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(A) According to Gauss's law,the total electric flux through a closed surface is given by $\phi_{total} = \frac{q}{\varepsilon_0}$.For the hollow cyli

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

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(A) According to Gauss's law,the total electric flux through a closed surface is given by $\phi_{total} = \frac{q}{\varepsilon_0}$.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 $(B)$: $\phi_A + \phi_C + \phi_B = \frac{q}{\varepsilon_0}$.Due to the symmetry of the cylinder,the electric flux passing through the two plane surfaces $A$ and $C$ is equal,i.e.,$\phi_A = \phi_C$.Given that the flux through the curved surface $B$ is $\phi_B = \phi$,we substitute this into the equation:$2\phi_A + \phi = \frac{q}{\varepsilon_0}$.Rearranging the equation to solve for $\phi_A$:$2\phi_A = \frac{q}{\varepsilon_0} - \phi$.$\phi_A = \frac{1}{2}\left(\frac{q}{\varepsilon_0} - \phi\right)$.

$A$ hollow cylinder has a charge '$q$' $C$ 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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