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 C , the flux linked with the plane surface ' \(A\) ' in unit of V-m will be [ \(\varepsilon_0=\) permittivity of free space]

• A. \(\frac{\varrho}{3}\)
• B. \((\frac{\mathrm{q}}{\epsilon _{0}}-\varrho )\)
• C. \(\frac{1}{2}(\frac{\mathrm{q}}{\epsilon _{0}}-\varrho )\)
• D. \(\frac{\mathrm{q}}{2\epsilon _{0}}\)

Hint:

Total flux is q over epsilon naught; the two end faces share what is left after the curved surface.

Answer 1

The Correct Option is C

Step 1: Total flux
$\Phi_{total}=\dfrac{q}{\varepsilon_0}$.

Step 2: Remaining flux
Flux through both ends together is $\dfrac{q}{\varepsilon_0}-\phi$.

Step 3: Each end
Equal sharing gives $\dfrac12\left(\dfrac{q}{\varepsilon_0}-\phi\right)$ for A.

Final Answer:
The flux through A is $\dfrac12\left(\dfrac{q}{\varepsilon_0}-\phi\right)$. \[ \boxed{\text{(C) }\dfrac{1}{2}\left(\dfrac{q}{\epsilon_{0}}-\phi\right)} \]