This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: oldmidterm 01 – KANUNGO, ANIL – Due: Sep 18 2007, 11:00 pm 1 Question 1, chap 22, sect 2. part 1 of 1 10 points Two spheres fastened to “pucks” are rid ing on a frictionless airtrack. Sphere “1” is charged with 1 nC, and sphere “2” is charged with 5 nC. Both objects have the same mass. 1 nC is equal to 1 × 10 − 9 C. As they repel, 1. sphere “2” accelerates 25 times as fast as sphere “1”. 2. sphere “1” accelerates 5 times as fast as sphere “2”. 3. sphere “2” accelerates 5 times as fast as sphere “1”. 4. sphere “1” accelerates 25 times as fast as sphere “2”. 5. they have the same magnitude of acceler ation. correct 6. they do not accelerate at all, but rather separate at constant velocity. Explanation: The force of repulsion exerted on each mass is determined by F = 1 4 π ǫ Q 1 Q 2 r 2 = ma where r is the distance between the centers of the two spheres. bardbl vector F 12 bardbl = bardbl vector F 21 bardbl m 1 a 1 = m 2 a 2 . Since both spheres have the same mass and are subject to the same force, they have the same acceleration. Question 2, chap 23, sect 1. part 1 of 1 10 points Three equal charges of 8 μ C are in the x y plane. One is placed at the origin, another is placed at (0 . , 16 cm), and the last is placed at (89 cm , . 0). Calculate the magnitude of the force on the charge at the origin. The value of the Coulomb constant is 9 × 10 9 N m 2 / C 2 . Correct answer: 22 . 5117 N (tolerance ± 1 %). Explanation: Let : q = 8 μ C = 8 × 10 − 6 C , ( x 1 , y 1 ) = (0 , 16 cm) = (0 , . 16 m) , ( x 2 , y 2 ) = (89 cm , 0) = (0 . 89 m , 0) , and ( x , y ) = (0 , 0) . The electric field along the yaxis is E y = k e q x 2 1 + y 2 1 = ( 9 × 10 9 N m 2 / C 2 )( 8 × 10 − 6 C ) 0 + (0 . 16 m) 2 = 2 . 8125 × 10 6 N / C , and along the xaxis E x = k e q x 2 2 + y 2 2 = ( 9 × 10 9 N m 2 / C 2 )( 8 × 10 − 6 C ) (0 . 89 m) 2 + 0 = 90897 . 6 N / C , so E = radicalBig E 2 x + E 2 y and F = q E = q radicalBig E 2 x + E 2 y = ( 8 × 10 − 6 C ) × radicalBig (2 . 8125 × 10 6 N / C) 2 + (90897 . 6 N / C) 2 = 22 . 5117 N . Question 3, chap 22, sect 2. part 1 of 1 10 points Two identical small charged spheres hang in equilibrium with equal masses as shown in oldmidterm 01 – KANUNGO, ANIL – Due: Sep 18 2007, 11:00 pm 2 the figure. The length of the strings are equal and the angles with the vertical are identical. The acceleration of gravity is 9 . 8 m / s 2 and the value of Coulomb’s constant is 8 . 98755 × 10 9 N m 2 / C 2 . . 9 m 6 ◦ . 03 kg . 03 kg Find the magnitude of the charge on each sphere. Correct answer: 3 . 48875 × 10 − 8 C (tolerance ± 1 %). Explanation: Let : L = 0 . 09 m , m = 0 . 03 kg , and θ = 6 ◦ ....
View
Full
Document
This homework help was uploaded on 04/02/2008 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Charge, Friction

Click to edit the document details