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Homework 02-solutions - sanne(as42476 Homework 02 Yao(59110...

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sanne (as42476) – Homework 02 – Yao – (59110) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The electron gun in a television tube is used to accelerate electrons (mass of 9 . 10939 × 10 - 31 kg and charge of 1 . 60218 × 10 - 19 C) from rest to 2 × 10 7 m / s within a distance of 6 . 7 cm. What electric field is required? Correct answer: 16972 N / C. Explanation: Let : m e = 9 . 10939 × 10 - 31 kg , q e = 1 . 60218 × 10 - 19 C , v = 2 × 10 7 m / s , and d = 6 . 7 cm . The magnitude of the force is F = q e E = m e a a = q e E m e The final velocity is v 2 f = v 2 i + 2 a d = 2 a d since v i = 0, so v 2 = 2 d q e E m e E = v 2 m e 2 d q e = parenleftBig 2 × 10 7 m / s 2 parenrightBig ( 9 . 10939 × 10 - 31 kg ) 2 (6 . 7 cm) (1 . 60218 × 10 - 19 C) = 16972 N / C . 002 10.0 points A 124 cm diameter loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be 3 . 23 × 10 5 N · m 2 / C. What is the electric field strength? Correct answer: 2 . 67466 × 10 5 N / C. Explanation: Let : r = 62 cm = 0 . 62 m and Φ = 3 . 23 × 10 5 N · m 2 / C . By Gauss’ law, Φ = contintegraldisplay vector E · d vector A The position of maximum electric flux will be that position in which the plane of the loop is perpendicular to the electric field; i.e. , when vector E · d vector A = E dA . Since the field is constant, Φ = E A = Eπ r 2 E = Φ π r 2 = 3 . 23 × 10 5 N · m 2 / C π (0 . 62 m) 2 = 2 . 67466 × 10 5 N / C . 003 10.0 points A (4 . 13 m by 4 . 13 m) square base pyramid with height of 4 . 92 m is placed in a vertical electric field of 38 . 8 N / C. 4 . 13 m 4 . 92 m 38 . 8 N / C Calculate the total electric flux which goes out through the pyramid’s four slanted sur- faces. Correct answer: 661 . 808 N m 2 / C. Explanation: Let : s = 4 . 13 m , h = 4 . 92 m , and E = 38 . 8 N / C .
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sanne (as42476) – Homework 02 – Yao – (59110) 2 By Gauss’ law, Φ = vector E · vector A Since there is no charge contained in the pyra- mid, the net flux through the pyramid must be 0 N/C. Since the field is vertical, the flux through the base of the pyramid is equal and opposite to the flux through the four sides. Thus we calculate the flux through the base of the pyramid, which is Φ = E A = E s 2 = (38 . 8 N / C) (4 . 13 m) 2 = 661 . 808 N m 2 / C . 004 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 How much charge Q is inside the box? 1. insufficient information 2. Q encl = 0 3. Q encl = ǫ 0 E a 2 correct 4. Q encl = 3 ǫ 0 E a 2 5. Q encl = 2 ǫ 0 E a 2 6. Q encl = 1 2 ǫ 0 E a 2 Explanation: Electric flux through a surface S is, by con- vention, positive for electric field lines going out of the surface S and negative for lines going in. Here the surface is a cube and no flux goes through the vertical sides.
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