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Unformatted text preview: nguyen (jmn727) – oldhomework 06 – Turner – (59070) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A proton has an initial velocity of 8 . 7 × 10 6 m / s in the horizontal direction. It enters a uniform electric field of 28400 N / C directed vertically. Ignoring gravitational effects, find the time it takes the proton to travel 0 . 105 m horizon tally. Correct answer: 12 . 069 ns. Explanation: Let : v x = 8 . 7 × 10 6 m / s , E = 28400 N / C , and x = 0 . 105 m . The electric field vector E is in the vertical ( y ) di rection, so the electric force vector F elec = q vector E ex erted by the field on the proton is also in the ydirection, with no component in the x direction. Hence, the field can exert no force on the proton in the xdirection. This implies a constant speed in the xdirection. Conse quently, x = v x t t = x v x = . 105 m 8 . 7 × 10 6 m / s · 10 9 ns s = 12 . 069 ns . 002 (part 2 of 3) 10.0 points What is the vertical displacement of the pro ton after the electric field acts on it for that time? Correct answer: 0 . 198126 mm. Explanation: In the vertical direction, the proton experi ences an electric force with magnitude F elec = q E = ma y a y = q E m = ( 1 . 60218 × 10 19 C ) (28400 N / C) 1 . 67262 × 10 27 kg = 2 . 72039 × 10 12 m / s 2 . The vertical dispacement is Δ y = v t + 1 2 a t 2 = 1 2 a t 2 since v o = 0, so Δ y Δ y = 1 2 ( 2 . 72039 × 10 12 m / s 2 ) × (1 . 2069 × 10 8 s) 2 × 1000 mm 1 m = . 198126 mm ....
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 Spring '08
 Turner
 Work, Electric charge, 0.31 m, 0.55 M, 0.28 m

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