Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0 \times 10^{3} \; N m^{2} / C$.
$(a)$ What is the net charge inside the box?
$(b)$ If the net outward flux through the surface of the box were zero,could you conclude that there were no charges inside the box? Why or why not?

(A) Net outward flux through the surface of the box,$\phi = 8.0 \times 10^{3} \; N m^{2} / C$.
According to Gauss's Law,the flux $\phi$ is related to the net charge $q$ enclosed by the surface as $\phi = \frac{q}{\varepsilon_{0}}$.
Here,$\varepsilon_{0} = 8.854 \times 10^{-12} \; N^{-1} C^{2} m^{-2}$ is the permittivity of free space.
Therefore,$q = \varepsilon_{0} \phi = (8.854 \times 10^{-12}) \times (8.0 \times 10^{3}) \; C$.
$q = 7.0832 \times 10^{-8} \; C \approx 0.07 \; \mu C$.
Thus,the net charge inside the box is $0.07 \; \mu C$.
$(b)$ No.
If the net outward flux through the surface is zero,it implies that the net charge enclosed by the surface is zero (since $\phi = q_{net} / \varepsilon_{0}$). This does not mean there are no charges inside; it only means that the total algebraic sum of all charges inside the box is zero. The box could contain equal amounts of positive and negative charges.