# Is this an Error in Griffiths Electrodynamics?

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 $$a. The answer provided by the book is: $$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.

• 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! Aug 22, 2020 at 7:22

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 $$l=0$$ ...
$$A_la^l+\frac{B_l}{a^{l+1}}=0$$
for $$l>0$$. It doesn’t imply that $$A_l=B_l=0$$.