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 related to the net charge $ q_{enc} $ enclosed by the surface by: $$\Phi_E = \frac{q_{enc}}{\epsilon_0}$$ where $ \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_{enc} = 1.8 \mu C = 1.8 \times 10^{-6} \text{ C} $ - Cube edge length is given but not needed for flux calculation since flux depends only on enclosed charge. 4. **Calculate the flux:** $$\Phi_E = \frac{1.8 \times 10^{-6}}{8.85 \times 10^{-12}}$$ 5. **Perform division:** $$\Phi_E = 2.034 \times 10^{5} \text{ N}\cdot\text{m}^2/\text{C}$$ --- 6. **Problem 10:** A cube of edge length 2.00 m lies in a region where the electric field is given by $$\mathbf{E} = (3.00 x + 4.00) \mathbf{j} + 6.00 \mathbf{j} + 7.00 \mathbf{k} = (3.00 x + 10.00) \mathbf{j} + 7.00 \mathbf{k}$$ with $ x $ in meters. Find the net charge contained by the cube. 7. **Gauss' law and divergence:** The net flux through the cube is related to the charge enclosed by $$\Phi_E = \frac{q_{enc}}{\epsilon_0}$$ The net flux can also be found by integrating the divergence of $ \mathbf{E} $ over the volume: $$\Phi_E = \iiint (\nabla \cdot \mathbf{E}) dV$$ 8. **Calculate divergence:** $$\nabla \cdot \mathbf{E} = \frac{\partial}{\partial x}(0) + \frac{\partial}{\partial y}(3.00 x + 10.00) + \frac{\partial}{\partial z}(7.00)$$ Since $ E_x = 0 $, $ E_y = 3.00 x + 10.00 $, and $ E_z = 7.00 $, partial derivatives are: $$\frac{\partial E_x}{\partial x} = 0$$ $$\frac{\partial E_y}{\partial y} = 0$$ (because $ E_y $ depends on $ x $, not $ y $) $$\frac{\partial E_z}{\partial z} = 0$$ (constant) So, $$\nabla \cdot \mathbf{E} = 0 + 0 + 0 = 0$$ 9. **Interpretation:** Since divergence is zero everywhere, the net flux through the cube is zero. 10. **Calculate net charge:** $$q_{enc} = \epsilon_0 \Phi_E = \epsilon_0 \times 0 = 0$$ **Final answers:** - For problem 7: $ \Phi_E = 2.03 \times 10^{5} \text{ N}\cdot\text{m}^2/\text{C} $ - For problem 10: $ q_{enc} = 0 \text{ C} $ (no net charge enclosed)