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# oldhw 05 - nguyen(jmn727 oldhomework 05 Turner(59070 This...

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nguyen (jmn727) – oldhomework 05 – Turner – (59070) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An electron, starting from rest, is accelerated by a uniform electric field of 80000 N / C that extends over a distance of 1 cm. The charge and mass of an electron are 1 . 6 × 10 19 C and 9 . 11 × 10 31 kg respectively. Find the speed of the electron after it leaves the region of uniform electric field. Correct answer: 1 . 67633 × 10 7 m / s. Explanation: Let e = 1 . 6 × 10 19 C , m e = 9 . 11 × 10 31 kg , E = 80000 N / C , and Δ x = 1 cm = 0 . 01 m . Because of the constant acceleration, v 2 = v 2 0 + 2 a Δ x . Since v 0 = 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) (80000 N / C) 9 . 11 × 10 31 kg × 0 . 01 m = 1 . 67633 × 10 7 m / s . 002 (part 1 of 3) 10.0 points An electron traveling at 4 × 10 6 m / s enters a 0 . 1 m region with a uniform electric field of 235 N / C , as in the figure. The mass of an electron is 9 . 10939 × 10 31 kg and the charge on an electron is 1 . 60218 × 10 19 C . - - - - - - - - - 0 . 1 m + + + + + + + + + ˆ ı ˆ 4 × 10 6 m / s Find the magnitude of the acceleration of the electron while in the electric field. Correct answer: 4 . 13323 × 10 13 m / s 2 . Explanation: Let : q e = 1 . 60218 × 10 19 C , m e = 9 . 10939 × 10 31 kg , and E = 235 N / C . F = m a = q E , so vectora = - q e E m e ˆ = - (1 . 60218 × 10 19 C)(235 N / C) 9 . 10939 × 10 31 kg ˆ = - (4 . 13323 × 10 13 m / s 2  , and the magnitude of the acceleration of the electron is 4 . 13323 × 10 13 m / s 2 .

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