A charge $q$ is placed at the centre of the open end of a cylindrical vessel. The flux of the electric field…
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
zero
$\mathrm{q} / \varepsilon_{\mathrm{o}}$
$\mathrm{q} / 2 \varepsilon_{\mathrm{o}}$
$2 q / \varepsilon_{0}$
Solution
Given that, A charge q is placed at the center of open end Q a cylindrical vessel, we have to find the flux through the surface of the vessel. So, when charge \(\mathrm{Q}\) is placed at the center of open end of a cylindrical vessel then only half of the charge will contribute to the flux, because half will lie inside the surface and half will lie outside the surface. so, flux through the surface of vessel is \(\frac{q}{2 \varepsilon_{0}}\)
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