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Homework 3-solutions

# Homework 3-solutions - gonzales(pag757 Homework 3...

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gonzales (pag757) – Homework 3 – Hoffmann – (56745) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In a particular region of the earth’s atmo- sphere, the electric field above the earth’s sur- face has been measured to be 151 N / C down- ward at an altitude of 253 m and 174 . 5 N / C downward at an altitude of 419 m. The permittivity of free space is 8 . 85 × 10 12 C 2 / N · m 2 . Neglecting the curvature of the earth, cal- culate the volume charge density of the at- mosphere assuming it to be uniform between 253 m and 419 m. Correct answer: - 1 . 25286 × 10 12 C / m 3 . Explanation: Let : 1 = 253 m , 2 = 419 m , E l 1 = 151 N / C , E l 2 = 174 . 5 N / C , and ǫ 0 = 8 . 85 × 10 12 C 2 / N · m 2 . Choose a closed cylinderical surface for this portion of the atmosphere. The total flux is integraldisplay S E n dA = Q net ǫ 0 ( E l 1 - E l 2 ) A = Q net ǫ 0 Q net = ( E l 1 - E l 2 ) A ǫ 0 . Thus the volume charge density is ρ = Q net V = ( E l 1 - E l 2 ) A ǫ 0 A l 1 = ( E l 1 - E l 2 ) ǫ 0 2 - 1 = 151 N / C - 174 . 5 N / C 419 m - 253 m × (8 . 85 × 10 12 C 2 / N · m 2 ) = - 1 . 25286 × 10 12 C / m 3 . The curvature of the earth can be neglected because the maximum helight is < 0 . 01% of the radius of the earth. 002 10.0 points A charge of 8 . 3 μ C is 20 cm above the center of a square of side length 40 cm. The permittivity of free space is 8 . 85 × 10 12 C 2 / N · m 2 . Find the flux through the square. Correct answer: 1 . 56309 × 10 5 N · m 2 / C. Explanation: Let : q = 8 . 3 μ C = 8 . 3 × 10 6 C , = 20 cm = 0 . 2 m , s = 40 cm = 0 . 4 m , and ǫ 0 = 8 . 85 × 10 12 C 2 / N · m 2 . The point charge is at the center of the cube. Applying Gauss’ law, the flux through one face of the square is Φ square = 1 6 Φ total = 1 6 q inside ǫ 0 = 1 6 parenleftbigg 8 . 3 × 10 6 C 8 . 85 × 10 12 C 2 / N · m 2 parenrightbigg = 1 . 56309 × 10 5 N · m 2 / C . keywords: 003 10.0 points A point charge 1 . 4 μ C is located at the center of a uniform ring having linear charge density 10 . 3 μ C / m and radius 5 . 79 m.

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