Question

A hollow cylinder has a charge $q$ coulomb within it. If $\varphi$ is the electric flux in unit of volt-meter associated with the curved surface $B$, the flux linked with the plane surface $A$ in unit of volt-metre will be

• A. $\frac{q}{{\epsilon }_0}-\varphi$
• B. $\frac{q}{2{\epsilon }_0}$
• C. $\frac{\varphi }{3}$
• D. $\frac{1}{2}\left(\frac{q}{{\epsilon }_0}-\varphi \right)$

Answer 1

Answer: (D)

Gauss's law states that the net electric flux through any closed surface is equal to the net charge inside the surface divided by ${\epsilon }_0$.
i.e., ${\varphi }_{\text{total }}=\frac{q}{{\epsilon }_0}$
Let electric flux linked with surfaces $A,B$ and $C$ are ${\varphi }_A,{\varphi }_B$ and ${\varphi }_C$, respectively. That is
${\varphi }_{\text{total }}={\varphi }_A+{\varphi }_B+{\varphi }_C$
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, ${\varphi }_A=\frac{1}{2}\left(\frac{q}{{\epsilon }_0}-\varphi \right)$