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(C) According to Gauss's Law, the total electric flux $\phi_{total}$ through a closed surface is given by $\phi_{total} = \frac{q}{\epsilon_0}$.For th

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$\frac{1}{2}\left(\frac{q}{\epsilon_0}-\phi\right)$

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(C) According to Gauss's Law, the total electric flux $\phi_{total}$ through a closed surface is given by $\phi_{total} = \frac{q}{\epsilon_0}$.For the hollow cylinder, the total flux is the sum of the flux through the two plane surfaces ($A$ and $B$) and the curved surface $(C)$: $\phi_{total} = \phi_A + \phi_B + \phi_C$.Given that $\phi_C = \phi$ and by symmetry, the flux through the two plane surfaces is equal, i.e., $\phi_A = \phi_B$.Substituting these into the Gauss's Law equation: $\frac{q}{\epsilon_0} = \phi_A + \phi_A + \phi$.$\frac{q}{\epsilon_0} - \phi = 2\phi_A$.Therefore, the flux linked with the plane surface $A$ is $\phi_A = \frac{1}{2}\left(\frac{q}{\epsilon_0} - \phi\right)$.

$A$ hollow cylinder has a charge $q$ coulomb within it. If $\phi$ is the electric flux in unit of $V-m$ associated with the curved surface $C$, the flux linked with the plane surface $A$ in unit of $V-m$ will be. $[\epsilon_0 = \text{permittivity of free space}]$

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Assertion: Four point charges $q_1, q_2, q_3$ and $q_4$ are as shown in the figure. The flux over the shown Gaussian surface depends only on charges $q_1$ and $q_2$.
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