Solution 5 - homework 05 KANUNGO, ANIL Due: Sep 13 2007,...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 .--------- . 07 m + + + + + + + + + 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 =- (1 . 60218 10 19 C)(243 N / C) 9 . 10939 10 31 kg =- (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 doesnt hit the side walls....
View Full Document

Page1 / 3

Solution 5 - homework 05 KANUNGO, ANIL Due: Sep 13 2007,...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online