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 ${\Phi }_{total}=\frac{q}{{ϵ}_0}$.

For a point charge placed at the center of a cube, the electric flux is symmetrically distributed through all six faces of the cube. Therefore, the flux through one face is ${\Phi }_{one\_face}=\frac{{\Phi }_{total}}{6}=\frac{q}{6{ϵ}_0}$. Substituting the given values: ${\Phi }_{one\_face}=\frac{2\times 10^{-9}\text{ C}}{6\times 8.85\times 10^{-12}{\text{ C}}^2/{\text{Nm}}^2}=\frac{2}{53.1}\times 10^3{\text{ Nm}}^2/\text{C}\approx 0.03766\times 10^3{\text{ Nm}}^2/\text{C}\approx 37.66{\text{ Nm}}^2/\text{C}$. Rounding to the nearest integer, the magnitude of the electric flux through one face is $38{\text{ Nm}}^2/\text{C}$.