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(A) According to Gauss's Law,the total electric flux through the closed surface of the cylinder is $\phi_{	ext{Total}} = \frac{q}{\epsilon_0}$.The to

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

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(A) According to Gauss's Law,the total electric flux through the closed surface of the cylinder is $\phi_{	ext{Total}} = \frac{q}{\epsilon_0}$.The total flux is the sum of the flux through the two plane surfaces ($A$ and $C$) and the curved surface $(B)$: $\phi_{	ext{Total}} = \phi_A + \phi_C + \phi_B = \frac{q}{\epsilon_0}$.Given that the charge is inside the cylinder,by symmetry,the flux through the two plane ends $A$ and $C$ must be equal,so let $\phi_A = \phi_C = \phi'$.We are given that the flux through the curved surface $B$ is $\phi_B = \phi$.Substituting these into the equation: $2\phi' + \phi = \frac{q}{\epsilon_0}$.Solving for $\phi'$: $2\phi' = \frac{q}{\epsilon_0} - \phi$,which gives $\phi' = \frac{1}{2} \left( \frac{q}{\epsilon_0} - \phi ight)$.

$A$ hollow cylinder has a charge $q$ coulomb within it. If $\phi$ is the electric flux in units of volt-meter associated with the curved surface $B$,the flux linked with the plane surface $A$ in units of volt-meter will be:

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