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\begin{document}

Assume a corner of angle \( \beta  \) in a conductor.

\( \nabla ^{2}\Psi =\frac{1}{\rho }\frac{\partial }{\partial \rho }(\rho \frac{\partial \psi }{\partial \rho })+\frac{1}{\rho ^{2}}\frac{\partial }{\partial \phi }\frac{\partial \phi }{\partial \phi }+\frac{\partial ^{2}\phi }{\partial z^{2}}=0 \).

This seperates into 2 equations:\( \nu ^{2}=\frac{\rho }{A}\frac{\partial }{\partial \rho }(\rho \frac{\partial A}{\partial \rho }) \) and \( -\nu ^{2}=\frac{1}{B}\frac{\partial }{\partial \phi }\frac{\partial \phi }{\partial \phi } \). The second equation has solution
of exponentials or linear term for \( \nu =0 \). The first equation is solved
by \( \rho ^{\nu } \), \( \rho ^{-\nu } \), or for \( \nu =0 \), \( a_{0}+b_{0}ln\rho  \). The general solution is a superposition of all
these solutions. \( \phi =(a_{0}+b_{0}ln\rho )(A_{0}+B_{0}\phi )+\sum _{\nu }(a_{\nu }\rho ^{v}+b_{\nu }\rho ^{-\nu })(A_{\nu }e^{i\nu \phi }+B_{\nu }e^{-i\nu \phi }) \)

\( \phi =0 \): \( b_{o}=0 \) and \( a_{0}B_{0}=0 \)so we want \( \nu _{n}=\frac{n\pi }{\beta } \) to satisfy \( sin(\nu \beta )=0 \) \( \Phi =a_{o}A_{0}+\sum _{n}(a_{n}\rho ^{\nu _{n}}+b_{n}\rho ^{-\nu _{n}})sin(\frac{n\pi }{\beta }\phi ) \). Must have \( b_{n}=0 \) in order to avoid
the singularity. so\( \Phi =a_{o}A_{0}+\sum _{n}a_{n}\rho ^{\nu _{n}}sin(\frac{n\pi }{\beta }\phi ) \). Very near to the cornet, you will get only the
lowest order terms contributing. Then \( E^{i}=-\nabla ^{i}\phi =-(\frac{\partial \phi }{\partial r},\frac{1}{\rho }\frac{\partial \Phi }{\partial \phi }) \) \( =-\frac{\pi a_{1}}{\beta }\rho ^{\frac{\pi }{\beta }-1}(sin(\frac{\pi }{\beta }\phi ),cos(\frac{\pi }{\beta }\phi )) \).

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