chapter 12.

Step 2:Uses of the Green function:

The green function is a function that is used to solve a corresponding partial differential equation in three dimensions, namely Poisson’s equation.

$\begin{array}{c}{\mathbf{\nabla }}^{\mathbf{2}}\mathbf{u}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\\ \mathbf{=}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\end{array}$

Use the formula given in section 5 of chapter 12:

The summing the series in equation (8.20) from the book:

$V=\frac{q}{\sqrt{r^2-2ar\cos \theta +a^2}}-q\sum _l\frac{R^{2l+1}{r}^{-l-1}{P}_l\left(\cos \theta \right)}{a^{l+1}}$

Modify the equation to bring the equation mentioned below.

$V=\frac{q}{\sqrt{r^2-2ar\cos \theta +a^2}}-\frac{\left(\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right)q}{\sqrt{r^2+{\left(\raisebox{1ex}{$R^2$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right)}^2-2r\left(\raisebox{1ex}{$R^2$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right)\cos \theta }}$

(In every sum in this problem, goes from 0 to $\infty$).

Calculate the sum using the equations mentioned below.

$V=\frac{K}{R}{[1-\frac{2r}{R}\cos \theta +{\left(\frac{r}{R}\right)}^2]}^{-1/2}$

$\begin{array}{c}V=\frac{K}{R}\sum _l\frac{r^l{P}_l\left(\cos \theta \right)}{R^l}\\ =\frac{K}{R}\sum _l{\left(\frac{r}{R}\right)}^l{P}_l\left(\cos \theta \right)\end{array}$

Prove both the equation to be equal.

Rewrite the term with the sum in the equation.

$q\sum _l\frac{R^{2l+1}{r}^{-l-1}{P}_l\left(\cos \theta \right)}{a^{l+1}}=\frac{qR}{ar}\sum _l{\left(\frac{R^2}{ar}\right)}^l{P}_l\left(\cos \theta \right)$ …….(1)

The factor $\left(\frac{R^2}{ar}\right)$in the sum in the last equation plays the same role as the factor $\left(\frac{r}{R}\right)$in the sum of equation (5.17). Thus, comparing (5.17) with (5.12) and seeing where the factor $\left(\frac{r}{R}\right)$is present inside the brackets of (5.12), It can be concluded what will be the result of the sum from equation (1) - it is equal to the bracket in (5.12) where $\left(\frac{r}{R}\right)$change with $\left(\frac{R^2}{ar}\right)$:

$\begin{array}{c}q\sum _l\frac{R^{2l+1}{r}^{-l-1}{P}_l\left(\cos \theta \right)}{a^{l+1}}=\frac{qR}{ar}{[1-2\left(\frac{R^2}{ar}\right)\cos \theta +{\left(\frac{R^2}{ar}\right)}^2]}^{-1/2}\\ =\frac{\left(\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right)q}{\sqrt{r^2+{\left(\raisebox{1ex}{$R^2$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right)}^2-2r\left(\raisebox{1ex}{$R^2$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right)\cos \theta }}\end{array}$

Hence, it has been proved that summing the series in (8.20) gives us (8.21).