# Separation of variables and choosing solutions

On Griffiths E&M page 145 he does an example of an uncharged metal sphere of radius R in a uniform E-field $$E=E_0\hat{z}$$. Now with the general solution $$V(r,\theta)=\sum_{l=0}^{\infty}\left(A_lr^l+\frac{B_l}{r^{l+1}}\right)P_l(cos(\theta))$$ and noting that $$V=-E_0z+C$$ And V=0 at r=R we get $$A_l R^l+\frac{B_l}{r^{l+1}}=0$$ This is fine. But then it says that for the solution as r$$\rightarrow \infty$$, our $$B_l$$ Becomes negligible and therefore we only deal with the $$A_l$$ terms. This is breaking my head a little bit because I thought that we always chose the non-problematic term as the solution i.e non-diverging or blowing up at zero. But the $$A_l r^l$$ terms will diverge as r$$\rightarrow \infty$$. What is going on here? Any help would be greatly appreciated!

• Do you mean r instead of v? – Jake Rose Mar 29 '19 at 14:57
• If we ignore the B terms, then we require the A terms to be 0. (You’ve actually written all the working to deduce this) – Jake Rose Mar 29 '19 at 15:29

## 2 Answers

You've written it out yourself: "$$V= -E_0z+ C$$". $$z= r\cos\theta$$, so $$V\sim -E_0 r$$ as $$r\rightarrow \infty$$. Then we do have to deal with the $$A_l$$ terms, but it doesn't mean all of them. We want to match the coefficients of $$A_l$$ to $$V$$, which means $$A_1 \neq 0$$ and $$A_l = 0$$ for $$l\neq1$$.

A point of clarification: As long as the series is converging, it is a well defined function.

For example, $$e^{-x}=\sum_{n=0}^\infty \frac{(-x)^n}{n!}$$

Here $$A_n=\frac{(-1)^n}{n!}$$ and the function $$\rightarrow0$$ as $$x\rightarrow \infty$$