Careful measurement of the electric field at the surface of a black box indicates
Question:

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 $8.0 \times$ $10^{3} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}$.

(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?

Solution:

(a) Net outward flux through the surface of the box, $\Phi=8.0 \times 10^{3} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}$

For a body containing net charge q, flux is given by the relation,

$\phi=\frac{q}{\epsilon_{0}}$

$\epsilon_{0}=$ Permittivity of free space

$=8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^{2} \mathrm{~m}^{-2}$

$q=\epsilon_{0} \Phi$

$=8.854 \times 10^{-12} \times 8.0 \times 10^{3}$

$=7.08 \times 10^{-8}$

$=0.07 \mu \mathrm{C}$

Therefore, the net charge inside the box is $0.07 \mu \mathrm{C}$.

(b) No

Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.