Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 16 - Electric Charge and Electric Field - General Problems - Page 471: 60

Answer

$1.02\times10^{-7}N/C$

Work Step by Step

Set the magnitudes of the gravitational and electric forces equal to each other; solve for the field strength. $$F_{electric}=F_{gravitational}$$ $$qE=mg$$ $$E=\frac{mg}{q}$$ $$E=\frac{(1.67\times10^{-27}kg)(9.80m/s^2)}{1.602\times10^{-19}C}=1.02\times10^{-7}N/C$$ Gravity pulls down, so the electric field must point up so that the positively charged proton feels an upward force.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.