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 $8.0\times10^{3}\,\text{N m}^{2}/\text{C}$.
(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?

Hint:

Use Gauss's law q = epsilon_0 * flux with epsilon_0 = 8.85e-12. For part (b), zero net flux only means the net enclosed charge is zero, not that there is no charge (equal +q and -q give zero flux).

Answer 1

The net outward flux through the closed surface of the box is \(\phi = 8.0\times10^{3}\,\text{N m}^{2}/\text{C}\).

Step 1: Concept (Gauss's law). The net flux through any closed surface equals the enclosed charge divided by \(\varepsilon_0\):

\[\phi = \frac{q}{\varepsilon_0}, \qquad \varepsilon_0 = 8.85\times10^{-12}\,\text{C}^{2}\text{N}^{-1}\text{m}^{-2}.\]

Step 2: Part (a) net charge inside. Rearrange for \(q\):

\[q = \varepsilon_0\,\phi\]\[q = (8.85\times10^{-12})(8.0\times10^{3})\]\[q = 7.08\times10^{-8}\,\text{C} = 0.07\,\mu\text{C}.\]

The positive flux means the enclosed net charge is positive.

\[\boxed{q = 7.08\times10^{-8}\,\text{C} \approx 0.07\,\mu\text{C}}\]

Step 3: Part (b) zero flux. No. Zero net flux only tells us the net enclosed charge is zero. The box could still contain equal amounts of positive and negative charge whose total is zero (for example \(+q\) and \(-q\) together). So we cannot conclude that there are no charges inside, only that they sum to zero.

Answer 2

Unit-and-dimension check plus reasoning approach:

Step 1: Gauss's law in the form \(\phi = q/\varepsilon_0\) can be inverted as \(q = \varepsilon_0\phi\). Check the units: \(\varepsilon_0\) carries \(\text{C}^{2}\text{N}^{-1}\text{m}^{-2}\) and \(\phi\) carries \(\text{N m}^{2}\text{C}^{-1}\); their product gives \(\text{C}^{2}\text{N}^{-1}\text{m}^{-2}\times\text{N m}^{2}\text{C}^{-1} = \text{C}\), confirming a charge.

Step 2: Evaluate with \(\varepsilon_0 = 8.85\times10^{-12}\):

\[q = (8.85\times10^{-12})(8.0\times10^{3}) = 7.08\times10^{-8}\,\text{C}.\]

Step 3 (part b, by counterexample): Place a dipole inside the box, charges \(+5\,\mu\text{C}\) and \(-5\,\mu\text{C}\). The enclosed charge is \(0\), so Gauss's law gives \(\phi = 0\), yet charges clearly exist inside. Hence zero flux never proves an empty box; it only fixes the algebraic sum of charges at zero.

\[\boxed{q = 7.08\times10^{-8}\,\text{C}; \ \text{zero flux} \Rightarrow \text{only net charge} = 0}\]