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(C) According to Gauss's Law,the total electric flux through a closed surface is $\frac{q_{enclosed}}{\varepsilon_0}$.To calculate the flux through th

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$\frac{q}{2 \varepsilon_0}$

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(C) According to Gauss's Law,the total electric flux through a closed surface is $\frac{q_{enclosed}}{\varepsilon_0}$.To calculate the flux through the cylindrical vessel,we can imagine an identical cylinder placed adjacent to the first one such that the charge $q$ is now enclosed within a larger closed cylindrical surface formed by the two vessels.The total flux through this combined closed surface is $\Phi_{total} = \frac{q}{\varepsilon_0}$.Since the charge $q$ is placed symmetrically at the center of the common open end,the flux is distributed equally between the two identical cylinders.Therefore,the flux through the surface of one cylindrical vessel is $\Phi = \frac{\Phi_{total}}{2} = \frac{q}{2 \varepsilon_0}$.

$A$ charge $q$ is placed at the centre of the open end of a cylindrical vessel. The flux of the electric field through the surface of the vessel is:

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