Exam 2-solutions - Version 041 Exam 2 markert(55325 This...

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Version 041 – Exam 2 – markert – (55325) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points The current in a wire decreases with time according to the relationship I = (3 . 03 mA) e a t where a = 0 . 13328 s 1 . Determine the total charge that passes through the wire from t = 0 to the time the current has diminished to zero. 1. 0.0222839 2. 0.00877851 3. 0.0234094 4. 0.0282113 5. 0.0181573 6. 0.0164316 7. 0.0143307 8. 0.0252101 9. 0.0208583 10. 0.0227341 Correct answer: 0 . 0227341 C. Explanation: I = d q dt q = integraldisplay t t =0 I dt = integraldisplay t =0 (0 . 00303 A) e ( 0 . 13328 s - 1 ) t dt = (0 . 00303 A) e (0 . 13328 s - 1 ) t - 0 . 13328 s 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 0 = 0 . 0227341 C . 002 10.0points Two parallel conducting plates separated by a distance d are connected to a battery of voltage E . What is correct if the plate separation is doubled while the battery remains connected? 1. The potential difference between the plates is doubled. 2. The capacitance is unchanged. 3. The electric charge on the plates is halved. correct 4. The potential difference between the plates is halved. 5. The electric charge on the plates is dou- bled. Explanation: The capacitance of the two parallel con- ducting plates is given by C = ǫ 0 A d , so C = ǫ 0 A d = ǫ 0 A 2 d = 1 2 ǫ 0 A d = 1 2 C and when the separation d is doubled, the capacitance is halved. The battery remains connected during the whole process, so the potential difference re- mains the same throughout. Q = C V, so the electric charge on the plates is also halved. 003 10.0points Consider two coaxial rings of 17 . 3 cm radius and separated by 12 . 2 cm . Find the electric potential at a point on their common axis midway between the two rings, assuming that each ring carries a uniformly distributed charge of 6 . 97 μ C . The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 1. 325817.0 2. 272568.0 3. 342272.0 4. 346022.0 5. 372068.0 6. 378302.0 7. 289750.0 8. 321695.0 9. 682986.0 10. 257074.0
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Version 041 – Exam 2 – markert – (55325) 2 Correct answer: 6 . 82986 × 10 5 V. Explanation: Let : d = 12 . 2 cm = 0 . 122 m , q = 6 . 97 μ C = 6 . 97 × 10 6 C , r = 17 . 3 cm = 0 . 173 m , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . With d the distance between the rings, con- sider the potential due only to one ring of radius r and charge Q . The point of measure- ment is a distance L = radicalBigg r 2 + parenleftbigg d 2 parenrightbigg 2 = radicalBigg (0 . 173 m) 2 + parenleftbigg 0 . 122 m 2 parenrightbigg 2 = 0 . 183439 m from any point on the ring. If the charge in an infinitesimal section of the ring is dq , then the potential due to this section is dV = k e dq L , and the potential due to the entire ring is V = integraldisplay dV = k e Q L . Note: All infinitesimal sections are the same distance from the point of measurement and contribute equally. For the two rings, V = 2 k e q L = 2 ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × 6 . 97 × 10 6 C 0 . 183439 m = 6 . 82986 × 10 5 V . 004 10.0points A 0 . 8 V potential difference is maintained across a 2 . 3 m length of tungsten wire that has a cross-sectional area of 0 . 9 mm 2 .
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