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Question

(A) Let the electric flux linked with the plane surfaces $A$ and $C$ be $\phi_A$ and $\phi_C$ respectively, and the flux linked with the curved surfac

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the electric flux linked with the plane surfaces $A$ and $C$ be $\phi_A$ and $\phi_C$ respectively, and the flux linked with the curved surface $B$ be $\phi_B = \phi$.By symmetry, the flux through the two identical plane ends is equal, so $\phi_A = \phi_C$.According to Gauss's Law, the total electric flux $\phi_{	ext{total}}$ through the closed cylinder is given by $\phi_{	ext{total}} = \frac{q}{\varepsilon_0}$.The total flux is the sum of the fluxes through all surfaces: $\phi_{	ext{total}} = \phi_A + \phi_C + \phi_B$.Substituting the known values: $\frac{q}{\varepsilon_0} = \phi_A + \phi_A + \phi$.$\frac{q}{\varepsilon_0} = 2\phi_A + \phi$.Rearranging for $\phi_A$: $2\phi_A = \frac{q}{\varepsilon_0} - \phi$.Therefore, $\phi_A = \frac{1}{2}\left( \frac{q}{\varepsilon_0} - \phi ight)$.

$A$ hollow cylinder has a charge $q$ inside it. If $\phi$ is the electric flux in units of $\text{V}\cdot\text{m}$ associated with the curved surface $B$, what is the flux linked with the plane surface $A$ in units of $\text{V}\cdot\text{m}$?

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