The total electric flux leaving a closed surface can be calculated using Gauss's law, which states:

$\Phi = \frac{q_{\text{enclosed}}}{\epsilon_0},$

where $q_{\text{enclosed}}$ is the total charge enclosed within the surface and $\epsilon_0$ is the permittivity of free space ($\epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2)$).

Problem Setup

The surface is a cube with bounds $x, y, z = \pm 5$. We'll evaluate the charge enclosed for each scenario:

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(a) Two point charges:

The charges are:

• $q_1 = 0.1 \, \mu\text{C} = 0.1 \times 10^{-6} \, \text{C}$, located at $(1, -2, 3)$.
• $q_2 = 17 \, \mu\text{C} = 17 \times 10^{-6} \, \text{C}$, located at $(-1, 2, -2)$.

Both charges lie inside the cube since their coordinates $x, y, z$ are within $[-5, 5]$. Therefore, the total charge enclosed is:

$q_{\text{enclosed}} = q_1 + q_2 = 0.1 \times 10^{-6} + 17 \times 10^{-6} = 17.1 \times 10^{-6} \, \text{C}.$

The flux is then:

$\Phi = \frac{q_{\text{enclosed}}}{\epsilon_0} = \frac{17.1 \times 10^{-6}}{8.854 \times 10^{-12}} \, \text{N} \cdot \text{m}^2/\text{C}.$

$\Phi \approx 1.93 \times 10^6 \, \text{N} \cdot \text{m}^2/\text{C}.$

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(b) Uniform line charge:

The line charge density is $\lambda = \pi \, \mu\text{C/m} = \pi \times 10^{-6} \, \text{C/m}$, located at $x = -2, y = 3$.

The cube spans $z = -5$ to $z = 5$, so the line charge enclosed is along this range ($z_{\text{length}} = 10 \, \text{m}$).

$q_{\text{enclosed}} = \lambda \cdot z_{\text{length}} = \pi \times 10^{-6} \cdot 10 = 10\pi \times 10^{-6} \, \text{C}.$

The flux is:

$\Phi = \frac{q_{\text{enclosed}}}{\epsilon_0} = \frac{10\pi \times 10^{-6}}{8.854 \times 10^{-12}} \, \text{N} \cdot \text{m}^2/\text{C}.$

$\Phi \approx 3.55 \times 10^6 \, \text{N} \cdot \text{m}^2/\text{C}.$

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(c) Uniform surface charge:

The surface charge density is $\sigma = 0.1 \, \mu\text{C/m}^2 = 0.1 \times 10^{-6} \, \text{C/m}^2$, located on the plane $y = 3x$.

This plane passes through the cube, but the enclosed area of the charge within the cube is complex. Let’s assume it covers the entire cube’s side facing the plane (area $A = 10 \times 10 = 100 \, \text{m}^2$).

$q_{\text{enclosed}} = \sigma \cdot A = (0.1 \times 10^{-6}) \cdot 100 = 10 \times 10^{-6} \, \text{C}.$

The flux is:

$\Phi = \frac{q_{\text{enclosed}}}{\epsilon_0} = \frac{10 \times 10^{-6}}{8.854 \times 10^{-12}} \, \text{N} \cdot \text{m}^2/\text{C}.$

$\Phi \approx 1.13 \times 10^6 \, \text{N} \cdot \text{m}^2/\text{C}.$

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Summary of Flux Results

• (a): $\Phi \approx 1.93 \times 10^6 \, \text{N} \cdot \text{m}^2/\text{C}$
• (b): $\Phi \approx 3.55 \times 10^6 \, \text{N} \cdot \text{m}^2/\text{C}$
• (c): $\Phi \approx 1.13 \times 10^6 \, \text{N} \cdot \text{m}^2/\text{C}$

Would you like step-by-step explanations for any part?

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Related Questions

1. How does Gauss's law simplify electric flux calculations for symmetric charge distributions?
2. What if the cube’s boundaries were changed? How would this affect the enclosed charge?
3. Can you compute flux if the charges are outside the cube but nearby?
4. How does the permittivity of the medium affect the electric flux?
5. What are the practical applications of Gauss’s law in real-world scenarios?

Tip: For non-symmetric setups, break the