HW#3-solutions - nanni(arn437 HW#3 Erskine(56905 This...

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nanni (arn437) – HW#3 – Erskine – (56905) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points ±our charges are fxed at the corners oF a square centered at the origin as Follows: q at ( a, + a ); 2 q at (+ a, + a ); 3 q at (+ a, a ); and 6 q at ( a, a ). A fFth charge + q with mass m is placed at the origin and released From rest. ±ind the speed when it is a great distance From the origin, where the potential energy oF the fFth charge due to the Four point charges is negligible. 1. b vV b = q r 2 5 k ma 2. b b = q r 3 5 k 3. b b = q r 6 6 k 4. b b = q r 2 2 k 5. b b = q r 3 2 k 6. b b = q r 6 2 k correct 7. b b = q r 6 5 k 8. b b = q r 6 3 k 9. b b = q r 3 3 k 10. b b = q r 3 6 k Explanation: x a +6 q a 3 q +2 q + q + q, m 2 a The initial energy oF the charge is E i = K i + U i = U i = q p k q 2 a + 2 k q 2 a + ( 3 q ) k 2 a + 6 k q 2 a P = 6 k q 2 2 a . The fnal energy is E f = 1 2 mV 2 . ±rom energy conversation, we have E i = E f 6 k q 2 2 a = 1 2 2 V = q r 6 2 k . 002 10.0 points Two insulating spheres having radii 0 . 2 cm and 0 . 4 cm, masses 0 . 14 kg and 0 . 55 kg, and charges 4 μ C and 2 μ C are released From rest when their centers are separated by 0 . 9 m . How Fast is the smaller sphere moving when they collide? Correct answer: 11 . 6441 m / s. Explanation: Let : k e = 8 . 99 × 10 9 N · m 2 / C 2 , r 1 = 0 . 2 cm = 0 . 002 m , r 2 = 0 . 4 cm = 0 . 004 m , m 1 = 0 . 14 kg , m 2 = 0 . 55 kg , q 1 = 4 μ C = 4 × 10 6 C , q 2 = 2 μ C = 2 × 10 6 C , and d = 0 . 9 m .
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nanni (arn437) – HW#3 – Erskine – (56905) 2 By conservation of momentum 0 = m 1 v 1 m 2 v 2 v 2 = m 1 v 1 m 2 By conservation of energy ( K + U ) i = ( K + U ) f 0 + k e q 1 q 2 d = 1 2 m 1 v 2 1 + 1 2 m 2 v 2 2 + k e q 1 q 2 r 1 + r 2 k e q 1 q 2 d = 1 2 m 1 v 2 1 + 1 2 m 2 p m 1 v 1 m 2 P 2 + k e q 1 q 2 r 1 + r 2 k e q 1 q 2 p 1 r 1 + r 2 1 d P = m 1 v 2 1 2 m 2 ( m 1 + m 2 ) . So v 2 1 = 2 m 2 k e q 1 q 2 m 1 ( m 1 + m 2 ) p 1 r 1 + r 2 1 d P = 2 (0 . 55 kg) ( 8 . 99 × 10 9 N · m 2 / C 2 ) (0 . 14 kg) (0 . 14 kg + 0 . 55 kg) × ( 4 × 10 6 C) (2 × 10 6 C) × p 1 0 . 002 m + 0 . 004 m 1 0 . 9 m P = 135 . 584 m 2 / s 2 . and v 1 = r 135 . 584 m 2 / s 2 = 11 . 6441 m / s . 003 10.0 points The work needed to carry a 6 C charge from point A to point B is 23 J. Calculate the potential diFerence between point A and B. Correct answer: 3 . 83333 V. Explanation: Let : q = 6 C and W = 23 J . The work is W = q V V = W q = 23 J 6 C = 3 . 83333 V . 004 10.0 points Assume we are given an electric ±eld set up by an unknown charge distribution. U 0 is the amount of work needed to bring a point charge of charge q 0 in from in±nity to a point P . If the charge q 0 is returned to in±nity, how much work would it take to bring a new charge of 4 q 0 from in±nity to point P ? 1. 8 U 0 2. U 0 4 3. U 0 4. U 0 2 5. 0 6. 2 U 0 7. 4 U 0 correct 8. The correct answer is not given.
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HW#3-solutions - nanni(arn437 HW#3 Erskine(56905 This...

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