Hello,

The net electric flux through a closed Gaussian surface depends on the enclosed charge. By Gauss's law, the electric flux (Φ) is given by:

\[

\Phi = \frac{q}{\epsilon_0}

\]

where \( q \) is the enclosed charge and \( \epsilon_0 \) is the permittivity of free space (\(8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2\)).

For a 2 C charge:

\[

\Phi = \frac{2}{8.85 \times 10^{-12}} \approx 2.26 \times 10^{11} \, \text{Nm}^2/\text{C}

\]

The net flux depends only on the charge, not the cube size.

Hope this helps you,

Thank you