37. find the electric flux through the plane surface shown in figure p23.37 if $\theta = 60.0^{\circ}$, $e = 350\text{ n/c}$, and $d = 5.00\text{ cm}$. the electric field is uniform over the entire area of the surface.

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{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Electric Flux", "Flux Through Inclined Surface" ], "new_concepts": [], "current_concepts": [ "Electric Flux", "Flux Through Inclined Surface" ] } ## Calculate the area of the square surface $$ \begin{aligned} d &= 5.00\text{ cm} = 0.0500\text{ m} \ A &= d^2 = (0.0500\text{ m})^2 = 2.50 \times 10^{-3}\text{ m}^2 \end{aligned} $$ ## Apply the electric flux formula $$ \begin{aligned} \Phi_E &= E A \cos\theta \ E &= 350\text{ N/C} \ \theta &= 60.0^\circ \ \Phi_E &= (350\text{ N/C}) \times (2.50 \times 10^{-3}\text{ m}^2) \times \cos(60.0^\circ) \end{aligned} $$ ## Compute the final numerical value $$ \begin{aligned} \cos(60.0^\circ) &= 0.5 \ \Phi_E &= 350 \times 0.00250 \times 0.5 \ \Phi_E &= 0.4375\text{ N}\cdot\text{m}^2/\text{C} \approx 0.438\text{ N}\cdot\text{m}^2/\text{C} \end{aligned} $$ $0.438\text{ N}\cdot\text{m}^2/\text{C}$ { "subject": "Natural Science", "question_type": "Free Response", "knowledge_point": [ "Natural Science", "Physics", "Electric Flux" ] }

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Explanation:

Calculate the area of the square surface

Apply the electric flux formula

Compute the final numerical value

Answer: