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Unformatted text preview: terry (ect328) homework 09 Turner (59130) 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Three identical point charges, each of mass 150 g and charge + q , hang from three strings, as in the figure. The acceleration of gravity is 9 . 8 m / s 2 , and the Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 . 9 . 8 m / s 2 1 c m 150 g + q 1 c m 150 g + q 45 150 g + q If the lengths of the left and right strings are each 10 cm, and each forms an angle of 45 with the vertical, determine the value of q . Correct answer: 0 . 80885 C. Explanation: Let : = 45 , m = 150 g = 0 . 15 kg , L = 10 cm = 0 . 1 m , g = 9 . 8 m / s 2 , and k e = 8 . 98755 10 9 N m 2 / C 2 . Newtons 2nd law: summationdisplay F = ma. Electrostatic force between point charges q 1 and q 2 separated by a distance r F = k e q 1 q 2 r 2 . All three charges are in equilibrium, so for each holds summationdisplay F = 0 . Consider the forces acting on the charge on the right. There must be an electrostatic force F acting on this charge, keeping it balanced against the force of gravity mg . The electro static force is due to the other two charges and is therefore horizontal. In the xdirection F T sin = 0 . In the ydirection T cos mg = 0 . These can be rewritten as F = T sin and mg = T cos . Dividing the former by the latter, we find F mg = tan , or F = mg tan (1) = (0 . 15 kg) (9 . 8 m / s 2 ) tan45 = 1 . 47 N . The distance between the right charge and the middle charge is L sin , and the distance to the left one is twice that. Since all charges are of the same sign, both forces on the right charge are repulsive (pointing to the right). We can add the magnitudes F = k e q q ( L sin ) 2 + k e q q (2 L sin ) 2 or F = 5 k e q 2 4 L 2 sin 2 . (2) We have already found F , and the other quan tities are given, so we solve for the squared charge q 2 q 2 = 4 F L 2 sin 2 5 k e (3) terry (ect328) homework 09 Turner (59130) 2 or, after taking the square root (we know q > 0) and substituting F from Eq. 1 into Eq. 3 and solving for q , we have q = 2 L sin radicalbigg mg tan 5 k e = 2 (0 . 1 m) sin 45 radicalBigg (0 . 15 kg) (9 . 8 m / s 2 ) tan 45 5 (8 . 98755 10 9 N m 2 / C 2 ) = 8 . 0885 10 7 C = . 80885 C . 002 10.0 points An electron moves at 3 . 5 10 6 m / s into a uniform electric field of magnitude 1194 N / C. The charge on an electron is 1 . 60218 10 19 C and the mass of an electron is 9 . 10939 10 31 kg . The field is parallel to the electrons velocity and acts to decelerate the electron....
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 Fall '09
 Turner

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