## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 16 - Problems - Page 614: 30

#### Answer

The magnitude of the acceleration is $1.07\times 10^{15}~m/s^2$ and the direction of the acceleration is south.

#### Work Step by Step

The electric field is directed north. Since the charge on the electron is negative, the force exerted on the electron will be directed in the opposite direction as the electric field. Therefore, the direction of the force exerted on the electron, and thus the direction of the acceleration, is south. We can find the force exerted on the electron: $F = E~q$ $F = (6100~N/C)(-1.6\times 10^{-19}~C)$ $F = -9.76\times 10^{-16}~N$ We can use the magnitude of the force to find the electron's acceleration: $F = ma$ $a = \frac{F}{m}$ $a = \frac{9.76\times 10^{-16}~N}{9.109\times 10^{-31}~kg}$ $a = 1.07\times 10^{15}~m/s^2$ The magnitude of the acceleration is $1.07\times 10^{15}~m/s^2$ and the direction of the acceleration is south.

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