Question

A hollow cylinder has a charge $q$ coulomb within it. If $\varphi$ 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 through any closed surface is equal to the net charge inside the surface divided by ${\epsilon }_0$ . That is
ie, ${\varphi }_{total}=\frac{q}{{\epsilon }_0}$
Let electric flux linked with surfaces $A$, $B$ and $C$ are ${\varphi }_A$, ${\varphi }_B$ and ${\varphi }_C$ respectively. That is
${\varphi }_{tatal}={\varphi }_A+{\varphi }_B+{\varphi }_C$
Since ${\varphi }_C={\varphi }_A$
$\therefore 2{\varphi }_A+{\varphi }_B={\varphi }_{total}=\frac{q}{{\epsilon }_0}$
or ${\varphi }_A=\frac{1}{2}\left(\frac{q}{{\epsilon }_0}-{\varphi }_B\right)$
But ${\varphi }_B=\varphi$ (given)
Hence, ${\varphi }_A=\frac{1}{2}\left(\frac{q}{{\epsilon }_0}-\varphi \right)$