Answer
$8.8\times10^5\;N/C\;up$.
Work Step by Step
Use equation 16–3 to calculate the direction and magnitude of the electric field.
$$\vec{E}=\frac{\vec{F}}{q}$$
$$\vec{E}=\frac{6.4\;N \;down}{-7.3\times10^{-6}C }$$
$$= 8.8\times10^5\;N/C\;up$$
Physics: Principles with Applications (7th Edition)
by Giancoli, Douglas C.
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: 22
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