Question

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

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

Hint:

Gauss's Law gives the "Total" budget of lines of force. Just subtract what is already used by one surface and split the rest between symmetric parts.

Answer 1

The Correct Option is D

Step 1: Concept
According to Gauss's Law, the total flux through a closed surface is $\Phi_{total} = q/\varepsilon_0$.

Step 2: Meaning
The cylinder has three surfaces: two flat ends (A and C) and one curved surface (B). $\Phi_{total} = \phi_A + \phi_B + \phi_C$.

Step 3: Analysis
Due to symmetry, the flux through the two plane ends is equal ($\phi_A = \phi_C$).
$q/\varepsilon_0 = 2\phi_A + \phi$
$2\phi_A = \frac{q}{\varepsilon_0} - \phi \implies \phi_A = \frac{1}{2}\left(\frac{q}{\varepsilon_0} - \phi\right)$.

Step 4: Conclusion
The flux linked with plane surface A is $\frac{1}{2}\left(\frac{q}{\varepsilon_0} - \phi\right)$. Final Answer: (D)