This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: hopkins (tlh982) HW03 criss (4908) 1 This printout should have 19 questions. Multiplechoice 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 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 Coulombs 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 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 N 1 1 . 2 3 4 4 N 3 9 . 3 2 5 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 ....
View Full
Document
 Fall '08
 RobertR.Criss
 Physics, Charge

Click to edit the document details