A charge q is placed at the centre of the open end of a cylindrical vessel (figure 30-Q3). The flux of the electric field through the surface of the vessel is
(a) zero
(b) $q/{\epsilon }_{\nu }$
(c) $q/2{\epsilon }_{\nu }$
(d) $2q/{\epsilon }_{\nu }$.

(C) q/2${\in }_0$

According to Gauss's Law, the flux through a closed cylindrical Gaussian surface is q/${\in }_0$. But the question is about an open cylindrical vessel. Now, take another identical vessel and make a closed Gaussian surface enclosing the charge, as shown in the figure.

Total flux linked with the closed Gaussian surface,
${\varphi }_T=\frac{q}{{\in }_0}$

Flux linked with the surface of a open ended cylindrical vessel,
$\varphi =\frac{{\varphi }_T}{2}=\frac{q}{2{\in }_0}$