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Unformatted text preview: klarin (sjk772) – 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 7 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 . 43517 × 10 6 m / s. Explanation: Let : e = 1 . 6 × 10 − 19 C , m e = 9 . 11 × 10 − 31 kg , E = 800 N / C , and Δ x = 7 cm = 0 . 07 m . Because of the constant acceleration, v 2 = v 2 + 2 a Δ x . Since v = 0 and a = F net m e = e E m e , v = radicalbigg 2 e E Δ x m e = radicalBigg 2 (1 . 6 × 10 − 19 C) (800 N / C) 9 . 11 × 10 − 31 kg × √ . 07 m = 4 . 43517 × 10 6 m / s . 002 10.0 points A particle of mass 0 . 000149 g and charge 44 mC moves in a region of space where the electric field is uniform and is 4 . 6 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 . 1 × 10 5 m / s, v x = v z = 0, what is the speed of the particle at 0 . 5 s? Correct answer: 9 . 1291 × 10 5 m / s. Explanation: Let : m = 0 . 000149 g = 1 . 49 × 10 − 7 kg , E x = 4 . 6 N / C , E y = E z = 0 , v y = 6 . 1 × 10 5 m / s , v x = v z = 0 , and t = 0 . 5 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 . 044 C) (4 . 6 N / C)(0 . 5 s) (1 . 49 × 10 − 7 kg) = 6 . 79195 × 10 5 m / s . No external force acts on the particle in the y direction so v yi = v yf = 6 . 1 × 10 5 m / s. Hence the final speed is given by v f = radicalBig v 2 yf + v 2 xf = bracketleftbigg ( 6 . 1 × 10 5 m / s ) 2 + ( 6 . 79195 × 10 5 m / s ) 2 bracketrightbigg 1 / 2 = 9 . 1291 × 10 5 m / s ....
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This note was uploaded on 10/10/2010 for the course PHY 56705 taught by Professor Turner during the Fall '10 term at University of Texas.
 Fall '10
 Turner
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