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 - Problems - Page 469: 20

Answer

$1.16\times10^5\;N/C\;south$.

Work Step by Step

Use equation 16–3 to calculate the direction and magnitude of the electric field. $$\vec{E}=\frac{\vec{F}}{q}$$ $$\vec{E}=\frac{1.86\times10^{-14}\;N \;south}{1.602\times10^{-19}C }$$ $$= 1.16\times10^5\;N/C\;south$$
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