Question

A hollow cylinder has a charge $q$ coulomb within it. If $\phi$ is the electric flux, in unit of $V m$, associated with the curved surface $B$, the flux linked with the plane surface $A$, in unit of $V m$, will be

• A. $\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)$
• B. $\frac{q}{2 \varepsilon_0}$
• C. $\frac{\phi}{3}$
• D. $\frac{q}{\varepsilon_0}-\phi$

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 $\varepsilon_0$.
i.e., $\phi_{\text {total }}=\frac{q}{\varepsilon_0}$
Let electric flux linked with surfaces $A, B$ and $C$ are $\phi_A, \phi_B$ and $\phi_C$, respectively. That is
$\phi_{\text {total }}=\phi_A+\phi_B+\phi_C $
Since, $ \phi_C=\phi_A $
$\Rightarrow 2 \phi_A+\phi_B=\phi_{\text {total }}=\frac{q}{\varepsilon_0}$
or $ \phi_A=\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi_B\right) $
But $\phi_B=\phi $ (given)
Hence, $\phi_A=\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)$