The potential V0(θ) is again specified on the surface of a sphere of radius R, but this time we are asked to find the potential outside, assuming there is no charge there.

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In this case it’s the Al ’s that must be zero (or else V would not go to zero at∞), so

[latex]V(r, θ) =\sum\limits_{l=0}^{\infty }{\frac{B_{l}}{r^{l+1}}P_{l}(\cos heta) } [/latex] (3.72)

At the surface of the sphere, we require that

[latex]V(r, θ) =\sum\limits_{l=0}^{\infty }{\frac{B_{l}}{r^{l+1}}P_{l}(\cos heta) } =V_{0}( heta).[/latex]

Multiplying by [latex]P_{\acute{l}} (\cos heta) \sin heta[/latex] and integrating—exploiting, again, the orthogonality relation 3.68—we have

[latex]\int_{-1}^{1}{P_{l} (x)P_{\acute{l}} (x) dx}=\int_{0}^{\pi}{P_{l} (\cos{ heta})P_{\acute{l}} (\cos{ heta}) \sin{ heta} d{ heta}} \ =\begin{cases} 0, if \acute{l} eq l, \ \frac{2}{2l+1} , if \acute{l} = l\end{cases} [/latex] (3.68)

[latex]\frac{B_{\acute{l}}}{R^{\acute{l}+1}}\frac{2}{2\acute{l}+1}=\int_{0}^{\pi}{V_{0}( heta)P_{\acute{l}} (\cos heta) \sin heta d heta,} [/latex]

or

[latex]B_{l}=\frac{2l+1}{2}R^{l+1} \int_{0}^{\pi}{V_{0}( heta)P_{{l}} (\cos heta) \sin heta d heta,} .[/latex] (3.73)

Equation 3.72, with the coefficients given by Eq. 3.73, is the solution to our problem.

Question 3.7: The potential V0(θ) is again specified on the surface of a s...

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