Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 2 / b / Solution

Boundedness at the origin allows only $r^n$ for nonconstant interior modes; decay at infinity allows only $r^{-n}$ outside. Write the common surface potential in its Fourier series as ${\sum }_{n\ge 1}{v}_n(\theta )$, where $v_n=A_n\cos n\theta +B_n\sin n\theta$. Continuity then makes the corresponding interior and exterior factors $(r/R{)}^n$ and $(R/r{)}^n$.

Their outward radial derivatives at the surface are $n{v}_n/R$ inside and $-n{v}_n/R$ outside. Thus the jump is $-2n{v}_n/R$. Equating it to minus the prescribed surface-charge mode givesTherefore the potential of a charged cylindrical shell with angular Fourier data isEither branch gives the same value at $R$. The mean angular mode is zero: outside, neither a constant nor a logarithm can survive the condition at infinity; continuity then forces the interior constant to zero. Consequently $\Phi (0)=0$.

The absence of a mean term in the surface charge density is essential. A nonzero mean would require a logarithmic exterior potential and would be incompatible with decay at infinity in this two-dimensional reduction. The normalization follows the stated derivative jump, with no additional permittivity factor introduced.