Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
16
Electric Charge and Electric Field
Change chapter

16-5 and 16-6: Coulomb's Law
16-7 and 16-8: Electric Field, Field Lines
16-10: DNA
16-12: Gauss's Law

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 20
Q

A proton is released in a uniform electric field, and it experiences an electric force of 1.86×1014 N1.86 \times 10^{-14} \textrm{ N} toward the south. Find the magnitude and direction of the electric field.

A
1.16×105 N/C, South1.16 \times 10^5 \textrm{ N/C, South}
Giancoli 7th Edition, Chapter 16, Problem 20 solution video poster
Padlock

In order to watch this solution you need to have a subscription.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The electric field points in the same direction as the force it applies on a positive charge. Because this proton experiences a force towards the South, that means electric field must be to the South, because electric field points in the direction of force on a positive charge. The electric field strength is the force divided by the charge. That's gonna be 1.86 times 10 to the minus 14 Newtons divided by 1.602 times 10 to the minus 19 Coulombs, magnitude of a charge in a proton. This gives us 1.16 times 10 to the five Newtons per Coulomb electric field strength which is directed South.

Find us on:

Facebook iconTrustpilot icon
Giancoli Answers, including solutions and videos, is copyright © 2009-2024 Shaun Dychko, Vancouver, BC, Canada. Giancoli Answers is not affiliated with the textbook publisher. Book covers, titles, and author names appear for reference purposes only and are the property of their respective owners. Giancoli Answers is your best source for the 7th and 6th edition Giancoli physics solutions.