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Unformatted text preview: lim (kl9356) – Homework 3 – Weathers – (22202) 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 A particle of mass 6 . 6 × 10 − 5 g and charge 42 mC moves in a region of space where the electric field is uniform and is 6 . 7 N / C in the x direction and zero in the y and z direction. If the initial velocity of the particle is given by v y = 6 . 8 × 10 5 m / s, v x = v z = 0, what is the speed of the particle at 0 . 4 s? Correct answer: 1 . 83602 × 10 6 m / s. Explanation: Let : m = 6 . 6 × 10 − 5 g = 6 . 6 × 10 − 8 kg , E x = 6 . 7 N / C , E y = E z = 0 , v y = 6 . 8 × 10 5 m / s , v x = v z = 0 , and t = 0 . 4 s . According to Newton’s second law and the definition of an electric field, vector F = mvectora = q vector E . Since the electric field has only an x compo nent, the particle accelerates only in the x direction a x = q E x m . To determine the x component of the final velocity, v xf , use the kinematic relation v xf = v xi + a ( t f t i ) = a t f . Since t i = 0 and v xi = 0 , v xf = q E x t f m v xf = (0 . 042 C) (6 . 7 N / C)(0 . 4 s) (6 . 6 × 10 − 8 kg) = 1 . 70545 × 10 6 m / s . No external force acts on the particle in the y direction so v yi = v yf = 6 . 8 × 10 5 m / s. Hence the final speed is given by v f = radicalBig v 2 yf + v 2 xf = bracketleftbigg ( 6 . 8 × 10 5 m / s ) 2 + ( 1 . 70545 × 10 6 m / s ) 2 bracketrightbigg 1 / 2 = 1 . 83602 × 10 6 m / s . Note: This is analogous to a particle in a gravitational field with the coordinates ro tated clockwise by π 2 (90 ◦ )....
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This note was uploaded on 09/07/2010 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
 Spring '00
 Littler
 Charge, Magnetism, Mass, Work

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