Question

An infinitely long cylindrical surface of circular cross-section is uniformly charged lengthwise with the surface density $\sigma ={\sigma }_0\cos \phi$, where $\phi$ is the polar angle of the cylindrical coordinate system whose $z$ axis coincides and direction of the electric field strength vector on the $z$ axis.

Answer 1

Take a section of the cylinder perpendicular to its axis through the point where the electric field is to be calculated. (All points on the axis are equivalent.) Consider an element $S$ with azimuthal angle $\phi$. The length of the element is $Rd\phi ,R$ being the radius of cross section of the cylinder. The element itself is a section of an infinite strip. The electric field at $O$ due to this strip is

$E=\frac{{\sigma }_0\cos \phi \left(Rd\phi \right)}{2\pi {ϵ}_{0}R}$ along $SO$

This can be resolved into

$E_o=\frac{{\sigma }_0\cos \phi d\phi }{2\pi {ϵ}_0}\{\begin{array}{c}\cos \phi alongOXtowardsO\\ \sin \phi alongYO\end{array}$

On integration the component along $YO$ vanishes. What remains is

$E_o={\int }_0^{2\pi }\frac{{\sigma }_0{\cos }^2\phi d\phi }{2\pi {ϵ}_0}=\frac{{\sigma }_0}{2{ϵ}_0}$ along $XO$ i.e. along the direction $\phi =\pi$.