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: 110

Answer

(a) $2.1\times 10^4N/C$ (b) $14.4^{\circ}$

Work Step by Step

First, we solve part(b) to find the new angle. (b) We know that $T=\frac{mg}{cos\theta}$ $\implies cos\theta=\frac{mg}{T}$ $\implies \theta=cos^{-1}(\frac{mg}{T})$ We plug in the known values to obtain: $\theta=cos^{-1}(\frac{(0.025Kg)(9.8m/s^2)}{0.253N})$ $\theta=14.4^{\circ}$ (a) We know that $E=\frac{Tsin\theta}{q}$ We plug in the known values to obtain: $E=\frac{(0.253N)sin(14.4^{\circ})}{3\times 10^{-6}C}$ $\implies E=2.1\times 10^4N/C$
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