Question

A hollow cylinder has a charge $q\,\, C$ within it. If $\Phi$ is the electric flux in unit of voltmeter associated with the curved surface $B$, the flux linked with the plane surface $A$ in unit of voltmeter will be

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

Answer 1

Answer: (A)

$\frac{1}{2}\left(\frac{q}{\varepsilon_{0}}-\phi\right)$

Answer 2

Apply Gauss's law to calculate the charge associated with plane surface $A$. Gauss's law states that the net electric flux thorugh any closed surface is equal to the net charge inside the surface divided by $\varepsilon_{0}$. ie, $\phi_{\text {total }}=\frac{q}{\varepsilon_{0}}$ Let electric flux linked with surfaces $A, B$ and $C$ are $\phi_{A}, \phi_{B}$ and $\phi_{C}$ respectively. That is $\phi_{\text {total }} =\phi_{A}+\phi_{B}+\phi_{C}$ $\phi_{C} =\phi_{A}$ 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, $\phi_{A}=\frac{1}{2}\left(\frac{q}{\varepsilon_{0}}-\phi\right)$