Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 19 - Electric Charges, Forces, and Fields - Problems and Conceptual Exercises - Page 689: 109

Answer

(a) greater than (b) $-4.50\mu C$

Work Step by Step

(a) We know that $q=\frac{mgsin\theta}{Ecos\theta}$ $\implies q=\frac{mg}{E} tan\theta$ Thus, the magnitude of new $q$ is greater than the previous values of $q$ because the angle $\theta$ is greater than the previous case. (b) We know that $q=\frac{mg}{E}tan\theta$ We plug in the known values to obtain: $q=\frac{(0.025Kg)(9.8m/s^2)}{1.46\times 10^4N/C}tan15^{\circ}$ $\implies q=-4.5\times 10^6C=-4.50\mu C$
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