Question

How does the electric flux due to a point charge enclosed by a spherical gaussian surface get affected when its radius is increased?

Answer 1

1. The flux through a closed surface is given by the Gauss law.
2. According to the law, $\varphi =\frac{q}{{\epsilon }_0}$ (where $\varphi$ is the flux, $q$is the charge enclosed by the Gaussian surface and ${\epsilon }_0$ is the permittivity of free space.)
3. As we can observe in the formula, the electric flux does not depend on the area and will remain unchanged with a change in the area.

Hence, the electric flux due to a point charge enclosed by a spherical Gaussian surface does not get affected when its radius is increased.