Physics - Electrostatics Question with Solution | TestHub

According to Gauss's Law, the total electric flux through a closed surface enclosing a charge $q$ is given by:

${\Phi }_{\text{total}}=\frac{q}{{ϵ}_0}$

The total flux through the hollow cylinder is the sum of the fluxes through its three surfaces: the two plane surfaces A and C, and the curved surface B.

${\Phi }_{\text{total}}={\Phi }_A+{\Phi }_B+{\Phi }_C$

Since the charge $q$ is placed exactly at the centre of the cylinder, the setup is symmetric with respect to the two plane ends. Therefore, the flux linked with surface A is equal to the flux linked with surface C:

${\Phi }_A={\Phi }_C$

We are given that the flux associated with the curved surface B is $\varphi$, so ${\Phi }_B=\varphi$. Substituting these into the total flux equation, we get:

${\Phi }_A+\varphi +{\Phi }_A=\frac{q}{{ϵ}_0}$

$2{\Phi }_A+\varphi =\frac{q}{{ϵ}_0}$

$2{\Phi }_A=\frac{q}{{ϵ}_0}-\varphi$

${\Phi }_A=\frac{1}{2}\left(\frac{q}{{ϵ}_0}-\varphi \right)$

Answer: $\frac{1}{2}\left(\frac{q}{{ϵ}_0}-\varphi \right)$