Solution 5

# Solution 5 - homework 05 KANUNGO ANIL Due 4:00 am Question...

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homework 05 – KANUNGO, ANIL – Due: Sep 13 2007, 4:00 am 1 Question 1, chap 23, sect 4. part 1 of 3 10 points An electron traveling at 5 × 10 6 m / s enters a 0 . 07 m region with a uniform electric field of 243 N / C , as in the figure. The mass of an elec- tron is 9 . 10939 × 10 31 kg and its charge is 1 . 60218 × 10 19 C . - - - - - - - - - 0 . 07 m + + + + + + + + + ˆ ı ˆ 5 × 10 6 m / s Find the magnitude of the acceleration of the electron while in the electric field. Correct answer: 4 . 27393 × 10 13 m / s 2 (toler- ance ± 1 %). Explanation: Let : q e = 1 . 60218 × 10 19 C , m e = 9 . 10939 × 10 31 kg , and E = 243 N / C . F = m a = q E , so vectora = - q e E m e ˆ = - (1 . 60218 × 10 19 C)(243 N / C) 9 . 10939 × 10 31 kg ˆ = - (4 . 27393 × 10 13 m / s 2  , and the magnitude of the acceleration of the electron is 4 . 27393 × 10 13 m / s 2 . Question 2, chap 23, sect 4. part 2 of 3 10 points Find the time it takes the electron to travel through the region of the electric field, assum- ing it doesn’t hit the side walls. Correct answer: 1 . 4 × 10 8 s (tolerance ± 1 %). Explanation: Let : = 0 . 07 m , and v 0 = 5 × 10 6 m / s .

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• Spring '08
• Turner
• Mass, Work, Correct Answer, Electric charge, Fundamental physics concepts, Anil

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