As shown in the figure, a charge q is positioned at the centre of...

To find the electric flux passing through the surface of the cylindrical vessel, we can use Gauss's law.

Imagine placing an identical cylindrical vessel symmetrically over the open end of the given vessel to form a closed Gaussian surface (a complete cylinder).

The charge $q$ is now located exactly at the center of this closed cylindrical surface.

According to Gauss's law, the total electric flux passing through this entire closed surface is:

$\Phi_{\text{total}} = \dfrac{q}{\varepsilon_0}$

Due to the symmetry of the configuration, exactly half of the electric field lines will pass through the upper vessel and the other half will pass through the lower vessel.

Therefore, the electric flux passing through the surface of the given (lower) cylindrical vessel is:

$\Phi = \dfrac{\Phi_{\text{total}}}{2} = \dfrac{q}{2\varepsilon_0}$