A hollow cylinder has a charge q within it. I | vedclass

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(D) Let $\phi_A, \phi_B,$ and $\phi_C$ be the electric flux linked with the surfaces $A, B,$ and $C$ respectively.According to Gauss's Law,the total e

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

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(D) Let $\phi_A, \phi_B,$ and $\phi_C$ be the electric flux linked with the surfaces $A, B,$ and $C$ respectively.According to Gauss's Law,the total electric flux through the closed surface is $\phi_{total} = \phi_A + \phi_B + \phi_C = \frac{q}{\varepsilon_0}$.Due to the symmetry of the cylinder,the flux linked with the two plane surfaces $A$ and $C$ is equal,so $\phi_A = \phi_C$.Substituting this into the Gauss's Law equation,we get $2\phi_A + \phi_B = \frac{q}{\varepsilon_0}$.Given that the flux through the curved surface $B$ is $\phi_B = \phi$,we have $2\phi_A + \phi = \frac{q}{\varepsilon_0}$.Rearranging for $\phi_A$,we get $2\phi_A = \frac{q}{\varepsilon_0} - \phi$.Therefore,$\phi_A = \frac{1}{2}\left(\frac{q}{\varepsilon_0} - \phi\right)$.

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

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