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give answer in 2 step with explanation at the end of each step and final answer at the end:1. (The figure below shows a Gaussian surface in the shape of a cube with edge length 1.5m. (a) (10pt) What is the net flux # enclosed by the surface if E=(~4.02{)N/C. with r in meters? (b) (10pt) What is the net charge g,,. enclosed by the surface if £= [-3.0i+5.0k] N/C?

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give answer in 2 step with explanation at the end of each step and final answer at the end:Uploaded Image1. (The figure below shows a Gaussian surface in the shape of a cube with edge length 1.5m. (a) (10pt) What is the net flux # enclosed by the surface if E=(~4.02{)N/C. with r in meters? (b) (10pt) What is the net charge g,,. enclosed by the surface if £= [-3.0i+5.0k] N/C?

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Asked by: Rahul
Created at: 2025-11-12 09:53:11

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## Step-by-Step Solution --- ### **Step 1: Net Flux Enclosed by the Surface** #### **Given:** - Cube edge length, \( L = 1.5 \) m - Electric field for part (a): \( \vec{E} = (-4.x \hat{i}) \) N/C (where \( x \) is in meters) #### **To Find:** Net flux \( \Phi \) through the cube. #### **Calculation:** The net flux through a closed surface is given by Gauss’s Law: \[ \Phi = \oint_S \vec{E} \cdot d\vec{A} \] Since \( \vec{E} \) only has an \( x \)-component and varies with \( x \), only the faces perpendicular to the \( x \)-axis contribute. - At \( x = \): \( E_x = \) - At \( x = L \): \( E_x = -4.L \) Flux through the face at \( x=L \) (outward normal \( +\hat{i} \)): \[ \Phi_{x=L} = E_x \cdot A = (-4.L) \cdot (L^2) \] Flux through the face at \( x= \) (outward normal \( -\hat{i} \)): \[ \Phi_{x=} = () \cdot (L^2) = \] Total flux: \[ \Phi = \Phi_{x=L} + \Phi_{x=} = -4.L^3 \] Substitute \( L = 1.5 \) m: \[ \Phi = -4. \times (1.5)^3 = -4. \times 3.375 = -13.5~\text{N}\cdot\text{m}^2/\text{C} \] **Explanation:** Only the two faces perpendicular to the \( x \)-axis contribute, since the electric field is only in the \( x \)-direction and depends on \( x \). The flux at \( x= \) is zero because \( E= \) there. --- ### **Step 2: Net Charge Enclosed by the Surface** #### **Given:** - Electric field for part (b): \( \vec{E} = [-3.\hat{i} + 5.\hat{k}] \) N/C (constant) #### **To Find:** Net charge \( q_{\text{enc}} \) enclosed by the surface. #### **Calculation:** The total flux through a closed surface with a **constant** electric field is zero (since the flux leaving one side is exactly balanced by the flux entering the opposite side): \[ \Phi = \oint_S \vec{E} \cdot d\vec{A} = \] From Gauss’s Law: \[ \Phi = \frac{q_{\text{enc}}}{\epsilon_} \] \[ q_{\text{enc}} = \Phi \cdot \epsilon_ = \] **Explanation:** For a constant electric field, the net flux through a closed surface is always zero, so no net charge is enclosed. --- ## **Final Answers** **(a) Net flux enclosed by the surface:** \[ \boxed{-13.5~\text{N}\cdot\text{m}^2/\text{C}} \] **(b) Net charge enclosed by the surface:** \[ \boxed{} \]

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