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

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 .
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern