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Unformatted text preview: yang (ey942) oldhomework 05 Turner (56705) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points An electron begins at rest, and then is accel erated by a uniform electric field of 800 N / C that extends over a distance of 6 cm. Find the speed of the electron after it leaves the region of uniform electric field. The elementary charge is 1 . 6 10 19 C and the mass of the electron is 9 . 11 10 31 kg. Correct answer: 4 . 10617 10 6 m / s. Explanation: Let : e = 1 . 6 10 19 C , m e = 9 . 11 10 31 kg , E = 800 N / C , and x = 6 cm = 0 . 06 m . Because of the constant acceleration, v 2 = v 2 + 2 a x . Since v = 0 and a = F net m e = e E m e , v = radicalbigg 2 e E x m e = radicalBigg 2 (1 . 6 10 19 C) (800 N / C) 9 . 11 10 31 kg . 06 m = 4 . 10617 10 6 m / s . 002 10.0 points A particle of mass 9 . 9 10 5 g and charge 95 mC moves in a region of space where the electric field is uniform and is 5 . 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 = 9 . 8 10 5 m / s, v x = v z = 0, what is the speed of the particle at 0 . 2 s? Correct answer: 1 . 46871 10 6 m / s. Explanation: Let : m = 9 . 9 10 5 g = 9 . 9 10 8 kg , E x = 5 . 7 N / C , E y = E z = 0 , v y = 9 . 8 10 5 m / s , v x = v z = 0 , and t = 0 . 2 s . According to Newtons 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 . 095 C) (5 . 7 N / C)(0 . 2 s) (9 . 9 10 8 kg) = 1 . 09394 10 6 m / s . No external force acts on the particle in the y direction so v yi = v yf = 9 . 8 10 5 m / s. Hence the final speed is given by v f = radicalBig v 2 yf + v 2 xf = bracketleftbigg ( 9 . 8 10 5 m / s ) 2 + ( 1 . 09394 10 6 m / s ) 2 bracketrightbigg 1 / 2 = 1 . 46871 10 6 m / s ....
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This note was uploaded on 11/21/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.
 Spring '10
 Turner

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