Capa2 - 1. [1 point, 10 tries] An electron is projected at...

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

View Full Document Right Arrow Icon
1 . [1 point, 10 tries] An electron is projected at an angle of 29.3° above the horizontal at a speed of 8.11×10 5 m/s in a region where the electric field is E = 396 j N/C. Neglecting the effects of gravity, calculate the time it takes the electron to return to its initial height. Answer: 1.14e-08 s 1 2 . [1 point, 10 tries] Calculate the maximum height it reaches. Answer: 1.13e-03 m 1 3 . [1 point, 10 tries] Calculate its horizontal displacement when it reaches its maximum height. Answer: 4.03e-03 m 1 4 . [1 point, 10 tries] A flat surface having an area of 3.48 m 2 is rotated in a uniform electric field of magnitude E = 5.88×10 5 N/C. Determine the electric flux through this area when the electric field is perpendicular to the surface. Answer: 2.05e+06 N*m^2/C 1 5 . [1 point, 10 tries] Determine the electric flux through this area when the electric field is parallel to the surface. Answer:
Background image of page 1

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

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

This note was uploaded on 12/03/2009 for the course CHEMISTRY 1E03 taught by Professor Britz during the Spring '09 term at McMaster University.

Page1 / 4

Capa2 - 1. [1 point, 10 tries] An electron is projected at...

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