Question

A hollow cylinder has a charge q placed at its centre. If $\phi$ is the electric flux (in $V \, m$ ) associated with the curved surface B, the flux linked with the plane surface A is

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

Answer 1

Answer: (A)

According to Gauss law, $\phi_{t o t a l}=\frac{q}{\epsilon _{0}}$
Let electric flux linked with surfaces $A, \, B \, and \, C$ are $ \, \phi_{A},\phi_{B} \, $ and $ \, \phi_{C}$ respectively. That is
$\phi_{t o t a l}=\phi_{A}+\phi_{B}+\phi_{C}$
Since $\phi_{C}=\phi_{A}$
$\therefore 2\phi_{A}+\phi_{B}=\phi_{t o t a l}=\frac{q}{\epsilon _{0}}$
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) $