Homework 3-solutions

Homework 3-solutions - kumar(vk4433 Homework 3...

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kumar (vk4433) – Homework 3 – markert – (55325) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points The point charge Q shown is at the center of a metal box that is isolated, ungrounded, and uncharged. a Q Which of the following is true? 1. The electric field outside the box is zero everywhere. 2. The electric field inside the box is zero. 3. The net charge on the outside surface of the box is Q . correct 4. The electric field inside the box is the same strength everywhere. 5. The electric field outside the box is the same as if only the point charge (and not the box) were there. Explanation: Consider a Gaussian surface between the outside and inside surface of the box and ap- ply Gauss’s law. The electric field on such surface is zero, because the box is a conductor and there is no current, so the charge on the inside surface of the box is Q . Since the box is neutral, the charge on the outside surface of the box is Q . 002 10.0points A point charge of 0 . 05481 μ C is inside a pyra- mid. Determine the total electric flux through the surface of the pyramid. The permittivity of a vacuum is 8 . 85419 × 10 12 C 2 / N · m 2 . Correct answer: 6190 . 29 N · m 2 / C. Explanation: Let : q = 0 . 05481 μ C = 5 . 481 × 10 8 C and ǫ 0 = 8 . 85419 × 10 12 C 2 / N · m 2 . By Gauss’ Law, Φ = contintegraldisplay vector E · d vector A = q ǫ 0 = 5 . 481 × 10 8 C 8 . 85419 × 10 12 C 2 / N · m 2 = 6190 . 29 N · m 2 / C . 003 10.0points A charge of 6 . 9 μ C is 20 cm above the center of a square with sides of length 40 cm. Find the flux through the square. The value of the permittivity of a vacuum is 8 . 8542 × 10 12 C 2 / N · m 2 . Correct answer: 1 . 29882 × 10 5 Nm 2 / C. Explanation: Let : q = 6 . 9 μ C = 6 . 9 × 10 6 C , = 20 cm = 0 . 2 m , s = 40 cm = 0 . 4 m , and ǫ 0 = 8 . 8542 × 10 12 C 2 / N · m 2 . To solve this problem we will use Gauss’ law and exploit symmetry and the fact that the point charge centered above the center of the square is 1 / 6 th of a point charge placed at the center of a cubic box with the same dimensions. The point charge is at the center of a cube with sides of length 0 . 2 m ; applying Gauss’ law, the flux through one face of that cube is Φ square = 1 6 Φ total = 1 6 q inside ǫ 0 = 1 6 parenleftbigg 6 . 9 × 10 6 C 8 . 8542 × 10 12 C 2 / N · m 2 parenrightbigg = 1 . 29882 × 10 5 N · m 2 / C .
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kumar (vk4433) – Homework 3 – markert – (55325) 2 004 10.0points A closed surface with dimensions a = b = 0 . 368 m and c = 0 . 6992 m is located as in the figure. The electric field throughout the region is nonuniform and defined by vector E = ( α + β x 2 ı where x is in meters, α = 3 N / C, and β = 6 N / (C · m 2 ). E y x z a c b a What is the magnitude of the net charge enclosed by the surface? Correct answer: 7 . 21954 × 10 12 C. Explanation: Let : a = b = 0 . 368 m , c = 0 . 6992 m , α = 3 N / C , and β = 6 N / (C · m 2 ) .
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