## College Physics (4th Edition)

$t = 3.0\times 10^{-9}~s$
We can find the acceleration of the electron: $F = ma$ $\vert q \vert~E = ma$ $a = \frac{\vert q \vert~E}{m}$ $a = \frac{(1.6\times 10^{-19}~C)(5.0\times 10^4~N/C)}{9.1\times 10^{-31}~kg}$ $a = 8.79\times 10^{15}~m/s^2$ We can find the time it takes to travel 4.0 cm: $y = \frac{1}{2}at^2$ $t^2 = \frac{2y}{a}$ $t = \sqrt{\frac{2y}{a}}$ $t = \sqrt{\frac{(2)(4.0\times 10^{-2}~m)}{8.79\times 10^{15}~m/s^2}}$ $t = 3.0\times 10^{-9}~s$