PhyIIHW3Solutions - hopkins(tlh982 – HW03 – criss...

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Unformatted text preview: hopkins (tlh982) – HW03 – criss – (4908) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An uncharged molecule of DNA (deoxyri- bonucleic acid) is 2 . 17 μ m long. The ends of the molecule become singly ionized so that there is a charge of- 1 . 6 × 10 − 19 C on one end and +1 . 6 × 10 − 19 C on the other. The heli- cal molecule acts like a spring and compresses 1 . 2% upon becoming charged. The value of Coulomb’s constant is 8 . 98755 × 10 9 N · m 2 / C 2 and the acceleration due to gravity is 9 . 8 m / s 2 . Find the effective spring constant of the molecule. Correct answer: 1 . 92224 × 10 − 9 N / m. Explanation: Let : L = 2 . 17 μ m = 2 . 17 × 10 − 6 m , q 1 =- q e =- 1 . 6 × 10 − 19 C , q 2 = q e = 1 . 6 × 10 − 19 C , x = 0 . 012 L , r = 0 . 988 L , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . The compressed length of the molecule is . 988 L . The compression force is supplied by the electric force: F electric = k e q 1 q 2 r 2 =- k x k = k e q 1 q 2- x r 2 = k e ( q e ) (- q e )- (0 . 012 L )(0 . 988 L ) 2 = k e q 2 e . 0117137 L 3 = (8 . 98755 × 10 9 N · m 2 / C 2 ) × (1 . 6 × 10 − 19 C) 2 . 0117137(2 . 17 × 10 − 6 m) 3 = 1 . 92224 × 10 − 9 N / m . 002 (part 1 of 2) 10.0 points Three positive point charges are arranged in a triangular pattern in a plane, as shown below. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . + 2 nC + 5 nC + 7 nC 2m 2m 2 m Find the magnitude of the net electric force on the 5 nC charge. Correct answer: 4 . 0894 × 10 − 8 N. Explanation: Let : q 1 = +2 nC = +2 × 10 − 9 C , q 2 = +5 nC = +5 × 10 − 9 C , q 3 = +7 nC = +7 × 10 − 9 C , ( x 1 , y 1 ) = (0 m , 2 m) , ( x 2 , y 2 ) = (2 m , 0 m) , ( x 3 , y 3 ) = (0 m ,- 2 m) , and k C = 8 . 98755 × 10 9 N · m 2 / C 2 . + 2 nC + 5 nC + 7 nC 2m 2m 2 m 2 9 . 5 4 6 ◦ 4 . 8 9 4 μ N 1 1 . 2 3 4 4 μ N 3 9 . 3 2 5 μ N The distances between the 5 nC charge and q 1 and q 2 are r 2 1 = ( x 2- x 1 ) 2 + ( y 2- y 1 ) 2 = ( x 2- 0) 2 + (0- y 1 ) 2 hopkins (tlh982) – HW03 – criss – (4908) 2 = x 2 2 + y 2 1 = (2 m) 2 + (- 2 m) 2 = 8 m 2 , and r 2 3 = ( x 3- x 2 ) 2 + ( y 3- y 2 ) 2 = (0- x 2 ) 2 + ( y 3- 0) 2 = x 2 2 + y 2 3 = (- 2 m) 2 + (- 2 m) 2 = 8 m 2 . vector F net = vector F 1 + vector F 3 F net = radicalBig F 2 net,x + F 2 net,y F electric = k C q 1 q 2 r 2 Consider the magnitudes of the forces. The repulsive force F 1 = k C q 2 q 1 r 2 1 = (8 . 98755 × 10 9 N · m 2 / C 2 ) × ( 5 × 10 − 9 C )( 2 × 10 − 9 C ) (8 m 2 ) = 1 . 12344 × 10 − 8 N acts downward and to the right along the line connecting q 2 and q 1 , and the repulsive force F 3 = k C q 2 q 3 r 2 3 = (8 . 98755 × 10 9 N · m 2 / C 2 ) × ( 5 × 10 − 9 C )( 7 × 10 − 9 C ) (8 m 2 ) = 3 . 93205 × 10 − 8 N acts upward and to the right along the line connecting q 2 and q 3 ....
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This note was uploaded on 09/03/2009 for the course PHY PHY 2049 taught by Professor Robertr.criss during the Fall '08 term at University of South Florida.

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PhyIIHW3Solutions - hopkins(tlh982 – HW03 – criss...

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