College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 16 - Problems - Page 615: 52


At $t = 2.30~ns$: $x = -1.08\times 10^{-4}~m$ $y=0$

Work Step by Step

Since the direction of the electric field is in the positive x-direction, the force exerted on the electron is in the negative x-direction. We can find the force exerted on the electron: $F = E~q = (232~N/C)(-1.6\times 10^{-19}~C) = -3.712\times 10^{-17}~N$ We can find the acceleration: $a = \frac{F}{m} = \frac{-3.712\times 10^{-17}~N}{9.109\times 10^{-31}~kg} = -4.075\times 10^{13}~m/s^2$ We can find the x-coordinate at $t = 2.30~ns$: $x = \frac{1}{2}at^2$ $x = \frac{1}{2}(-4.075\times 10^{13}~m/s^2)(2.30\times 10^{-9}~s)^2$ $x = -1.08\times 10^{-4}~m$ Since there is no electric force in the y-direction, the y-component of the position is zero. At $t = 2.30~ns$: $x = -1.08\times 10^{-4}~m$ $y=0$
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.