Question

A hollow cylinder has a charge $qC$ 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. $\frac{1}{2}\left(\frac{q}{{\epsilon }_0}-\varphi \right)$
• B. $\frac{q}{2{\epsilon }_0}$
• C. $\frac{\varphi }{3}$
• D. $\frac{q}{{\epsilon }_0}-\varphi$

Answer 1

Answer: (A)

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 ${\epsilon }_0$.
ie, ${\varphi }_{\text{total }}=\frac{q}{{\epsilon }_0}$
Let electric flux linked with surfaces $A,B$ and $C$ are ${\varphi }_A,{\varphi }_B$ and ${\varphi }_C$ respectively. That is
${\varphi }_{\text{total }}={\varphi }_A+{\varphi }_B+{\varphi }_C$
${\varphi }_C={\varphi }_A$
Since, ${\varphi }_C={\varphi }_A$
$\therefore 2{\varphi }_A+{\varphi }_B={\varphi }_{\text{total }}=\frac{q}{{\epsilon }_0}$
or ${\varphi }_A=\frac{1}{2}\left(\frac{q}{{\epsilon }_0}-{\varphi }_B\right)$
But ${\varphi }_B=\varphi$ (given)
Hence, ${\varphi }_A=\frac{1}{2}\left(\frac{q}{{\epsilon }_0}-\varphi \right)$