Question

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

• A. $\frac{q}{\varepsilon_{0}}-\phi$
• B. $\frac{q}{2 \varepsilon_{0}}$
• C. $\frac{\phi}{3}$
• D. $\frac{1}{2}\left(\frac{q}{\varepsilon_{0}}-\phi\right)$

Answer 1

Answer: (D)

Gauss's law states that the net electric flux through any closed surface is equal to the net charge inside the surface divided by $\varepsilon_{0}$.
i.e., $\phi_{\text {total }}=\frac{q}{\varepsilon_{0}}$
Let electric flux linked with surfaces $A, B$ and $C$ are $\phi_{A}, \phi_{B}$ and $\phi_{C}$, respectively. That is
$\phi_{\text {total }}=\phi_{A}+\phi_{B}+\phi_{C}$
Since $\phi_{C}=\phi_{A}$
$\therefore 2 \phi_{A}+\phi_{B}=\phi_{\text {total }}=\frac{q}{\varepsilon_{0}}$
or $\phi_{A}=\frac{1}{2}\left(\frac{q}{\varepsilon_{0}}-\phi_{B}\right)$
But $\phi_{B}=\phi $ (given)
Hence, $\phi_{A}=\frac{1}{2}\left(\frac{q}{\varepsilon_{0}}-\phi\right)$