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Check Problem 3.43 in Griffiths Introduction to Electrodynamics

A conducting sphere of radius $a$, at potential $V_0$, is surrounded by a thin concentric spherical shell of radius $b$, over which someone has glued a surface charge $\sigma(\theta)=k\cos(\theta)$ where $k$ is a constant and $\theta$ is the polar angle.

It then asks to find the potential in the $r>b$ and $a<r<b$. The answer provided by the book is: $V(r, θ) = \frac{aV_0}{r} + \frac{(b^3 − a^3)k \cos θ}{3r^2 \epsilon_0}, r ≥ b,$

$V(r, θ) = \frac{aV_0}{r} + \frac{(r^3 − a^3)k \cos θ}{3r^2 \epsilon_0}, r ≤ b$

Credit: Griffiths, David J.. Introduction to Electrodynamics (p. 162). Cambridge University Press. Kindle Edition.

Solving for the region in between the discs: using the boundary condition $V(a,\theta)=V_0$ we find that: $$V(a,\theta)=\sum_{l=0}^{\infty}{(A_la^{l}+\frac{B_l}{a^{l+1}})P_l(\cos\theta)}=V_0$$ Since $V_0$ is a constant, and thus has no $\theta$ dependence, we conclude the only term in the series must be the one with $l=0$ to ensure that the left side of the equation has no terms in $\cos \theta$. Thus in summary we find that: $$A_0+\frac{B_0}{a}=V_0$$ and thus the potential has the form: $$V(r,\theta)=A_0+\frac{a(V_0-A_0)}{r}$$ but this will obviously not satisfy the form given by the answer in the book, as it must have a $\cos \theta$ term and of course the $A_l$'s and $B_l$'s are constants. I am probably at fault, but I don't see where exactly. Please help.

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    $\begingroup$ Whenever in doubt, you can check the errata for the book. You can find all the corrected errors directly on the website of Griffiths himself! $\endgroup$
    – Quiver
    Aug 22, 2020 at 7:22

1 Answer 1

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Since $V_0$ is a constant, and thus has no $\theta$ dependence, we conclude the only term in the series must be the one with $l=0$ ...

This implies that

$$A_la^l+\frac{B_l}{a^{l+1}}=0$$

for $l>0$. It doesn’t imply that $A_l=B_l=0$.

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