Q.
A hollow cylinder has a charge q coulomb 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) $\frac{q}{\varepsilon_{0}} - \phi$
Explanation
(a) According to Gauss law, $\phi_{total} = \frac{q}{\varepsilon_{0}}$
Let electric flux linked with surfaces $A,\ B\ and\ C$ are$\ \phi_{A},\phi_{B}\ $and$\ \phi_{C}$ respectively. That is
$\phi_{total} = \phi_{A} + \phi_{B} + \phi_{C}$
Since $\phi_{C} = \phi_{A}$
$\therefore 2\phi_{A} + \phi_{B} = \phi_{total} = \frac{q}{\varepsilon_{0}}$
$or\ \ \ \ \ \phi_{A} = \frac{1}{2}\left( \frac{q}{\varepsilon_{0}} - \phi_{B} \right)$
$But\ \ \ \ \phi_{B} = \phi\ \ \ \ \ (given)$
Hence,
$\phi_{A} = \frac{1}{2}\left( \frac{q}{\varepsilon_{0}} - \phi \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.