Question

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 $

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 $ε_0$ . That is
ie, $\phi_{total}=\frac{q}{\varepsilon_{0}}$
Let electric flux linked with surfaces $A$, $B$ and $C$ are $\phi_{A}$, $\phi_{B}$ and $\phi_{C}$ respectively. That is
$\phi_{tatal}=\phi_{A}+\phi_{B}+\phi_{C}$
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, $\phi_{A}=\frac{1}{2}\left(\frac{q}{\varepsilon_{0}}-\phi\right)$