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(B) According to Gauss's Law,the total electric flux $\phi_{total}$ through the closed surface of the cylinder is $\frac{q}{\epsilon_0}$.This total fl

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$\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)$

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(B) According to Gauss's Law,the total electric flux $\phi_{total}$ through the closed surface of the cylinder is $\frac{q}{\epsilon_0}$.This total flux is the sum of the flux through the two plane surfaces ($A$ and $C$) and the curved surface $(B)$: $\phi_A + \phi_B + \phi_C = \frac{q}{\epsilon_0}$.Due to the symmetry of the cylinder and the central position of the charge,the flux through the two plane surfaces is equal,i.e.,$\phi_A = \phi_C$.Given that the flux through the curved surface $B$ is $\phi$,we have: $2\phi_A + \phi = \frac{q}{\epsilon_0}$.Solving for $\phi_A$,we get: $\phi_A = \frac{1}{2}\left(\frac{q}{\epsilon_0} - \phi\right)$.

$A$ hollow cylinder has a charge $q$ at its center. If $\phi$ is the electric flux associated with the curved surface $B,$ the flux linked with the plane surface $A$ will be $:-$

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