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# Homework - mehta(snm425 – Homework 4 – erskine –(58715 1 This print-out should have 15 questions Multiple-choice questions may continue on

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Unformatted text preview: mehta (snm425) – Homework 4 – erskine – (58715) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A charge Q is spread uniformly along the circumference of a circle of radius R . A point charge q is placed at the center of this circle. What is the total force exerted on q as calculated by Coulomb’s law? 1. Use 2 π R for the distance. 2. Use 2 R for the distance. 3. None of these. 4. The result of the calculation is zero. cor- rect 5. Use R for the distance. Explanation: By symmetry and using the fact that the charge is uniformly distributed along the cir- cumference, the total force on q is zero. For example, consider a small element of charge δQ on the circle. The force on q is along the vector connecting δQ and q . But on the ex- act opposite side there is another element of charge δQ which exerts an equal but opposite force on q . This is true for every point on the circle, so the net force is zero. 002 10.0 points A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radius r with r < R is used to calculate the magnitude of the electric field E at a distance r from the center of the sphere. Which of the following equations results from a correct application of Gauss’s law for this situation? 1. E (4 π r 2 ) = Q ǫ 3 r 3 4 π R 2. E (4 π r 2 ) = Q ǫ 3. E (4 π R 2 ) = Q ǫ 4. E (4 π r 2 ) = Q ǫ r 3 R 3 correct 5. E (4 π r 2 ) = 0 Explanation: Applying Gauss’s law, contintegraldisplay S vector E · vectorndA = Q inside ǫ ; i . e ., contintegraldisplay S E dA = Q inside ǫ . Because charge Q is uniformly distributed, so Q inside is just Q V olume ( r ) V olume ( total ) = Q r 3 R 3 , Where V olume ( r ) = 4 π 3 r 3 , and V olume ( total ) = 4 π 3 R 3 . So Gauss’s law gives E (4 π r 2 ) = Q ǫ r 3 R 3 . 003 10.0 points A cubic box of side a , oriented as shown, con- tains an unknown charge. The vertically di- rected electric field has a uniform magnitude E at the top surface and 2 E at the bottom surface. a E 2 E mehta (snm425) – Homework 4 – erskine – (58715) 2 How much charge Q is inside the box? 1. Q encl = 0 2. Q encl = 2 ǫ E a 2 3. Q encl = 3 E ǫ a 2 4. Q encl = 2 E ǫ a 2 5. Q encl = 1 2 ǫ E a 2 6. Q encl = 3 ǫ E a 2 7. Q encl = 6 ǫ E a 2 8. Q encl = ǫ E a 2 correct 9. insufficient information 10. Q encl = E ǫ a 2 Explanation: Electric flux through a surface S is, by con- vention, positive for electric field lines going out of the surface...
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## This note was uploaded on 01/20/2010 for the course PHY 1 taught by Professor Erskine during the Spring '10 term at University of Texas at Austin.

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Homework - mehta(snm425 – Homework 4 – erskine –(58715 1 This print-out should have 15 questions Multiple-choice questions may continue on

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