Given Data:

• The electric field $E$ is vertically downward.
• At an altitude of $290 \, \text{m}$, $E = 78.8 \, \text{N/C}$.
• At an altitude of $188 \, \text{m}$, $E = 152 \, \text{N/C}$.
• The cube has an edge length of $102 \, \text{m}$.
• The cube's horizontal faces are at altitudes of $188 \, \text{m}$ and $290 \, \text{m}$.

Solution:

Using Gauss's Law, the net charge $Q$ enclosed in the cube is:

$\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$

Since the electric field is vertical, only the top and bottom faces contribute to the flux. The net electric flux is:

$\Phi = E_{\text{top}} A - E_{\text{bottom}} A$

Where:

$A = \text{102 m} \times \text{102 m} = 10404 \, \text{m}^2$

Substituting values:

$\Phi = (78.8 \times 10404) - (152 \times 10404)$

$\Phi = 819715.2 - 1581408$

$\Phi = -761692.8 \, \text{N} \cdot \text{m}^2/\text{C}$

Using Gauss's Law to find charge:

$Q_{\text{enclosed}} = \varepsilon_0 \Phi$

Where $\varepsilon_0$ (permittivity of free space) is:

$\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/ \text{N} \cdot \text{m}^2$

$Q_{\text{enclosed}} = (8.85 \times 10^{-12}) (-761692.8)$

$Q_{\text{enclosed}} \approx -6.74 \times 10^{-6} \, \text{C}$

Final Answer:

$\mathbf{-6.74 \, \mu C}$

Would you like a detailed explanation of any step?

Related Questions:

1. How does Gauss's Law apply to different geometries?
2. What happens if the electric field is not uniform in the given region?
3. How do changes in altitude affect electric field strength?
4. What are practical applications of electric flux calculations?
5. How does permittivity affect charge distribution calculations?

Tip:

Always verify whether the given electric field is uniform or varies with position when applying Gauss's Law.