A charge q is enclosed by a hollow cylinder as shown. If the electric flux associated with the curved surface C be \( \phi \). The flux through the surface A will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{1}{2}\left(\displaystyle\frac{q}{\varepsilon_0}-\phi\right) \)
By Gauss’s law the total electric flux through a closed surface is \( q/\varepsilon_0 \). The closed surface here is the curved surface C together with the two flat ends A and B.
The charge is at the middle of the cylinder, as drawn, so by symmetry the flux through A equals the flux through B.
\( \displaystyle \phi_A + \phi_B + \phi = \frac{q}{\varepsilon_0} \)
\( \displaystyle 2\phi_A = \frac{q}{\varepsilon_0} - \phi \)
\( \displaystyle \phi_A = \frac{1}{2}\left(\frac{q}{\varepsilon_0} - \phi\right) \)