1. **Problem 7:** A particle of charge 1.8 μC is at the center of a Gaussian cube 55 cm on edge. Find the net electric flux through the surface. 2. **Formula:** According to Gauss' law, the net electric flux $ \Phi_E $ through a closed surface is given by: $$\Phi_E = \frac{q_{enc}}{\epsilon_0}$$ where $ q_{enc} $ is the net charge enclosed and $ \epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2 $ is the permittivity of free space. 3. **Given:** - Charge $ q = 1.8 \mu C = 1.8 \times 10^{-6} \text{ C} $ - Cube edge length is irrelevant for flux calculation since flux depends only on enclosed charge. 4. **Calculate flux:** $$\Phi_E = \frac{1.8 \times 10^{-6}}{8.85 \times 10^{-12}} = 2.03 \times 10^{5} \text{ N}\cdot\text{m}^2/\text{C}$$ --- 5. **Problem 10:** A cube of edge length 2.00 m lies in a region where the electric field is: $$\vec{E} = (3.00x + 4.00)\hat{j} + 6.00\hat{k} + 7.00\hat{k} = (3.00x + 4.00)\hat{j} + 13.00\hat{k}$$ with $ x $ in meters. 6. **Goal:** Find the net charge contained by the cube. 7. **Gauss' law:** $$\Phi_E = \frac{q_{enc}}{\epsilon_0}$$ and $$\Phi_E = \oint \vec{E} \cdot d\vec{A}$$ 8. **Cube dimensions:** edge length $ L = 2.00 \text{ m} $, assume cube extends from $ x=0 $ to $ x=2.00 $, similarly for y and z. 9. **Calculate flux through each face:** - The cube has 6 faces: two each perpendicular to x, y, and z axes. - $ \vec{E} $ has components only in $ \hat{j} $ and $ \hat{k} $ directions, so flux through faces perpendicular to $ \hat{i} $ (x-faces) is zero because $ \vec{E} \cdot \hat{n} = 0 $ there. - For faces perpendicular to y-axis ($ \hat{j} $): - At $ y=0 $, normal vector $ \hat{n} = -\hat{j} $ - At $ y=2.00 $, normal vector $ \hat{n} = +\hat{j} $ - For faces perpendicular to z-axis ($ \hat{k} $): - At $ z=0 $, normal vector $ \hat{n} = -\hat{k} $ - At $ z=2.00 $, normal vector $ \hat{n} = +\hat{k} $ 10. **Flux through y-faces:** $$\Phi_{y} = \int_0^{2} \int_0^{2} \vec{E} \cdot \hat{n} \, dx \, dz$$ - At $ y=2 $, $ \hat{n} = +\hat{j} $, $ \vec{E} = (3.00x + 4.00)\hat{j} + 13.00\hat{k} $ $$\Phi_{y=2} = \int_0^{2} \int_0^{2} (3.00x + 4.00) \, dx \, dz$$ - At $ y=0 $, $ \hat{n} = -\hat{j} $, $ \vec{E} = (3.00x + 4.00)\hat{j} + 13.00\hat{k} $ $$\Phi_{y=0} = - \int_0^{2} \int_0^{2} (3.00x + 4.00) \, dx \, dz$$ 11. **Calculate integrals:** - $ \int_0^{2} (3.00x + 4.00) dx = \left[1.5x^2 + 4x\right]_0^2 = 1.5 \times 4 + 8 = 6 + 8 = 14 $ - Area in $ z $ direction is 2 m, so total integral over $ x,z $ is: $$14 \times 2 = 28$$ - Therefore, $$\Phi_{y} = 28 - 28 = 0$$ 12. **Flux through z-faces:** - $ \vec{E} \cdot \hat{n} = 13.00 $ at $ z=2 $ (normal $ +\hat{k} $) - $ \vec{E} \cdot \hat{n} = -13.00 $ at $ z=0 $ (normal $ -\hat{k} $) - Area of each face is $ 2 \times 2 = 4 \text{ m}^2 $ - Flux through z-faces: $$\Phi_z = 13.00 \times 4 - 13.00 \times 4 = 52 - 52 = 0$$ 13. **Flux through x-faces:** - $ \vec{E} $ has no $ \hat{i} $ component, so flux is zero. 14. **Total flux:** $$\Phi_E = 0 + 0 + 0 = 0$$ 15. **Net charge enclosed:** $$q_{enc} = \epsilon_0 \Phi_E = 8.85 \times 10^{-12} \times 0 = 0$$ **Final answers:** - Problem 7: $ \Phi_E = 2.03 \times 10^{5} \text{ N}\cdot\text{m}^2/\text{C} $ - Problem 10: $ q_{enc} = 0 \text{ C} $