The plane shown the charge on the inner cylinder per

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the plane shown. The charge on the inner cylinder per unit length is λ > 0 and the corresponding charge on the outer cylindrical shell per unit length is λ/ 2 (half the magni- tude and same sign) . λ R 1 R 2 R 3 λ/ 2
Version 036 – Final 58010 Fall 14 – yeazell – (58010) 11 3. | E | = 3 λ R 3 4 π ǫ 0 r 2 4. | E | = 0 5. | E | = λ 2 R 3 4 π ǫ 0 r 2 6. | E | = λ R 3 3 π ǫ 0 r 2 7. | E | = 3 λ 2 π ǫ 0 R 3 8. | E | = λ 3 π ǫ 0 r 9. | E | = 3 λ 2 π ǫ 0 r 10. | E | = 2 λ 3 π ǫ 0 r Explanation: Pick a cylindrical Gaussian surface with the radius r that is outside both cylinders and apply Gauss’s law: | E | · · 2 π r = (3 / 2) λℓ ǫ 0 | E | = 3 λ 4 π ǫ 0 r . 020 10.0points Consider charges placed at the corners of a rectangle. P 0.30 m 0.20 m + 8 . 0 μC 14 μC 6 . 0 μC Find the electric potential at point P due to the grouping of charges at the other corners of the rectangle. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 and the acceleration due to gravity 9 . 8 m / s 2 . 1. -423250.0 2. -444576.0 3. -314908.0 4. -311256.0 5. -429564.0 6. -460673.0 7. -630346.0 8. -418378.0 9. -379040.0 10. -535536.0 Correct answer: 3 . 7904 × 10 5 V. Explanation: Let : q 1 = 8 . 0 μ C , q 2 = 6 . 0 μ C , q 3 = 14 μ C , r 1 ,P = a = 0 . 30 m , r 2 ,P = b = 0 . 20 m , and k e = 8 . 99 × 10 9 N · m 2 / C 2 . r 3 ,P = radicalbig a 2 + b 2 , so V tot = k e q 1 r 1 ,P + k e q 2 r 2 ,P + k e q 3 r 3 ,P = k e parenleftbigg q 1 a + q 2 b + q 3 a 2 + b 2 parenrightbigg = (8 . 98755 × 10 9 N · m 2 / C 2 ) × bracketleftbigg 8 C 0 . 3 m + 6 C 0 . 2 m + 14 C radicalbig (0 . 3 m) 2 + (0 . 2 m) 2 bracketrightBigg = 3 . 7904 × 10 5 V . 021 10.0points In the circuit shown, the capacitor is ini- tially uncharged. At t = 0, the switch S is moved to position a .
Version 036 – Final 58010 Fall 14 – yeazell – (58010) 12 C R 1 R 2 V 0 S b a Find the time when the potential difference across the capacitor is V 0 / 2. 1. t = R 1 C/ 2 2. t = ( R 1 + R 2 ) C/ 2 3. t = R 1 C 4. t = ln (2) R 2 C 5. t = R 2 C 6. t = ( R 1 + R 2 ) C 7. t = ln (2)( R 1 + R 2 ) C 8. t = R 2 C/ 2 9. t = ln (2) R 1 C correct Explanation: For an “ RC ” circuit, V cap = V 0 (1 e t/ ( R 1 C ) ) We want t when V cap = V 0 / 2 so, 1 / 2 = e t/ ( R 1 C ) t = ln (1 / 2) R 1 C = ln (2) R 1 C 022 10.0points A fully charged capacitor stores 9 J of energy. How much energy remains when its charge has decreased to half its original value? 5. 2.125 6. 4.05 7. 2.325 8. 2.3 9. 3.25 10. 2.25 Correct answer: 2 . 25 J. Explanation: Let : U = 9 J . Q = C V , so the energy is U = 1 2 C V 2 = 1 2 Q 2 C . If Q is halved, U falls to one-quarter: U f = 1 2 parenleftbigg Q 2 parenrightbigg 2 C = 1 2 Q 2 4 C = 1 4 parenleftbigg 1 2 Q 2 C parenrightbigg = 1 4 U = 1 4 (9 J) = 2 . 25 J . 023 10.0points A thin film of cryolite ( n c = 1 . 21 ) is applied to a camera lens ( n g = 1 . 67 ). The coating is designed to reflect wavelengths at the blue end of the spectrum and transmit wavelengths in the near infrared. What minimum thickness gives high trans- mission at λ = 952 nm? 1. 194.262 2. 183.712 3. 198.8 4. 196.694 5. 187.591 6. 213.223 7. 190.625 8. 223.361 9. 206.4 10. 202.963

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