Question 3.45: A long cylindrical shell of radius R carries a uniform surfa...

Use separation of variables in cylindrical coordinates (Prob. 3.24):

V(s, \phi)=a_{0}+b_{0} \ln s+\sum_{k=1}^{\infty}\left[s^{k}\left(a_{k} \cos k \phi+b_{k} \sin k \phi\right)+s^{-k}\left(c_{k} \cos k \phi+d_{k} \sin k \phi\right)\right] .

s<R: \quad V(s, \phi)=\sum_{k=1}^{\infty} s^{k}\left(a_{k} \cos k \phi+b_{k} \sin k \phi\right) \quad\left(\ln s \text { and } s^{-k} \text { blow up at } s=0\right) ;

s>R: V(s, \phi)=\sum_{k=1}^{\infty} s^{-k}\left(c_{k} \cos k \phi+d_{k} \sin k \phi\right) \quad\left(\ln s \text { and } s^{k} \text { blow up as } s \rightarrow \infty\right) .

(We may as well pick constants so V → 0 as s → ∞ , and hence a_{0}=0.) Continuity at s = R ⇒

\sum R^{k}\left(a_{k} \cos k \phi+b_{k} \sin k \phi\right)=\sum R^{-k}\left(c_{k} \cos k \phi+d_{k} \sin k \phi\right), \text { so } c_{k}=R^{2 k} a_{k}, d_{k}=R^{2 k} b_{k} . Eq. 2.36 says:

\left.\frac{\partial V}{\partial s}\right|_{R^{+}}-\left.\frac{\partial V}{\partial s}\right|_{R^{-}}=-\frac{1}{\epsilon_{0}} \sigma . Therefore

\frac{\partial V_{ above }}{\partial n}-\frac{\partial V_{\text {below }}}{\partial n}=-\frac{1}{\epsilon_{0}} \sigma                    (2.36)

\sum \frac{-k}{R^{k+1}}\left(c_{k} \cos k \phi+d_{k} \sin k \phi\right)-\sum k R^{k-1}\left(a_{k} \cos k \phi+b_{k} \sin k \phi\right)=-\frac{1}{\epsilon_{0}} \sigma ,

or:

\sum 2 k R^{k-1}\left(a_{k} \cos k \phi+b_{k} \sin k \phi\right)=\left\{\begin{array}{ll}\sigma_{0} / \epsilon_{0} & (0<\phi<\pi) \\ -\sigma_{0} / \epsilon_{0} & (\pi<\phi<2 \pi)\end{array}\right\} .

Fourier’s trick: multiply by (\cos l \phi) d \phi and integrate from 0 to 2π, using

\int_{0}^{2 \pi} \sin k \phi \cos l \phi d \phi=0 ; \quad \int_{0}^{2 \pi} \cos k \phi \cos l \phi d \phi=\left\{\begin{array}{l}0, k \neq l \\\pi, k=l\end{array}\right\} .

Then

2 l R^{l-1} \pi a_{l}=\frac{\sigma_{0}}{\epsilon_{0}}\left[\int_{0}^{\pi} \cos l \phi d \phi-\int_{\pi}^{2 \pi} \cos l \phi d \phi\right]=\frac{\sigma_{0}}{\epsilon_{0}}\left\{\left.\frac{\sin l \phi}{l}\right|_{0} ^{\pi}-\left.\frac{\sin l \phi}{l}\right|_{\pi} ^{2 \pi}\right\}=0 ; \quad a_{l}=0 .

Multiply by (\sin l \phi) d \phi and integrate, using \int_{0}^{2 \pi} \sin k \phi \sin l \phi d \phi=\left\{\begin{array}{l}0, k \neq l \\\pi, k=l\end{array}\right\} :

=\left\{\begin{array}{lr}0, & \text { if } l \text { is even } \\ 4 \sigma_{0} / l \epsilon_{0}, & \text { if } l \text { is odd }\end{array}\right\} \Rightarrow b_{l}=\left\{\begin{array}{ll}0, & \text { if } l \text { is even } \\2 \sigma_{0} / \pi \epsilon_{0} l^{2} R^{l-1},& \text { if } l \text { is odd }\end{array}\right\} .

Conclusion:

V(s, \phi)=\frac{2 \sigma_{0} R}{\pi \epsilon_{0}} \sum_{k=1,3,5, \ldots} \frac{1}{k^{2}} \sin k \phi\left\{\begin{array}{l}(s / R)^{k} & (s<R) \\(R / s)^{k} &(s>R)\end{array}\right\} .