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(D) 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

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

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(D) 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 $C$) and the curved surface $(B)$.Let $\phi_A$,$\phi_C$,and $\phi_B$ be the flux through surfaces $A$,$C$,and $B$ respectively.Thus,$\phi_A + \phi_C + \phi_B = \frac{q}{\epsilon_0}$.Given that $\phi_B = \phi$,we have $\phi_A + \phi_C + \phi = \frac{q}{\epsilon_0}$.Due to the symmetry of the cylinder,the flux through the two plane ends $A$ and $C$ must be equal,so $\phi_A = \phi_C$.Substituting this into the equation: $2\phi_A + \phi = \frac{q}{\epsilon_0}$.Solving for $\phi_A$: $2\phi_A = \frac{q}{\epsilon_0} - \phi$.Therefore,$\phi_A = \frac{1}{2}\left(\frac{q}{\epsilon_0} - \phi\right)$.

$A$ hollow cylinder has a charge of $q$ $C$ within it. 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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