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(C) To apply Gauss's theorem,it is essential that the charge should be placed inside a closed surface. Imagine another similar cylindrical vessel plac

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

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(C) To apply Gauss's theorem,it is essential that the charge should be placed inside a closed surface. Imagine another similar cylindrical vessel placed inverted on top of the existing vessel to form a closed Gaussian surface (a larger cylinder). According to Gauss's law,the total electric flux $\phi_{total}$ through this closed surface is $\frac{q}{\varepsilon_0}$. Since the charge $q$ is placed at the interface of the two identical cylinders,by symmetry,the flux through each cylinder is equal. Therefore,the flux through the surface of the original 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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