## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

As the electric flux passing through a surface depends on the charge enclosed by the surface. That is, $\phi=E.dA$, where $E$ is electric field and $dA$ is surface area. So, flux depends on both electric field and surface area. In case of surface A its surface area is small but its electric field is large while in case of B its surface area is large but its electric field is small. Thus, $\phi=\frac{q}{\epsilon_{0}}$ and the charge is same for both surface A and surface B, therefore, the flux passing through them will also be same. Hence, Student 1 is correct.